1
Ch. 17 Kimball Spring 2019
Discussion Worksheet - Chapter 17- Part Deux
Additional Aspects of Aqueous Equilibria
Dr Cabirac Lecture
1. A 0.054 M HNO2 solution is titrated with a KOH solution. What is the [H+] and pH at the half-
equivalence point.
First, the weak acid (HNO2) will react to completion with the strong base (KOH)
HNO2 + KOH → K+ + NO2- + H2O buffer created
(weak acid) (conj base)
At the half-equivalence point, [HNO2] = [NO2-]
𝑝𝐻 = 𝑝𝐾𝑎+𝑙𝑜𝑔[𝐻𝑁𝑂2]
[𝑁𝑂2−] where [𝐻𝑁𝑂2]
[𝑁𝑂2−] = 1
𝑝𝐻 = 𝑝𝐾𝑎+log(1)
= -log(4.0 x 10-4) + 0
= 3.34
2. The Ka of a certain indicator is 2.0 x 10-6 . The color of HIn is green and that of In- is red. A few drops
of the indicator are added to an HCl solution, which is then titrated against a NaOH solution. At what
pH will the indicator change color?
The indicator is a weak acid, so establishes the following equilibrium:
HIn ⇌ H+ + In-
When [HIn] = [In-], half of the indicator is green, half red, so this is the “color change” point.
𝐾 = [𝐻+][𝐼𝑛−]
[𝐻𝐼𝑛]
[𝐻+]= 𝐾𝑎[𝐻𝐼𝑛]
[𝐼𝑛−] where [HIn] = [In-]
[H+] = Ka = 2.0 x 10-6
pH = pKa = -log(2.0 x 10-6) = 5.70
3. The solubility for SrF2 is 7.3 x 10-2 g/L. What is the solubility product, Ksp , for this compound?
SrF2 (s) ⇌ Sr2+(aq) + 2 F-(aq) Ksp = [Sr2+][F-]2
I - 0 0 = (s)(2s)2
C -s +s +2s = 4s3
E - s s = 4(5.8 x 10-4 M)3 = 7.8 x 10-10
𝑠 (𝑚𝑜𝑙
𝐿) = 7.3 𝑥 10−2 𝑔 𝑆𝑟𝐹2
1 𝐿 (1 𝑚𝑜𝑙
125.6 𝑔) = 5.8 𝑥 10−4 𝑚𝑜𝑙
𝐿
2
Ch. 17 Kimball Spring 2019
4. What is the molar solubility of silver sulfate at 25ºC? Ksp = 1.4 x 10-5
Ag2SO4 (s) ⇌ 2 Ag+(aq) + SO42-(aq) Ksp = [Ag+]2[SO42-]
I - 0 0 = (2s)2(s)
C -s+2s +s 1.4 x 10-5 = 4s3
E - 2s s s = 0.015 M
5. What is the pH of a saturated zinc hydroxide solution? (Ksp = 1.8 x 10-14)
Zn(OH)2 (s) ⇌ Zn2+(aq) + 2OH-(aq) Ksp = [Zn2+][OH-]2
I - 0 0 1.8 x 10-14 = (s)(2s)2
C -s+s +2s s = 1.7 x 10-5
E - s 2s [OH-] = 2s = 2(1.7 x 10-5) = 3.3 x 10-5
pOH = -log(3.3 x 10-5) = 4.48
pH = 14 – pOH = 14 – 4.48 = 9.52
6. A volume of 75 mL of 0.060 M NaF is mixed with 25 mL of 0.15 M Sr(NO3)2 . Calculate the final
concentrations of NO3- , Na+ , Sr2+ , and F- . (Ksp for SrF2 = 2.0 x 10-10 )
THIS ONE IS A LOT OF WORK!
NaF is soluble, so 0.060 M NaF [Na+] = 0.060 M, [F-] = 0.060 M
Sr(NO3)2 is soluble, so 0.15 M Sr(NO3)2[Sr2+] = 0.15 M, [NO3-] = 0.30 M
SrF2 is “insoluble,” so it will precipitate out upon mixing
moles Sr2+ = (0.025 L)(0.15 mol/L) = 0.00375 mol Sr2+ (mol NO3- = 2(0.00375) = 0.0075mol)
moles F- = (0.075 L)(0.060 mol/L) = 0.0045 mol F-
(also 0.0045 mol Na+)
Sr2+(aq) + 2 F-(aq) → SrF2 (s)
0.00375 0.0045 0
-0.00225 -0.0045 +0.00225
0.00150 0 0.00225 these are the moles of species after ppt rxn
xs mol Sr2+
[Sr2+] = 0.00150mol/0.10 L = 0.010 M
So now consider the solubility equilibrium:
SrF2 (s) ⇌ Sr2+(aq) + 2 F-(aq) Ksp = [Sr2+][F-]2
I - 0.0150 0 2.0 x 10-10 = (0.015 +s)(2s)2 s << 0.015
C -s+s +2s s = 5.8 x 10-5
E - 0.0150+s 2s
[𝑁𝑎+]=0.0045𝑚𝑜𝑙
0.10 𝐿 = 𝟎.𝟎𝟒𝟓𝑴
[𝑁𝑂3−]= 0.0075𝑚𝑜𝑙
0.10 𝐿 = 𝟎.𝟎𝟕𝟓𝑴
[𝑆𝑟2+]= 0.015𝑀 + 5.8𝑥10−5𝑀 = 𝟎.𝟎𝟏𝟓𝑴
[𝐹−]= 2(5.8𝑥10−5𝑀) = 𝟏.𝟐𝒙𝟏𝟎−𝟒𝑴
3
Ch. 17 Kimball Spring 2019
7. The molar solubility of AgCl in 6.5 x 10-3 M AgNO3 is 2.5 x 10-8 M. In deriving Ksp from these data,
which of the following assumptions are reasonable?
A. Ksp is the same as solubility NO
B. Ksp of AgCl is the same in 6.5 x 10-3 M AgNO3 as it is in pure water YES – Ksp is only
changed by ΔT
C. Solubility of AgCl is independent of the concentration of AgNO3 adding AgNO3 shifts eq ←
D. [Ag+] in solution does not change significantly upon the addition of AgCl to
6.5 x 10-3 M AgNO3 addition of more AgCl is addition of solid, so no effect
E. [Ag+] in solution after the addition of AgCl to 6.5 x 10-3 M AgNO3 is the same as
it would be after the addition of AgCl to pure water. [Ag+] less in AgNO3 soln than in
H2O, so eq shifts →
8. Which of the following ionic compounds will be more soluble in acidic solution than in water?
A. BaSO4 SO42- is a weak base C. Fe(OH)3 OH- is a base
B. PbCl2 Cl- negligible base D. CaCO3 CO32- is a weak base
9. How many grams of CaCO3 will dissolve in 3.0 x 102 mL of 0.050 M Ca(NO3)2 ? (Ksp = 8.7 x 10-9)
CaCO3 (s) ⇌ Ca2+(aq) + CO32-(aq) Ksp = [Ca2+][CO32-]
I - 0.050 0 = (0.050 + s)(s) where s << 0.050
C -s+s +s 8.7 x 10-5 = 0.050s
E - 0.050 + s s s = 1.7 x 10-7M
0.050 𝑀 𝐶𝑎𝐶𝑂3(1 𝐶𝑎2+
1 𝐶𝑎𝐶𝑂3) = 0.050 𝑀 𝐶𝑎2+
1.7𝑥10−7 𝑚𝑜𝑙 𝐶𝑎𝐶𝑂3
𝐿(0.30 𝐿)= 5.1𝑥10−8𝑚𝑜𝑙 𝐶𝑎𝐶𝑂3(100.1 𝑔
1 𝑚𝑜𝑙 ) = 𝟓.𝟐𝒙𝟏𝟎−𝟔𝒈 𝑪𝒂𝑪𝑶𝟑
10. A solution contains an unknown concentration of Pb2+ . HCl is added to this solution to make
[Cl- ] = 0.15 M and some PbCl2 precipitates. Calculate the concentration of Pb2+ remaining in
solution. (Ksp for PbCl2 =1.2 x 10-5
PbCl2 (s) ⇌ Pb2+(aq) + 2Cl-(aq) Ksp = [Pb2+][Cl-]2
2.4 x 10-4 = [Pb2+] (0.15)2
[𝑷𝒃𝟐+]=𝟐.𝟒𝒙𝟏𝟎−𝟒
(𝟎.𝟏𝟓)𝟐= 𝟎.𝟎𝟏𝟏 𝑴
4
Ch. 17 Kimball Spring 2019
11. BONUS PROBLEM
The molar mass of a certain metal carbonate, MCO3 , can be determined by adding an excess of HCl
to react with all the carbonate and then “back titrating” the remaining HCl with a NaOH solution.
MCO3 + 2HCl → MCl2 + CO2 + H2O
HCl + NaOH → NaCl + H2O
In a certain experiment, 18.68 mL of 5.653 M HCl were added to a 3.542 g sample of
MCO3 . The excess HCl required 12.06 mL of 1.789 M NaOH for neutralization. Calculate the molar
mass of the carbonate and identify M.
[Cl- ] = 0.15 M and some PbCl2 precipitates. Calculate the concentration of Pb2+ remaining in
solution. (Ksp for PbCl2 =1.2 x 10-5 )
Total HCl 0.01865 𝐿 𝐻𝐶𝑙(5.653𝑚𝑜𝑙
𝐿) = 0.1056 𝑚𝑜𝑙 𝐻𝐶𝑙
Excess HCl 0.01206 𝐿 𝐻𝐶𝑙(1.789𝑚𝑜𝑙
𝐿) = 0.0216 𝑚𝑜𝑙 𝐻𝐶𝑙
Mol HCl reacted with MCO3 = 0.1056 mol – 0.0216 mol = 0.0840 mol HCl
So according to the balanced equation:
0.0840 𝑚𝑜𝑙 𝐻𝐶𝑙(1 𝑚𝑜𝑙 𝑀𝐶𝑂3
2 𝑚𝑜𝑙 𝐻𝐶𝑙 ) = 0.0420 𝑚𝑜𝑙 𝑀𝐶𝑂3
𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 = 𝑔
𝑚𝑜𝑙=3.542 𝑔
0.0420 𝑚𝑜𝑙 =84.3 𝑔
𝑚𝑜𝑙
molar mass = massM + massCO3
84.3 g = massM + 60.01 g
MassM = 24.3 g Therefore, the metal must be Mg
Strong
Acids
K,>>4
Weak
Acids
1>K,>Ky
Acid
k,
Base
kK
ClO,(aq)_|
SMALL
Tag
~10
Br-(aq)
~10
Ct(aqg)
|
~10="
HS0,(aq) ~1057
No,(aq)
~1055
|
HO @
ixo+
$0,(aq)
8.3x10
HPO,(aq)
1.3x10*
Fag
1.5xi0
JC3H.02"(aq)
5.5x10-*
H 7
HCO,-(0q)
2.3104
H,PO,-(aq)_|
6.2x10-*
HPO,
(cq)
1.6x10-7
NHj(aq)
5.5x103*ff]
NH,(og)_|
1.8x10-*
HCN(aq)
4.9x10-|
a
HCO,-(0q)
5.5xi0
HPO,(aq)
42x10)
HO
ixio#
HS-(aq)
~107"
‘OH-(aq)
~102"
}
Bases
K,
<Ky
Weak
Bases
1>K,>K,
Strong
Bases
K,>>1
5
Ch. 17 Kimball Spring 2019
Powered by TCPDF (www.tcpdf.org)
