CHEM 105 - ELEMENTS OF
GENERAL CHEMISTRY - pH and
pOH calculations
Question Bank - Set 7
Liberty University
Question 1
Question
Calculate the pH of a solution with a hydroxide ion concentration of 1.0×10−4
M.
Solution
Step 1: Calculate the pOH of the solution using the formula pOH =−log[OH−].
pOH =−log1.0×10−4
=−log(1.0) −log10−4
=−0−(−4)
= 4
Step 2: Calculate the pH of the solution using the relation pH +pOH = 14.
pH + 4 = 14
pH = 14 −4
pH = 10
Therefore, the pH of the solution with a hydroxide ion concentration of
1.0×10−4M is 10.
Question 2
Question
Calculate the pH of a solution with a hydrogen ion concentration of 3.5×10−5
M.
Solution
Step 1: Recall that pH is defined as the negative logarithm (base 10) of the
hydrogen ion concentration.
pH = −log[H+]
Step 2: Given that the hydrogen ion concentration is 3.5×10−5M, we can
now substitute this value into the formula to find the pH.
pH = −log3.5×10−5
Step 3: Calculate the pH:
pH = −log3.5×10−5=−(−4.46) = 4.46
Step 4: Therefore, the pH of the solution with a hydrogen ion concentration
of 3.5×10−5M is 4.46.
Question 3
Question
Calculate the pH of a solution with a hydrogen ion concentration of 5.7×10−6
M.
Solution
Step 1: Recall that pH is calculated using the formula: pH =−log[H+]. Given
that the hydrogen ion concentration is 5.7×10−6M, we can substitute this
value into the formula:
pH = −log5.7×10−6
Step 2: Calculate the pH:
pH = −log5.7×10−6=−log(5.7) + log10−6=−log(5.7) −6 = −0.755 −6
Step 3: Finally, calculate the pH:
pH ≈ −6.755
Therefore, the pH of the solution with a hydrogen ion concentration of 5.7×
10−6M is approximately 6.755.
Question 4
Question
Calculate the pOH of a solution with a hydroxide ion concentration of 2.5×10−3
M.
2
Solution
Step 1: Write the expression for the ion product constant of water, Kw, and
for the relationship between [H+], [OH−], and Kw.
Kw= [H+][OH−]
Step 2: Since we are given the [OH−] concentration, we can rearrange the
equation from Step 1 to solve for [H+]. Then, calculate the [H+] concentration.
[H+] = Kw
[OH−]
[H+] = 1.0×10−14
2.5×10−3
[H+]=4.0×10−12 M
Step 3: To find the pOH of the solution, use the formula:
pOH =−log[OH−]
Substitute the given [OH−] concentration to find the pOH.
pOH =−log2.5×10−3
pOH =−log(2.5) −log10−3
pOH =−log(2.5) + 3
pOH ≈2.60
Therefore, the pOH of the solution is approximately 2.60.
Question 5
Question
Calculate the pH of a solution with a hydroxide ion concentration of 2.5×10−4
M.
Solution
Step 1: Calculate the pOH of the solution using the formula:
pOH = −log[OH−]
pOH = −log2.5×10−4
pOH ≈ −log(2.5) = 2 −log(2.5)
pOH ≈2−0.3979 = 1.6021
3
Step 2: Use the relationship between pH and pOH to calculate the pH:
pH + pOH = 14
pH = 14 −1.6021
pH ≈12.3979
Therefore, the pH of the solution with a hydroxide ion concentration of
2.5×10−4M is approximately 12.40.
Question 6
Question
A 0.025 M solution of hydrochloric acid, HCl, is prepared. Calculate the pH
and pOH of the solution.
Solution
Step 1: Write the balanced chemical equation for the dissociation of hydrochloric
acid in water.
HCl(aq)→H+(aq) + Cl−(aq)
Step 2: Calculate the concentration of H+ions in the solution. Since HCl is
a strong acid, it completely dissociates in water. Therefore, the concentration
of H+ions is equal to the initial concentration of the HCl solution.
[H+]=0.025 M
Step 3: Calculate the pH of the solution using the formula pH =−log[H+].
pH =−log(0.025) = −log2.5×10−2≈ −(−1.602) ≈1.602
Therefore, the pH of the solution is approximately 1.602.
Step 4: Calculate the pOH of the solution using the formula pOH =−log[OH−].
Since Kw= [H+][OH−]=1.0×10−14 at 25
°
C, we can find [OH−] using the
concentration of H+ions.
[OH−] = 1.0×10−14
0.025 = 4.0×10−13
pOH =−log4.0×10−13≈12.398
Therefore, the pOH of the solution is approximately 12.398.
4
Question 7
Question
Calculate the pH of a solution with a hydroxide ion concentration of 3.2×10−5
M.
Solution
Step 1: Write the expression for calculating the pOH of the solution. Step 2:
Use the pOH value to find the pH of the solution. Step 3: Calculate the final
pH value for the solution.
Step 1:
The pOH of a solution can be calculated using the formula:
pOH =−log[OH−]
Given that the hydroxide ion concentration is 3.2×10−5M, we can substitute
this value into the formula:
pOH =−log3.2×10−5
Step 2:
To find the pH of the solution, we can use the relationship:
pH +pOH = 14
Substitute the calculated pOH value to solve for pH:
pH + (−log3.2×10−5) = 14
Step 3:
Solve for pH:
pH = 14 + log3.2×10−5
pH = 14 + (−4.49485)
pH = 9.50515
Therefore, the pH of the solution with a hydroxide ion concentration of
3.2×10−5M is approximately 9.51.
Question 8
Question
Calculate the pH of a solution with a hydrogen ion concentration of 3.2×10−5
M.
5
Solution
Step 1: Recall that the pH of a solution is calculated using the formula:
pH = −log[H+]
Step 2: Substitute the given hydrogen ion concentration into the formula:
pH = −log3.2×10−5
Step 3: Calculate the pH:
pH = −log3.2×10−5=−log(3.2) −log10−5
Step 4: Recall that log(10−x) = −x, so:
pH = −log(3.2) −(−5) = −log(3.2) + 5
Step 5: Use a calculator to find the value of −log(3.2):
pH ≈ −0.5052 + 5 ≈4.4948
Step 6: Therefore, the pH of the solution with a hydrogen ion concentration
of 3.2×10−5M is approximately 4.49.
Question 9
Question
Calculate the pH of a solution with a hydroxide ion concentration of 3.5×10−5
M.
Solution
Step 1: Use the relation between pOH and OH−concentration, which is pOH =
−log[OH−].
Given that [OH−]=3.5×10−5M, we can calculate pOH:
pOH = −log3.5×10−5
Step 2: Calculate pOH.
pOH = −log3.5×10−5=−(−4.4559) ≈4.46
Step 3: Use the relation between pH and pOH, which is pH + pOH = 14, to
find pH.
pH = 14 −pOH = 14 −4.46
Step 4: Calculate the pH of the solution.
pH = 14 −4.46 = 9.54
Therefore, the pH of a solution with a hydroxide ion concentration of 3.5×
10−5M is 9.54.
6
Question 10
Question
Calculate the pH of a solution with a hydroxide ion concentration of 1.5×10−4
M.
Solution
Step 1: Write the equilibrium expression for the autoionization of water:
H2O⇌H++ OH−
Step 2: Write the expression for the equilibrium constant, Kw, for the au-
toionization of water:
Kw= [H+][OH−]=1.0×10−14
Step 3: Given the hydroxide ion concentration, we can find the hydrogen ion
concentration using the equilibrium constant:
[H+] = Kw
[OH−]=1.0×10−14
1.5×10−4= 6.67 ×10−11 M
Step 4: Calculate the pH of the solution using the hydrogen ion concentra-
tion:
pH = −log[H+]=−log6.67 ×10−11= 10.18
Therefore, the pH of the solution is 10.18.
Question 11
Question
Calculate the pH of a solution with a hydronium ion concentration of 1.5×10−3
M.
Solution
Step 1: Recall that pH is defined as −log[H3O+], where [H3O+] is the concen-
tration of hydronium ions in the solution.
Step 2: Substitute the given hydronium ion concentration into the pH for-
mula:
pH = −log1.5×10−3
Step 3: Calculate the pH using a calculator:
pH = −log1.5×10−3=−(−2.823) = 2.823
Step 4: Therefore, the pH of the solution with a hydronium ion concentration
of 1.5×10−3M is 2.823.
7
Question 12
Question
Calculate the pH of a solution with a hydronium ion concentration of 3.2×10−5
M.
Solution
Step 1: Recall that the pH of a solution is defined as −log[H3O+], where [H3O+]
represents the concentration of hydronium ions in moles per liter.
Step 2: Given that [H3O+]=3.2×10−5M, we can calculate the pH as:
pH = −log3.2×10−5
Step 3: Using the properties of logarithms, we get:
pH = −log(3.2) −log10−5
Step 4: Further simplifying, we find:
pH = −log(3.2) −(−5)
Step 5: Using a calculator to evaluate −log(3.2), we obtain −0.5051.
Step 6: Therefore, the pH of the solution is:
pH = −0.5051 −(−5) = 4.4949
Step 7: Thus, the pH of the solution with a hydronium ion concentration of
3.2×10−5M is approximately 4.49.
Question 13
Question
Calculate the pH of a solution that has a hydroxide ion concentration of 2.5×
10−4M.
Solution
Step 1: Recall that the pH of a solution can be calculated using the formula:
pH =−log[H+]. Since pH +pOH = 14, we can first find the pOH of the
solution.
Step 2: Given that [OH−] = 2.5×10−4M, we can calculate the pOH using
the formula pOH =−log[OH−].
pOH =−log2.5×10−4
pOH ≈ −log(2.5) −log10−4
8
pOH ≈ −log(2.5) + 4
pOH ≈ −0.3979 + 4
pOH ≈3.6021
Step 3: Now, we can find the pH using the relationship pH +pOH = 14.
pH = 14 −pOH
pH = 14 −3.6021
pH ≈10.3979
Therefore, the pH of the solution is approximately 10.40.
Question 14
Question
Calculate the pH of a solution with a hydronium ion concentration of 1.5×
10−3M.
Solution
Step 1: Recall that the pH is defined as −log[H3O+]. We are given that
[H3O+]=1.5×10−3M.
Step 2: Substitute the given concentration into the formula for pH:
pH = −log1.5×10−3
Step 3: Calculate the pH using a calculator:
pH = −log1.5×10−3≈ −log(0.0015) ≈2.82
Step 4: Therefore, the pH of the solution is approximately 2.82.
Question 15
Question
Calculate the pH of a solution with a hydroxide ion concentration of 1.5×10−9
M.
9
Solution
Step 1: Write the expression for the ion product constant of water.
Kw= [H+]×[OH−]=1.0×10−14
Step 2: Since the solution is basic, we know that pOH = −log[OH−].
pOH = −log1.5×10−9
Step 3: Use the relationship between pH and pOH to find the pH of the
solution.
pH = 14 −pOH
pH = 14 −(−log1.5×10−9)
Step 4: Calculate the pH.
pH = 14 −(−log1.5×10−9) = 14 + log1.5×10−9
pH ≈14 + 8.823
pH ≈22.823
Therefore, the pH of the solution is approximately 22.823.
Question 16
Question
Calculate the pH of a solution with a hydroxide ion concentration of 3.5×10−4
M.
Solution
Step 1: Write the expression for calculating the pOH of the solution.
pOH = −log[OH−]
Step 2: Plug in the given hydroxide ion concentration to calculate pOH.
pOH = −log3.5×10−4≈ −(−3.455) ≈3.455
Step 3: Use the relationship pH + pOH = 14 to calculate the pH of the
solution.
pH = 14 −pOH = 14 −3.455 ≈10.545
Therefore, the pH of the solution is approximately 10.545.
10
Question 17
Question
Calculate the pH of a solution with a hydroxide ion concentration of 2.5×10−10
M.
Solution
Step 1: Recall that the relationship between pOH and pH is given by:
pOH =−log[OH−]
Step 2: We are given the hydroxide ion concentration as 2.5×10−10 M. Let’s
calculate the pOH:
pOH =−log2.5×10−10
pOH =−log(2.5) −log10−10
pOH =−log(2.5) −(−10)
pOH =−0.3979 + 10
pOH = 9.6021
Step 3: Since pH +pOH = 14, we can find the pH:
pH = 14 −pOH
pH = 14 −9.6021
pH = 4.3979
Step 4: Therefore, the pH of the solution with a hydroxide ion concentration of
2.5×10−10 M is approximately 4.40.
Question 18
Question
Calculate the pH of a 0.005 M hydrochloric acid solution.
Solution
Step 1: Write the chemical equation for the dissociation of hydrochloric acid:
HCl(aq)→H+(aq) + Cl−(aq)
Step 2: Use the concentration of hydrochloric acid to determine the con-
centration of hydrogen ions (H+) in the solution. Since hydrochloric acid is a
11
strong acid and completely dissociates, the concentration of H+ions is equal to
the concentration of the hydrochloric acid solution:
[H+]=0.005 M
Step 3: Calculate the pH using the formula:
pH = −log[H+]
Substitute the concentration of hydrogen ions into the formula:
pH = −log(0.005)
Step 4: Calculate the pH:
pH = −log(0.005) = −(−2.3) = 2.3
Therefore, the pH of the 0.005 M hydrochloric acid solution is 2.3.
Question 19
Question
Calculate the pH and pOH of a 0.005 M solution of perchloric acid (HClO4).
Solution
Step 1: Write the balanced equation for the dissociation of perchloric acid:
HClO4→H++ ClO−
4.
Step 2: Calculate the concentration of H+ions in the solution. Since perchlo-
ric acid is a strong acid, it dissociates completely. Therefore, the concentration
of H+ions is the same as the initial concentration of perchloric acid: 0.005 M.
Step 3: Calculate the pH using the formula: pH = −log[H+]. Substitute
the value of [H+] into the formula: pH = −log(0.005) ≈2.3.
Step 4: Calculate the pOH using the formula: pOH = −log[OH−]. Since
water autoionization is negligible in this case, [OH−]≈0. Therefore, pOH =
−log(0) →pOH = ∞.
Therefore, the pH of the 0.005 M solution of perchloric acid is approximately
2.3, and the pOH is infinity.
Question 20
Question
Calculate the pH of a solution with a hydroxide ion concentration of 6.3×10−9
M.
12
Solution
Step 1: Calculate the pOH of the solution using the hydroxide ion concentration.
Step 2: Use the relation between pOH and pH to calculate the pH of the
solution.
Step 1: Calculate the pOH. Given: [OH−]=6.3×10−9M.
The formula relating pOH to [OH−] is:
pOH =−log[OH−]
pOH =−log6.3×10−9
pOH =−log(6.3) −log10−9
pOH =−log(6.3) −(−9)
pOH =−0.799 −(−9)
pOH = 8.201
Step 2: Calculate the pH using the relation between pH and pOH. The
relation between pH and pOH is:
pH +pOH = 14
pH = 14 −pOH
pH = 14 −8.201
pH = 5.799
Therefore, the pH of the solution with a hydroxide ion concentration of
6.3×10−9M is 5.799.
Question 21
Question
Calculate the pH of a solution with a hydronium ion concentration of 1.5×10−6
M.
Solution
Step 1: Recall that pH is defined as the negative base-10 logarithm of the
hydronium ion concentration. The formula for pH is: pH =−log[H3O+].
Step 2: Substitute the given hydronium ion concentration into the formula:
pH =−log1.5×10−6.
Step 3: Calculate the pH value using a calculator:
pH =−log1.5×10−6
pH =−(−5.82)
pH ≈5.82
Therefore, the pH of the solution is approximately 5.82.
13
Question 22
Question
Calculate the pH of a solution with a hydroxide ion concentration of 2.5×
10−4M.
Solution
Step 1: Calculate the pOH of the solution using the formula
pOH =−log[OH−]
Given that the hydroxide ion concentration is 2.5×10−4M, we have
pOH =−log2.5×10−4
pOH =−log 2.5−log 10−4
pOH =−(log 2.5+(−4))
pOH =−(log 2.5−4)
pOH ≈ −(0.3979 −4)
pOH ≈ −(−3.6021)
pOH ≈3.6021
Therefore, the pOH of the solution is 3.6021.
Step 2: Calculate the pH using the formula
pH = 14 −pOH
pH = 14 −3.6021
pH ≈10.3979
Therefore, the pH of the solution is approximately 10.3979.
Question 23
Question
Calculate the pH of a solution with a hydronium ion concentration of 3.5×10−4
M.
14
Solution
Step 1: Recall that the pH of a solution is calculated using the formula:
pH = −log[H3O+]
Step 2: Substitute the given hydronium ion concentration into the formula:
pH = −log3.5×10−4
Step 3: Calculate the pH:
pH = −log3.5×10−4=−log(3.5) −log10−4=−log(3.5) −(−4)
Step 4: Use the property of logarithms (log(a)−log(b) = log(a/b)):
pH = −log(3.5) + 4 = −(0.5441) + 4
Step 5: Calculate the pH:
pH = 3.4559
Therefore, the pH of the solution with a hydronium ion concentration of
3.5×10−4M is 3.46.
Question 24
Question
Calculate the pH of a solution with a hydroxide ion concentration of 2.5×10−4
M.
Solution
Step 1: Write the expression for the ion product of water. Since water undergoes
autoionization to produce hydronium and hydroxide ions, the ion product of
water (Kw) is given by:
Kw= [H3O+][OH−]
Given that Kw= 1.0×10−14 at 25
°
C, we can use this value to find [H3O+].
Step 2: Calculate the hydronium ion concentration.
[H3O+] = Kw
[OH−]
[H3O+] = 1.0×10−14
2.5×10−4
[H3O+] = 4.0×10−11 M
15
Step 3: Calculate the pH of the solution. The pH is defined as the negative
logarithm of the hydronium ion concentration:
pH = −log[H3O+]
pH = −log4.0×10−11
pH = −(log 4.0 + log 10−11)
pH = −(0.6021 + (−11))
pH = 10.6021
Therefore, the pH of the solution with a hydroxide ion concentration of
2.5×10−4M is 10.60.
Question 25
Question
Calculate the pH of a solution with a hydroxide ion concentration of 2.5×10−3
M.
Solution
Step 1: Use the formula pOH =−log[OH−] to find the pOH of the solution.
pOH = −log2.5×10−3
pOH ≈2.60
Step 2: Since pH + pOH = 14, we can now find the pH of the solution.
pH = 14 −2.60
pH ≈11.40
Therefore, the pH of the solution is approximately 11.40.
Question 26
Question
Calculate the pH of a solution with a hydronium ion concentration of 3.2×10−9
M.
16
Solution
Step 1: Recall that pH is defined as the negative logarithm of the hydronium
ion concentration:
pH = −log[H3O+]
Step 2: Substitute the given hydronium ion concentration into the equation:
pH = −log3.2×10−9
Step 3: Use the properties of logarithms to simplify the expression:
pH = −log(3.2) −log10−9
pH = −log(3.2) + 9
Step 4: Calculate the value of −log(3.2) using a calculator:
−log(3.2) ≈ −0.505
Step 5: Add 9 to the value of −log(3.2) to find the pH:
pH ≈ −0.505 + 9
pH ≈8.495
Step 6: Therefore, the pH of the solution with a hydronium ion concentration
of 3.2×10−9M is approximately 8.495.
Question 27
Question
Calculate the pH of a solution with a hydroxide ion concentration of 1.0×10−4
M.
Solution
Step 1: Write the expression for the ion product of water.
Kw= [H+][OH−] = 1.0×10−14
Step 2: Since we know the hydroxide ion concentration, we can rearrange
the ion product of water expression to find the hydrogen ion concentration.
[H+] = Kw
[OH−]=1.0×10−14
1.0×10−4= 1.0×10−10
Step 3: Calculate the pH of the solution using the formula pH =−log [H+].
pH =−log 1.0×10−10 = 10
Therefore, the pH of the solution is 10.
17
Question 28
Question
Calculate the pH of a solution with a hydroxide ion concentration of 1.5×
10−4M.
Solution
Step 1: Calculate the pOH of the solution using the formula:
pOH =−log[OH−]
Given that [OH−]=1.5×10−4M, we have:
pOH =−log1.5×10−4
pOH ≈ −log(1.5) = −0.176
Step 2: Use the relation between pOH and pH to find the pH of the solution:
pOH +pH = 14
Substitute the calculated pOH value into the equation:
−0.176 + pH = 14
pH = 14 + 0.176
pH ≈14.176
Therefore, the pH of the solution with a hydroxide ion concentration of
1.5×10−4Mis approximately 14.176.
Question 29
Question
Calculate the pH of a solution with a hydrogen ion concentration of 3.5×10−9
M.
Solution
Step 1: Recall that the pH of a solution can be calculated using the formula:
pH = −log[H+]
Step 2: Given the hydrogen ion concentration is 3.5×10−9M, we can
substitute this value into the formula:
pH = −log3.5×10−9
18
Step 3: Calculate the pH using a calculator:
pH ≈ −log3.5×10−9≈ −(−8.46) ≈8.46
Step 4: Therefore, the pH of the solution with a hydrogen ion concentration
of 3.5×10−9M is approximately 8.46.
Question 30
Question
Calculate the pH of a solution with a hydroxide ion concentration of 2.5×10−3
M.
Solution
Step 1: Write the equation for the ionization of water:
H2O⇌H++ OH−
Step 2: Use the fact that in water, [H+] = [OH−]=1.0×10−14 at 25
°
C.
Therefore:
Kw= [H+]×[OH−]=1.0×10−14
[H+] = 1.0×10−14
2.5×10−3= 4 ×10−12 M
Step 3: Calculate the pH using the formula: pH =−log[H+]
pH =−log4×10−12=−log(4) −log10−12= 11.4
Therefore, the pH of the solution is 11.4.
Question 31
Question
Calculate the pH of a solution with a hydroxide ion concentration of 1.5×10−9
M.
Solution
Step 1: Recall the relationship between pOH and OH−concentration:
pOH =−log[OH−]
Step 2: Plug in the given hydroxide ion concentration to find pOH:
pOH =−log1.5×10−9
19
pOH ≈ −log(1.5) −log10−9
pOH ≈ −0.1761 −(−9)
pOH ≈8.8239 (rounded to 4 decimal places)
Step 3: Use the relationship between pOH and pH to find the pH:
pH +pOH = 14
Step 4: Substitute the calculated pOH value into the equation:
pH + 8.8239 = 14
pH = 14 −8.8239
pH ≈5.1761 (rounded to 4 decimal places)
The pH of the solution is approximately 5.1761.
Question 32
Question
A 0.025 M solution of sodium hydroxide (NaOH) is prepared. Calculate the
pOH of the solution.
Solution
Step 1: Write the balanced equation for the dissociation of sodium hydroxide.
Step 2: Calculate the concentration of hydroxide ions ([OH−]) in the solution.
Step 3: Use the concentration of hydroxide ions to calculate the pOH.
Step 1: Balanced equation for the dissociation of sodium hydroxide:
NaOH →Na++ OH−
Step 2: Calculate the concentration of hydroxide ions ([OH−]): Given:
M = 0.025 M for NaOH. Since NaOH dissociates into one Na+ion and one
OH−ion, the concentration of hydroxide ions is also 0.025 M.
Step 3: Calculate pOH:
pOH = −log[OH−]
pOH = −log(0.025)
pOH = −log2.5×10−2
pOH = −(−1.60)
pOH = 1.60
Therefore, the pOH of the 0.025 M solution of sodium hydroxide is 1.60.
20
Question 33
Question
Calculate the pH of a solution that has a hydronium ion concentration of 1.5×
10−3M.
Solution
Step 1: Recall that pH is defined as the negative logarithm of the hydronium
ion concentration:
pH = −log[H3O+]
Step 2: Substitute the given hydronium ion concentration into the formula
and solve for pH:
pH = −log1.5×10−3
Step 3: Calculate the pH:
pH = −log1.5×10−3=−log(1.5) + log10−3
Step 4: Simplify the logarithmic terms:
pH = −log(1.5) + (−3) = −0.176 + (−3) = −3.176
Step 5: Therefore, the pH of the solution is 3.176 .
Question 34
Question
Calculate the pH of a solution with a hydronium ion concentration of 1.5×10−4
M.
Solution
Step 1: Recall that the pH is calculated using the formula: pH =−log[H3O+].
Step 2: Substituting the given hydronium ion concentration into the formula,
we get:
pH =−log1.5×10−4
Step 3: Using the properties of logarithms, we simplify the expression to
find the pH:
pH =−log(1.5) −log10−4
Step 4: Remember that log10−4=−4, so we have:
pH =−log(1.5) −(−4)
21
Step 5: Calculating the value of −log(1.5) using a calculator, we find:
pH =−0.1761 −(−4)
Step 6: Simplifying further, we get:
pH = 4.0−0.1761
Step 7: Finally, we calculate the pH:
pH = 3.8239
Therefore, the pH of the solution is 3.8239.
Question 35
Question
A 0.050 M solution of acetic acid (Ka = 1.8×10−5) is prepared. Calculate the
pH of the solution.
Solution
Step 1: Write the equation for the dissociation of acetic acid:
CH3COOH +H2O⇌CH3COO−+H3O+
Step 2: Set up an ICE table to determine the equilibrium concentrations:
Species CH3COOH H2O CH3COO−
H3O+
Initial (M) 0.050 −0
0
Change (M) −x+x+x
+x
Equilibrium (M) 0.050 −x x x
x
Step 3: Write the equilibrium expression for the dissociation of acetic acid
and substitute the equilibrium concentrations into the expression:
Ka=[CH3COO−][H3O+]
[CH3COOH]
1.8×10−5=x·x
0.050 −x
Step 4: Since the value of x is much smaller than 0.050, we can assume that
0.050 −x≈0.050:
1.8×10−5=x·x
0.050
22
Question 7
Question
Calculate the pH of a solution with a hydroxide ion concentration of 3.2×10−5
M.
Solution
Step 1: Write the expression for calculating the pOH of the solution. Step 2:
Use the pOH value to find the pH of the solution. Step 3: Calculate the final
pH value for the solution.
Step 1:
The pOH of a solution can be calculated using the formula:
pOH =−log[OH−]
Given that the hydroxide ion concentration is 3.2×10−5M, we can substitute
this value into the formula:
pOH =−log3.2×10−5
Step 2:
To find the pH of the solution, we can use the relationship:
pH +pOH = 14
Substitute the calculated pOH value to solve for pH:
pH + (−log3.2×10−5) = 14
Step 3:
Solve for pH:
pH = 14 + log3.2×10−5
pH = 14 + (−4.49485)
pH = 9.50515
Therefore, the pH of the solution with a hydroxide ion concentration of
3.2×10−5M is approximately 9.51.
Question 8
Question
Calculate the pH of a solution with a hydrogen ion concentration of 3.2×10−5
M.
5
Solution
Step 1: Recall that the pH of a solution is calculated using the formula:
pH = −log[H+]
Step 2: Substitute the given hydrogen ion concentration into the formula:
pH = −log3.2×10−5
Step 3: Calculate the pH:
pH = −log3.2×10−5=−log(3.2) −log10−5
Step 4: Recall that log(10−x) = −x, so:
pH = −log(3.2) −(−5) = −log(3.2) + 5
Step 5: Use a calculator to find the value of −log(3.2):
pH ≈ −0.5052 + 5 ≈4.4948
Step 6: Therefore, the pH of the solution with a hydrogen ion concentration
of 3.2×10−5M is approximately 4.49.
Question 9
Question
Calculate the pH of a solution with a hydroxide ion concentration of 3.5×10−5
M.
Solution
Step 1: Use the relation between pOH and OH−concentration, which is pOH =
−log[OH−].
Given that [OH−]=3.5×10−5M, we can calculate pOH:
pOH = −log3.5×10−5
Step 2: Calculate pOH.
pOH = −log3.5×10−5=−(−4.4559) ≈4.46
Step 3: Use the relation between pH and pOH, which is pH + pOH = 14, to
find pH.
pH = 14 −pOH = 14 −4.46
Step 4: Calculate the pH of the solution.
pH = 14 −4.46 = 9.54
Therefore, the pH of a solution with a hydroxide ion concentration of 3.5×
10−5M is 9.54.
6
Question 10
Question
Calculate the pH of a solution with a hydroxide ion concentration of 1.5×10−4
M.
Solution
Step 1: Write the equilibrium expression for the autoionization of water:
H2O⇌H++ OH−
Step 2: Write the expression for the equilibrium constant, Kw, for the au-
toionization of water:
Kw= [H+][OH−]=1.0×10−14
Step 3: Given the hydroxide ion concentration, we can find the hydrogen ion
concentration using the equilibrium constant:
[H+] = Kw
[OH−]=1.0×10−14
1.5×10−4= 6.67 ×10−11 M
Step 4: Calculate the pH of the solution using the hydrogen ion concentra-
tion:
pH = −log[H+]=−log6.67 ×10−11= 10.18
Therefore, the pH of the solution is 10.18.
Question 11
Question
Calculate the pH of a solution with a hydronium ion concentration of 1.5×10−3
M.
Solution
Step 1: Recall that pH is defined as −log[H3O+], where [H3O+] is the concen-
tration of hydronium ions in the solution.
Step 2: Substitute the given hydronium ion concentration into the pH for-
mula:
pH = −log1.5×10−3
Step 3: Calculate the pH using a calculator:
pH = −log1.5×10−3=−(−2.823) = 2.823
Step 4: Therefore, the pH of the solution with a hydronium ion concentration
of 1.5×10−3M is 2.823.
7
Question 12
Question
Calculate the pH of a solution with a hydronium ion concentration of 3.2×10−5
M.
Solution
Step 1: Recall that the pH of a solution is defined as −log[H3O+], where [H3O+]
represents the concentration of hydronium ions in moles per liter.
Step 2: Given that [H3O+]=3.2×10−5M, we can calculate the pH as:
pH = −log3.2×10−5
Step 3: Using the properties of logarithms, we get:
pH = −log(3.2) −log10−5
Step 4: Further simplifying, we find:
pH = −log(3.2) −(−5)
Step 5: Using a calculator to evaluate −log(3.2), we obtain −0.5051.
Step 6: Therefore, the pH of the solution is:
pH = −0.5051 −(−5) = 4.4949
Step 7: Thus, the pH of the solution with a hydronium ion concentration of
3.2×10−5M is approximately 4.49.
Question 13
Question
Calculate the pH of a solution that has a hydroxide ion concentration of 2.5×
10−4M.
Solution
Step 1: Recall that the pH of a solution can be calculated using the formula:
pH =−log[H+]. Since pH +pOH = 14, we can first find the pOH of the
solution.
Step 2: Given that [OH−] = 2.5×10−4M, we can calculate the pOH using
the formula pOH =−log[OH−].
pOH =−log2.5×10−4
pOH ≈ −log(2.5) −log10−4
8
pOH ≈ −log(2.5) + 4
pOH ≈ −0.3979 + 4
pOH ≈3.6021
Step 3: Now, we can find the pH using the relationship pH +pOH = 14.
pH = 14 −pOH
pH = 14 −3.6021
pH ≈10.3979
Therefore, the pH of the solution is approximately 10.40.
Question 14
Question
Calculate the pH of a solution with a hydronium ion concentration of 1.5×
10−3M.
Solution
Step 1: Recall that the pH is defined as −log[H3O+]. We are given that
[H3O+]=1.5×10−3M.
Step 2: Substitute the given concentration into the formula for pH:
pH = −log1.5×10−3
Step 3: Calculate the pH using a calculator:
pH = −log1.5×10−3≈ −log(0.0015) ≈2.82
Step 4: Therefore, the pH of the solution is approximately 2.82.
Question 15
Question
Calculate the pH of a solution with a hydroxide ion concentration of 1.5×10−9
M.
9
Solution
Step 1: Write the expression for the ion product constant of water.
Kw= [H+]×[OH−]=1.0×10−14
Step 2: Since the solution is basic, we know that pOH = −log[OH−].
pOH = −log1.5×10−9
Step 3: Use the relationship between pH and pOH to find the pH of the
solution.
pH = 14 −pOH
pH = 14 −(−log1.5×10−9)
Step 4: Calculate the pH.
pH = 14 −(−log1.5×10−9) = 14 + log1.5×10−9
pH ≈14 + 8.823
pH ≈22.823
Therefore, the pH of the solution is approximately 22.823.
Question 16
Question
Calculate the pH of a solution with a hydroxide ion concentration of 3.5×10−4
M.
Solution
Step 1: Write the expression for calculating the pOH of the solution.
pOH = −log[OH−]
Step 2: Plug in the given hydroxide ion concentration to calculate pOH.
pOH = −log3.5×10−4≈ −(−3.455) ≈3.455
Step 3: Use the relationship pH + pOH = 14 to calculate the pH of the
solution.
pH = 14 −pOH = 14 −3.455 ≈10.545
Therefore, the pH of the solution is approximately 10.545.
10
Question 17
Question
Calculate the pH of a solution with a hydroxide ion concentration of 2.5×10−10
M.
Solution
Step 1: Recall that the relationship between pOH and pH is given by:
pOH =−log[OH−]
Step 2: We are given the hydroxide ion concentration as 2.5×10−10 M. Let’s
calculate the pOH:
pOH =−log2.5×10−10
pOH =−log(2.5) −log10−10
pOH =−log(2.5) −(−10)
pOH =−0.3979 + 10
pOH = 9.6021
Step 3: Since pH +pOH = 14, we can find the pH:
pH = 14 −pOH
pH = 14 −9.6021
pH = 4.3979
Step 4: Therefore, the pH of the solution with a hydroxide ion concentration of
2.5×10−10 M is approximately 4.40.
Question 18
Question
Calculate the pH of a 0.005 M hydrochloric acid solution.
Solution
Step 1: Write the chemical equation for the dissociation of hydrochloric acid:
HCl(aq)→H+(aq) + Cl−(aq)
Step 2: Use the concentration of hydrochloric acid to determine the con-
centration of hydrogen ions (H+) in the solution. Since hydrochloric acid is a
11
strong acid and completely dissociates, the concentration of H+ions is equal to
the concentration of the hydrochloric acid solution:
[H+]=0.005 M
Step 3: Calculate the pH using the formula:
pH = −log[H+]
Substitute the concentration of hydrogen ions into the formula:
pH = −log(0.005)
Step 4: Calculate the pH:
pH = −log(0.005) = −(−2.3) = 2.3
Therefore, the pH of the 0.005 M hydrochloric acid solution is 2.3.
Question 19
Question
Calculate the pH and pOH of a 0.005 M solution of perchloric acid (HClO4).
Solution
Step 1: Write the balanced equation for the dissociation of perchloric acid:
HClO4→H++ ClO−
4.
Step 2: Calculate the concentration of H+ions in the solution. Since perchlo-
ric acid is a strong acid, it dissociates completely. Therefore, the concentration
of H+ions is the same as the initial concentration of perchloric acid: 0.005 M.
Step 3: Calculate the pH using the formula: pH = −log[H+]. Substitute
the value of [H+] into the formula: pH = −log(0.005) ≈2.3.
Step 4: Calculate the pOH using the formula: pOH = −log[OH−]. Since
water autoionization is negligible in this case, [OH−]≈0. Therefore, pOH =
−log(0) →pOH = ∞.
Therefore, the pH of the 0.005 M solution of perchloric acid is approximately
2.3, and the pOH is infinity.
Question 20
Question
Calculate the pH of a solution with a hydroxide ion concentration of 6.3×10−9
M.
12
Solution
Step 1: Calculate the pOH of the solution using the hydroxide ion concentration.
Step 2: Use the relation between pOH and pH to calculate the pH of the
solution.
Step 1: Calculate the pOH. Given: [OH−]=6.3×10−9M.
The formula relating pOH to [OH−] is:
pOH =−log[OH−]
pOH =−log6.3×10−9
pOH =−log(6.3) −log10−9
pOH =−log(6.3) −(−9)
pOH =−0.799 −(−9)
pOH = 8.201
Step 2: Calculate the pH using the relation between pH and pOH. The
relation between pH and pOH is:
pH +pOH = 14
pH = 14 −pOH
pH = 14 −8.201
pH = 5.799
Therefore, the pH of the solution with a hydroxide ion concentration of
6.3×10−9M is 5.799.
Question 21
Question
Calculate the pH of a solution with a hydronium ion concentration of 1.5×10−6
M.
Solution
Step 1: Recall that pH is defined as the negative base-10 logarithm of the
hydronium ion concentration. The formula for pH is: pH =−log[H3O+].
Step 2: Substitute the given hydronium ion concentration into the formula:
pH =−log1.5×10−6.
Step 3: Calculate the pH value using a calculator:
pH =−log1.5×10−6
pH =−(−5.82)
pH ≈5.82
Therefore, the pH of the solution is approximately 5.82.
13
Question 22
Question
Calculate the pH of a solution with a hydroxide ion concentration of 2.5×
10−4M.
Solution
Step 1: Calculate the pOH of the solution using the formula
pOH =−log[OH−]
Given that the hydroxide ion concentration is 2.5×10−4M, we have
pOH =−log2.5×10−4
pOH =−log 2.5−log 10−4
pOH =−(log 2.5+(−4))
pOH =−(log 2.5−4)
pOH ≈ −(0.3979 −4)
pOH ≈ −(−3.6021)
pOH ≈3.6021
Therefore, the pOH of the solution is 3.6021.
Step 2: Calculate the pH using the formula
pH = 14 −pOH
pH = 14 −3.6021
pH ≈10.3979
Therefore, the pH of the solution is approximately 10.3979.
Question 23
Question
Calculate the pH of a solution with a hydronium ion concentration of 3.5×10−4
M.
14
Solution
Step 1: Recall that the pH of a solution is calculated using the formula:
pH = −log[H3O+]
Step 2: Substitute the given hydronium ion concentration into the formula:
pH = −log3.5×10−4
Step 3: Calculate the pH:
pH = −log3.5×10−4=−log(3.5) −log10−4=−log(3.5) −(−4)
Step 4: Use the property of logarithms (log(a)−log(b) = log(a/b)):
pH = −log(3.5) + 4 = −(0.5441) + 4
Step 5: Calculate the pH:
pH = 3.4559
Therefore, the pH of the solution with a hydronium ion concentration of
3.5×10−4M is 3.46.
Question 24
Question
Calculate the pH of a solution with a hydroxide ion concentration of 2.5×10−4
M.
Solution
Step 1: Write the expression for the ion product of water. Since water undergoes
autoionization to produce hydronium and hydroxide ions, the ion product of
water (Kw) is given by:
Kw= [H3O+][OH−]
Given that Kw= 1.0×10−14 at 25
°
C, we can use this value to find [H3O+].
Step 2: Calculate the hydronium ion concentration.
[H3O+] = Kw
[OH−]
[H3O+] = 1.0×10−14
2.5×10−4
[H3O+] = 4.0×10−11 M
15
Step 3: Calculate the pH of the solution. The pH is defined as the negative
logarithm of the hydronium ion concentration:
pH = −log[H3O+]
pH = −log4.0×10−11
pH = −(log 4.0 + log 10−11)
pH = −(0.6021 + (−11))
pH = 10.6021
Therefore, the pH of the solution with a hydroxide ion concentration of
2.5×10−4M is 10.60.
Question 25
Question
Calculate the pH of a solution with a hydroxide ion concentration of 2.5×10−3
M.
Solution
Step 1: Use the formula pOH =−log[OH−] to find the pOH of the solution.
pOH = −log2.5×10−3
pOH ≈2.60
Step 2: Since pH + pOH = 14, we can now find the pH of the solution.
pH = 14 −2.60
pH ≈11.40
Therefore, the pH of the solution is approximately 11.40.
Question 26
Question
Calculate the pH of a solution with a hydronium ion concentration of 3.2×10−9
M.
16
Solution
Step 1: Recall that pH is defined as the negative logarithm of the hydronium
ion concentration:
pH = −log[H3O+]
Step 2: Substitute the given hydronium ion concentration into the equation:
pH = −log3.2×10−9
Step 3: Use the properties of logarithms to simplify the expression:
pH = −log(3.2) −log10−9
pH = −log(3.2) + 9
Step 4: Calculate the value of −log(3.2) using a calculator:
−log(3.2) ≈ −0.505
Step 5: Add 9 to the value of −log(3.2) to find the pH:
pH ≈ −0.505 + 9
pH ≈8.495
Step 6: Therefore, the pH of the solution with a hydronium ion concentration
of 3.2×10−9M is approximately 8.495.
Question 27
Question
Calculate the pH of a solution with a hydroxide ion concentration of 1.0×10−4
M.
Solution
Step 1: Write the expression for the ion product of water.
Kw= [H+][OH−] = 1.0×10−14
Step 2: Since we know the hydroxide ion concentration, we can rearrange
the ion product of water expression to find the hydrogen ion concentration.
[H+] = Kw
[OH−]=1.0×10−14
1.0×10−4= 1.0×10−10
Step 3: Calculate the pH of the solution using the formula pH =−log [H+].
pH =−log 1.0×10−10 = 10
Therefore, the pH of the solution is 10.
17
Question 28
Question
Calculate the pH of a solution with a hydroxide ion concentration of 1.5×
10−4M.
Solution
Step 1: Calculate the pOH of the solution using the formula:
pOH =−log[OH−]
Given that [OH−]=1.5×10−4M, we have:
pOH =−log1.5×10−4
pOH ≈ −log(1.5) = −0.176
Step 2: Use the relation between pOH and pH to find the pH of the solution:
pOH +pH = 14
Substitute the calculated pOH value into the equation:
−0.176 + pH = 14
pH = 14 + 0.176
pH ≈14.176
Therefore, the pH of the solution with a hydroxide ion concentration of
1.5×10−4Mis approximately 14.176.
Question 29
Question
Calculate the pH of a solution with a hydrogen ion concentration of 3.5×10−9
M.
Solution
Step 1: Recall that the pH of a solution can be calculated using the formula:
pH = −log[H+]
Step 2: Given the hydrogen ion concentration is 3.5×10−9M, we can
substitute this value into the formula:
pH = −log3.5×10−9
18
Step 3: Calculate the pH using a calculator:
pH ≈ −log3.5×10−9≈ −(−8.46) ≈8.46
Step 4: Therefore, the pH of the solution with a hydrogen ion concentration
of 3.5×10−9M is approximately 8.46.
Question 30
Question
Calculate the pH of a solution with a hydroxide ion concentration of 2.5×10−3
M.
Solution
Step 1: Write the equation for the ionization of water:
H2O⇌H++ OH−
Step 2: Use the fact that in water, [H+] = [OH−]=1.0×10−14 at 25
°
C.
Therefore:
Kw= [H+]×[OH−]=1.0×10−14
[H+] = 1.0×10−14
2.5×10−3= 4 ×10−12 M
Step 3: Calculate the pH using the formula: pH =−log[H+]
pH =−log4×10−12=−log(4) −log10−12= 11.4
Therefore, the pH of the solution is 11.4.
Question 31
Question
Calculate the pH of a solution with a hydroxide ion concentration of 1.5×10−9
M.
Solution
Step 1: Recall the relationship between pOH and OH−concentration:
pOH =−log[OH−]
Step 2: Plug in the given hydroxide ion concentration to find pOH:
pOH =−log1.5×10−9
19
pOH ≈ −log(1.5) −log10−9
pOH ≈ −0.1761 −(−9)
pOH ≈8.8239 (rounded to 4 decimal places)
Step 3: Use the relationship between pOH and pH to find the pH:
pH +pOH = 14
Step 4: Substitute the calculated pOH value into the equation:
pH + 8.8239 = 14
pH = 14 −8.8239
pH ≈5.1761 (rounded to 4 decimal places)
The pH of the solution is approximately 5.1761.
Question 32
Question
A 0.025 M solution of sodium hydroxide (NaOH) is prepared. Calculate the
pOH of the solution.
Solution
Step 1: Write the balanced equation for the dissociation of sodium hydroxide.
Step 2: Calculate the concentration of hydroxide ions ([OH−]) in the solution.
Step 3: Use the concentration of hydroxide ions to calculate the pOH.
Step 1: Balanced equation for the dissociation of sodium hydroxide:
NaOH →Na++ OH−
Step 2: Calculate the concentration of hydroxide ions ([OH−]): Given:
M = 0.025 M for NaOH. Since NaOH dissociates into one Na+ion and one
OH−ion, the concentration of hydroxide ions is also 0.025 M.
Step 3: Calculate pOH:
pOH = −log[OH−]
pOH = −log(0.025)
pOH = −log2.5×10−2
pOH = −(−1.60)
pOH = 1.60
Therefore, the pOH of the 0.025 M solution of sodium hydroxide is 1.60.
20
Question 33
Question
Calculate the pH of a solution that has a hydronium ion concentration of 1.5×
10−3M.
Solution
Step 1: Recall that pH is defined as the negative logarithm of the hydronium
ion concentration:
pH = −log[H3O+]
Step 2: Substitute the given hydronium ion concentration into the formula
and solve for pH:
pH = −log1.5×10−3
Step 3: Calculate the pH:
pH = −log1.5×10−3=−log(1.5) + log10−3
Step 4: Simplify the logarithmic terms:
pH = −log(1.5) + (−3) = −0.176 + (−3) = −3.176
Step 5: Therefore, the pH of the solution is 3.176 .
Question 34
Question
Calculate the pH of a solution with a hydronium ion concentration of 1.5×10−4
M.
Solution
Step 1: Recall that the pH is calculated using the formula: pH =−log[H3O+].
Step 2: Substituting the given hydronium ion concentration into the formula,
we get:
pH =−log1.5×10−4
Step 3: Using the properties of logarithms, we simplify the expression to
find the pH:
pH =−log(1.5) −log10−4
Step 4: Remember that log10−4=−4, so we have:
pH =−log(1.5) −(−4)
21
Step 5: Calculating the value of −log(1.5) using a calculator, we find:
pH =−0.1761 −(−4)
Step 6: Simplifying further, we get:
pH = 4.0−0.1761
Step 7: Finally, we calculate the pH:
pH = 3.8239
Therefore, the pH of the solution is 3.8239.
Question 35
Question
A 0.050 M solution of acetic acid (Ka = 1.8×10−5) is prepared. Calculate the
pH of the solution.
Solution
Step 1: Write the equation for the dissociation of acetic acid:
CH3COOH +H2O⇌CH3COO−+H3O+
Step 2: Set up an ICE table to determine the equilibrium concentrations:
Species CH3COOH H2O CH3COO−
H3O+
Initial (M) 0.050 −0
0
Change (M) −x+x+x
+x
Equilibrium (M) 0.050 −x x x
x
Step 3: Write the equilibrium expression for the dissociation of acetic acid
and substitute the equilibrium concentrations into the expression:
Ka=[CH3COO−][H3O+]
[CH3COOH]
1.8×10−5=x·x
0.050 −x
Step 4: Since the value of x is much smaller than 0.050, we can assume that
0.050 −x≈0.050:
1.8×10−5=x·x
0.050
22
Question 7
Question
Calculate the pH of a solution with a hydroxide ion concentration of 3.2×10−5
M.
Solution
Step 1: Write the expression for calculating the pOH of the solution. Step 2:
Use the pOH value to find the pH of the solution. Step 3: Calculate the final
pH value for the solution.
Step 1:
The pOH of a solution can be calculated using the formula:
pOH =−log[OH−]
Given that the hydroxide ion concentration is 3.2×10−5M, we can substitute
this value into the formula:
pOH =−log3.2×10−5
Step 2:
To find the pH of the solution, we can use the relationship:
pH +pOH = 14
Substitute the calculated pOH value to solve for pH:
pH + (−log3.2×10−5) = 14
Step 3:
Solve for pH:
pH = 14 + log3.2×10−5
pH = 14 + (−4.49485)
pH = 9.50515
Therefore, the pH of the solution with a hydroxide ion concentration of
3.2×10−5M is approximately 9.51.
Question 8
Question
Calculate the pH of a solution with a hydrogen ion concentration of 3.2×10−5
M.
5
Solution
Step 1: Recall that the pH of a solution is calculated using the formula:
pH = −log[H+]
Step 2: Substitute the given hydrogen ion concentration into the formula:
pH = −log3.2×10−5
Step 3: Calculate the pH:
pH = −log3.2×10−5=−log(3.2) −log10−5
Step 4: Recall that log(10−x) = −x, so:
pH = −log(3.2) −(−5) = −log(3.2) + 5
Step 5: Use a calculator to find the value of −log(3.2):
pH ≈ −0.5052 + 5 ≈4.4948
Step 6: Therefore, the pH of the solution with a hydrogen ion concentration
of 3.2×10−5M is approximately 4.49.
Question 9
Question
Calculate the pH of a solution with a hydroxide ion concentration of 3.5×10−5
M.
Solution
Step 1: Use the relation between pOH and OH−concentration, which is pOH =
−log[OH−].
Given that [OH−]=3.5×10−5M, we can calculate pOH:
pOH = −log3.5×10−5
Step 2: Calculate pOH.
pOH = −log3.5×10−5=−(−4.4559) ≈4.46
Step 3: Use the relation between pH and pOH, which is pH + pOH = 14, to
find pH.
pH = 14 −pOH = 14 −4.46
Step 4: Calculate the pH of the solution.
pH = 14 −4.46 = 9.54
Therefore, the pH of a solution with a hydroxide ion concentration of 3.5×
10−5M is 9.54.
6
Question 10
Question
Calculate the pH of a solution with a hydroxide ion concentration of 1.5×10−4
M.
Solution
Step 1: Write the equilibrium expression for the autoionization of water:
H2O⇌H++ OH−
Step 2: Write the expression for the equilibrium constant, Kw, for the au-
toionization of water:
Kw= [H+][OH−]=1.0×10−14
Step 3: Given the hydroxide ion concentration, we can find the hydrogen ion
concentration using the equilibrium constant:
[H+] = Kw
[OH−]=1.0×10−14
1.5×10−4= 6.67 ×10−11 M
Step 4: Calculate the pH of the solution using the hydrogen ion concentra-
tion:
pH = −log[H+]=−log6.67 ×10−11= 10.18
Therefore, the pH of the solution is 10.18.
Question 11
Question
Calculate the pH of a solution with a hydronium ion concentration of 1.5×10−3
M.
Solution
Step 1: Recall that pH is defined as −log[H3O+], where [H3O+] is the concen-
tration of hydronium ions in the solution.
Step 2: Substitute the given hydronium ion concentration into the pH for-
mula:
pH = −log1.5×10−3
Step 3: Calculate the pH using a calculator:
pH = −log1.5×10−3=−(−2.823) = 2.823
Step 4: Therefore, the pH of the solution with a hydronium ion concentration
of 1.5×10−3M is 2.823.
7
Question 12
Question
Calculate the pH of a solution with a hydronium ion concentration of 3.2×10−5
M.
Solution
Step 1: Recall that the pH of a solution is defined as −log[H3O+], where [H3O+]
represents the concentration of hydronium ions in moles per liter.
Step 2: Given that [H3O+]=3.2×10−5M, we can calculate the pH as:
pH = −log3.2×10−5
Step 3: Using the properties of logarithms, we get:
pH = −log(3.2) −log10−5
Step 4: Further simplifying, we find:
pH = −log(3.2) −(−5)
Step 5: Using a calculator to evaluate −log(3.2), we obtain −0.5051.
Step 6: Therefore, the pH of the solution is:
pH = −0.5051 −(−5) = 4.4949
Step 7: Thus, the pH of the solution with a hydronium ion concentration of
3.2×10−5M is approximately 4.49.
Question 13
Question
Calculate the pH of a solution that has a hydroxide ion concentration of 2.5×
10−4M.
Solution
Step 1: Recall that the pH of a solution can be calculated using the formula:
pH =−log[H+]. Since pH +pOH = 14, we can first find the pOH of the
solution.
Step 2: Given that [OH−] = 2.5×10−4M, we can calculate the pOH using
the formula pOH =−log[OH−].
pOH =−log2.5×10−4
pOH ≈ −log(2.5) −log10−4
8
pOH ≈ −log(2.5) + 4
pOH ≈ −0.3979 + 4
pOH ≈3.6021
Step 3: Now, we can find the pH using the relationship pH +pOH = 14.
pH = 14 −pOH
pH = 14 −3.6021
pH ≈10.3979
Therefore, the pH of the solution is approximately 10.40.
Question 14
Question
Calculate the pH of a solution with a hydronium ion concentration of 1.5×
10−3M.
Solution
Step 1: Recall that the pH is defined as −log[H3O+]. We are given that
[H3O+]=1.5×10−3M.
Step 2: Substitute the given concentration into the formula for pH:
pH = −log1.5×10−3
Step 3: Calculate the pH using a calculator:
pH = −log1.5×10−3≈ −log(0.0015) ≈2.82
Step 4: Therefore, the pH of the solution is approximately 2.82.
Question 15
Question
Calculate the pH of a solution with a hydroxide ion concentration of 1.5×10−9
M.
9
Solution
Step 1: Write the expression for the ion product constant of water.
Kw= [H+]×[OH−]=1.0×10−14
Step 2: Since the solution is basic, we know that pOH = −log[OH−].
pOH = −log1.5×10−9
Step 3: Use the relationship between pH and pOH to find the pH of the
solution.
pH = 14 −pOH
pH = 14 −(−log1.5×10−9)
Step 4: Calculate the pH.
pH = 14 −(−log1.5×10−9) = 14 + log1.5×10−9
pH ≈14 + 8.823
pH ≈22.823
Therefore, the pH of the solution is approximately 22.823.
Question 16
Question
Calculate the pH of a solution with a hydroxide ion concentration of 3.5×10−4
M.
Solution
Step 1: Write the expression for calculating the pOH of the solution.
pOH = −log[OH−]
Step 2: Plug in the given hydroxide ion concentration to calculate pOH.
pOH = −log3.5×10−4≈ −(−3.455) ≈3.455
Step 3: Use the relationship pH + pOH = 14 to calculate the pH of the
solution.
pH = 14 −pOH = 14 −3.455 ≈10.545
Therefore, the pH of the solution is approximately 10.545.
10
Question 17
Question
Calculate the pH of a solution with a hydroxide ion concentration of 2.5×10−10
M.
Solution
Step 1: Recall that the relationship between pOH and pH is given by:
pOH =−log[OH−]
Step 2: We are given the hydroxide ion concentration as 2.5×10−10 M. Let’s
calculate the pOH:
pOH =−log2.5×10−10
pOH =−log(2.5) −log10−10
pOH =−log(2.5) −(−10)
pOH =−0.3979 + 10
pOH = 9.6021
Step 3: Since pH +pOH = 14, we can find the pH:
pH = 14 −pOH
pH = 14 −9.6021
pH = 4.3979
Step 4: Therefore, the pH of the solution with a hydroxide ion concentration of
2.5×10−10 M is approximately 4.40.
Question 18
Question
Calculate the pH of a 0.005 M hydrochloric acid solution.
Solution
Step 1: Write the chemical equation for the dissociation of hydrochloric acid:
HCl(aq)→H+(aq) + Cl−(aq)
Step 2: Use the concentration of hydrochloric acid to determine the con-
centration of hydrogen ions (H+) in the solution. Since hydrochloric acid is a
11
strong acid and completely dissociates, the concentration of H+ions is equal to
the concentration of the hydrochloric acid solution:
[H+]=0.005 M
Step 3: Calculate the pH using the formula:
pH = −log[H+]
Substitute the concentration of hydrogen ions into the formula:
pH = −log(0.005)
Step 4: Calculate the pH:
pH = −log(0.005) = −(−2.3) = 2.3
Therefore, the pH of the 0.005 M hydrochloric acid solution is 2.3.
Question 19
Question
Calculate the pH and pOH of a 0.005 M solution of perchloric acid (HClO4).
Solution
Step 1: Write the balanced equation for the dissociation of perchloric acid:
HClO4→H++ ClO−
4.
Step 2: Calculate the concentration of H+ions in the solution. Since perchlo-
ric acid is a strong acid, it dissociates completely. Therefore, the concentration
of H+ions is the same as the initial concentration of perchloric acid: 0.005 M.
Step 3: Calculate the pH using the formula: pH = −log[H+]. Substitute
the value of [H+] into the formula: pH = −log(0.005) ≈2.3.
Step 4: Calculate the pOH using the formula: pOH = −log[OH−]. Since
water autoionization is negligible in this case, [OH−]≈0. Therefore, pOH =
−log(0) →pOH = ∞.
Therefore, the pH of the 0.005 M solution of perchloric acid is approximately
2.3, and the pOH is infinity.
Question 20
Question
Calculate the pH of a solution with a hydroxide ion concentration of 6.3×10−9
M.
12
Solution
Step 1: Calculate the pOH of the solution using the hydroxide ion concentration.
Step 2: Use the relation between pOH and pH to calculate the pH of the
solution.
Step 1: Calculate the pOH. Given: [OH−]=6.3×10−9M.
The formula relating pOH to [OH−] is:
pOH =−log[OH−]
pOH =−log6.3×10−9
pOH =−log(6.3) −log10−9
pOH =−log(6.3) −(−9)
pOH =−0.799 −(−9)
pOH = 8.201
Step 2: Calculate the pH using the relation between pH and pOH. The
relation between pH and pOH is:
pH +pOH = 14
pH = 14 −pOH
pH = 14 −8.201
pH = 5.799
Therefore, the pH of the solution with a hydroxide ion concentration of
6.3×10−9M is 5.799.
Question 21
Question
Calculate the pH of a solution with a hydronium ion concentration of 1.5×10−6
M.
Solution
Step 1: Recall that pH is defined as the negative base-10 logarithm of the
hydronium ion concentration. The formula for pH is: pH =−log[H3O+].
Step 2: Substitute the given hydronium ion concentration into the formula:
pH =−log1.5×10−6.
Step 3: Calculate the pH value using a calculator:
pH =−log1.5×10−6
pH =−(−5.82)
pH ≈5.82
Therefore, the pH of the solution is approximately 5.82.
13
Question 22
Question
Calculate the pH of a solution with a hydroxide ion concentration of 2.5×
10−4M.
Solution
Step 1: Calculate the pOH of the solution using the formula
pOH =−log[OH−]
Given that the hydroxide ion concentration is 2.5×10−4M, we have
pOH =−log2.5×10−4
pOH =−log 2.5−log 10−4
pOH =−(log 2.5+(−4))
pOH =−(log 2.5−4)
pOH ≈ −(0.3979 −4)
pOH ≈ −(−3.6021)
pOH ≈3.6021
Therefore, the pOH of the solution is 3.6021.
Step 2: Calculate the pH using the formula
pH = 14 −pOH
pH = 14 −3.6021
pH ≈10.3979
Therefore, the pH of the solution is approximately 10.3979.
Question 23
Question
Calculate the pH of a solution with a hydronium ion concentration of 3.5×10−4
M.
14
Solution
Step 1: Recall that the pH of a solution is calculated using the formula:
pH = −log[H3O+]
Step 2: Substitute the given hydronium ion concentration into the formula:
pH = −log3.5×10−4
Step 3: Calculate the pH:
pH = −log3.5×10−4=−log(3.5) −log10−4=−log(3.5) −(−4)
Step 4: Use the property of logarithms (log(a)−log(b) = log(a/b)):
pH = −log(3.5) + 4 = −(0.5441) + 4
Step 5: Calculate the pH:
pH = 3.4559
Therefore, the pH of the solution with a hydronium ion concentration of
3.5×10−4M is 3.46.
Question 24
Question
Calculate the pH of a solution with a hydroxide ion concentration of 2.5×10−4
M.
Solution
Step 1: Write the expression for the ion product of water. Since water undergoes
autoionization to produce hydronium and hydroxide ions, the ion product of
water (Kw) is given by:
Kw= [H3O+][OH−]
Given that Kw= 1.0×10−14 at 25
°
C, we can use this value to find [H3O+].
Step 2: Calculate the hydronium ion concentration.
[H3O+] = Kw
[OH−]
[H3O+] = 1.0×10−14
2.5×10−4
[H3O+] = 4.0×10−11 M
15
Step 3: Calculate the pH of the solution. The pH is defined as the negative
logarithm of the hydronium ion concentration:
pH = −log[H3O+]
pH = −log4.0×10−11
pH = −(log 4.0 + log 10−11)
pH = −(0.6021 + (−11))
pH = 10.6021
Therefore, the pH of the solution with a hydroxide ion concentration of
2.5×10−4M is 10.60.
Question 25
Question
Calculate the pH of a solution with a hydroxide ion concentration of 2.5×10−3
M.
Solution
Step 1: Use the formula pOH =−log[OH−] to find the pOH of the solution.
pOH = −log2.5×10−3
pOH ≈2.60
Step 2: Since pH + pOH = 14, we can now find the pH of the solution.
pH = 14 −2.60
pH ≈11.40
Therefore, the pH of the solution is approximately 11.40.
Question 26
Question
Calculate the pH of a solution with a hydronium ion concentration of 3.2×10−9
M.
16
Solution
Step 1: Recall that pH is defined as the negative logarithm of the hydronium
ion concentration:
pH = −log[H3O+]
Step 2: Substitute the given hydronium ion concentration into the equation:
pH = −log3.2×10−9
Step 3: Use the properties of logarithms to simplify the expression:
pH = −log(3.2) −log10−9
pH = −log(3.2) + 9
Step 4: Calculate the value of −log(3.2) using a calculator:
−log(3.2) ≈ −0.505
Step 5: Add 9 to the value of −log(3.2) to find the pH:
pH ≈ −0.505 + 9
pH ≈8.495
Step 6: Therefore, the pH of the solution with a hydronium ion concentration
of 3.2×10−9M is approximately 8.495.
Question 27
Question
Calculate the pH of a solution with a hydroxide ion concentration of 1.0×10−4
M.
Solution
Step 1: Write the expression for the ion product of water.
Kw= [H+][OH−] = 1.0×10−14
Step 2: Since we know the hydroxide ion concentration, we can rearrange
the ion product of water expression to find the hydrogen ion concentration.
[H+] = Kw
[OH−]=1.0×10−14
1.0×10−4= 1.0×10−10
Step 3: Calculate the pH of the solution using the formula pH =−log [H+].
pH =−log 1.0×10−10 = 10
Therefore, the pH of the solution is 10.
17
Question 28
Question
Calculate the pH of a solution with a hydroxide ion concentration of 1.5×
10−4M.
Solution
Step 1: Calculate the pOH of the solution using the formula:
pOH =−log[OH−]
Given that [OH−]=1.5×10−4M, we have:
pOH =−log1.5×10−4
pOH ≈ −log(1.5) = −0.176
Step 2: Use the relation between pOH and pH to find the pH of the solution:
pOH +pH = 14
Substitute the calculated pOH value into the equation:
−0.176 + pH = 14
pH = 14 + 0.176
pH ≈14.176
Therefore, the pH of the solution with a hydroxide ion concentration of
1.5×10−4Mis approximately 14.176.
Question 29
Question
Calculate the pH of a solution with a hydrogen ion concentration of 3.5×10−9
M.
Solution
Step 1: Recall that the pH of a solution can be calculated using the formula:
pH = −log[H+]
Step 2: Given the hydrogen ion concentration is 3.5×10−9M, we can
substitute this value into the formula:
pH = −log3.5×10−9
18
Step 3: Calculate the pH using a calculator:
pH ≈ −log3.5×10−9≈ −(−8.46) ≈8.46
Step 4: Therefore, the pH of the solution with a hydrogen ion concentration
of 3.5×10−9M is approximately 8.46.
Question 30
Question
Calculate the pH of a solution with a hydroxide ion concentration of 2.5×10−3
M.
Solution
Step 1: Write the equation for the ionization of water:
H2O⇌H++ OH−
Step 2: Use the fact that in water, [H+] = [OH−]=1.0×10−14 at 25
°
C.
Therefore:
Kw= [H+]×[OH−]=1.0×10−14
[H+] = 1.0×10−14
2.5×10−3= 4 ×10−12 M
Step 3: Calculate the pH using the formula: pH =−log[H+]
pH =−log4×10−12=−log(4) −log10−12= 11.4
Therefore, the pH of the solution is 11.4.
Question 31
Question
Calculate the pH of a solution with a hydroxide ion concentration of 1.5×10−9
M.
Solution
Step 1: Recall the relationship between pOH and OH−concentration:
pOH =−log[OH−]
Step 2: Plug in the given hydroxide ion concentration to find pOH:
pOH =−log1.5×10−9
19
pOH ≈ −log(1.5) −log10−9
pOH ≈ −0.1761 −(−9)
pOH ≈8.8239 (rounded to 4 decimal places)
Step 3: Use the relationship between pOH and pH to find the pH:
pH +pOH = 14
Step 4: Substitute the calculated pOH value into the equation:
pH + 8.8239 = 14
pH = 14 −8.8239
pH ≈5.1761 (rounded to 4 decimal places)
The pH of the solution is approximately 5.1761.
Question 32
Question
A 0.025 M solution of sodium hydroxide (NaOH) is prepared. Calculate the
pOH of the solution.
Solution
Step 1: Write the balanced equation for the dissociation of sodium hydroxide.
Step 2: Calculate the concentration of hydroxide ions ([OH−]) in the solution.
Step 3: Use the concentration of hydroxide ions to calculate the pOH.
Step 1: Balanced equation for the dissociation of sodium hydroxide:
NaOH →Na++ OH−
Step 2: Calculate the concentration of hydroxide ions ([OH−]): Given:
M = 0.025 M for NaOH. Since NaOH dissociates into one Na+ion and one
OH−ion, the concentration of hydroxide ions is also 0.025 M.
Step 3: Calculate pOH:
pOH = −log[OH−]
pOH = −log(0.025)
pOH = −log2.5×10−2
pOH = −(−1.60)
pOH = 1.60
Therefore, the pOH of the 0.025 M solution of sodium hydroxide is 1.60.
20
Question 33
Question
Calculate the pH of a solution that has a hydronium ion concentration of 1.5×
10−3M.
Solution
Step 1: Recall that pH is defined as the negative logarithm of the hydronium
ion concentration:
pH = −log[H3O+]
Step 2: Substitute the given hydronium ion concentration into the formula
and solve for pH:
pH = −log1.5×10−3
Step 3: Calculate the pH:
pH = −log1.5×10−3=−log(1.5) + log10−3
Step 4: Simplify the logarithmic terms:
pH = −log(1.5) + (−3) = −0.176 + (−3) = −3.176
Step 5: Therefore, the pH of the solution is 3.176 .
Question 34
Question
Calculate the pH of a solution with a hydronium ion concentration of 1.5×10−4
M.
Solution
Step 1: Recall that the pH is calculated using the formula: pH =−log[H3O+].
Step 2: Substituting the given hydronium ion concentration into the formula,
we get:
pH =−log1.5×10−4
Step 3: Using the properties of logarithms, we simplify the expression to
find the pH:
pH =−log(1.5) −log10−4
Step 4: Remember that log10−4=−4, so we have:
pH =−log(1.5) −(−4)
21
Step 5: Calculating the value of −log(1.5) using a calculator, we find:
pH =−0.1761 −(−4)
Step 6: Simplifying further, we get:
pH = 4.0−0.1761
Step 7: Finally, we calculate the pH:
pH = 3.8239
Therefore, the pH of the solution is 3.8239.
Question 35
Question
A 0.050 M solution of acetic acid (Ka = 1.8×10−5) is prepared. Calculate the
pH of the solution.
Solution
Step 1: Write the equation for the dissociation of acetic acid:
CH3COOH +H2O⇌CH3COO−+H3O+
Step 2: Set up an ICE table to determine the equilibrium concentrations:
Species CH3COOH H2O CH3COO−
H3O+
Initial (M) 0.050 −0
0
Change (M) −x+x+x
+x
Equilibrium (M) 0.050 −x x x
x
Step 3: Write the equilibrium expression for the dissociation of acetic acid
and substitute the equilibrium concentrations into the expression:
Ka=[CH3COO−][H3O+]
[CH3COOH]
1.8×10−5=x·x
0.050 −x
Step 4: Since the value of x is much smaller than 0.050, we can assume that
0.050 −x≈0.050:
1.8×10−5=x·x
0.050
22
Step 5: Solve for x using the quadratic formula:
x=−b±√b2−4ac
2a
x=−0 + p0 + 4(0.050)(1.8×10−5)
2
x≈2.1×10−3
Step 6: Calculate the pH using the concentration of H3O+:
pH =−log[H3O+]
pH =−log2.1×10−3
pH ≈2.68
Therefore, the pH of the 0.050 M acetic acid solution is approximately 2.68.
23
