CHEMISTRY 150

5. (20 pts) Isotope dilution is a mass spectrometric technique in which a known amount of an unusual isotope (called the spike) is added to an unknown as an internal standard for quantitative analysis. The ratio of the isotopes is measured and from this ratio the quantity of the element in the original unknown can be calculated. For example, natural vanadium atom fractions 51V = 0.9975 and 50V = 0.0025. The atomic fraction is defined as:

atom fraction of 51V = atoms 51V/(atoms 51V + atoms 50V)

A spike enriched in 50V has atom fractions 51V = 0.6391 and 50V = 0.3609.

a) Let A be 51V and B be 50V. Let Ax be the atom fraction of A in an unknown. Let Bx be the atom fraction of B in the unknown. Let As and Bs be the corresponding atom fractions in a spike. Let Cx be the total concentration of all isotopes of vanadium ( mol/g) in the unknown, and let Cs be the concentration in the spike. After mixing mx grams of unknown with ms grams of the spike, show that the ratio of isotopes in the mixture (denoted R) is given by

𝑅 = 𝑚𝑜𝑙  𝐴 𝑚𝑜𝑙  𝐵

= 𝐴 𝐶 𝑚 + 𝐴 𝐶 𝑚 𝐵 𝐶 𝑚 + 𝐵 𝐶 𝑚

b) Solve the above equation for Cx to show that

𝐶 = 𝐶 𝑚 𝑚

𝐴 − 𝑅𝐵 𝑅𝐵 − 𝐴

c) A 0.40167 g sample of crude oil containing an unknown concentration of

natural vanadium was mixed with a 0.41946 g spike containing 2.2435 mol/g enriched with 50V (atom fractions: 51V = 0.6391 and 50V = 0.3609).

The measured isotope ratio by mass spectrometry was R = 51V /50V = 10.545. Determine the concentration of vanadium ( mol/g) in the crude oil.

6. (15 pts.) Consider the electrochemical cell described below.

𝐴𝑔(𝑠)|𝐴𝑔𝐶𝑙(𝑠)|𝐾𝐶𝑙(𝑎𝑞, 𝑠𝑎𝑡𝑢𝑟𝑎𝑡𝑒𝑑)||𝑐𝑒𝑙𝑙  𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛|𝐶𝑢(𝑠) The cell solution was made by mixing

25.0 mL of 4.00 mM 𝐾𝐶𝑁

25.0 mL of 4.00 mM 𝐾𝐶𝑢(𝐶𝑁)

25.0 mL of 0.400 M acid, 𝐻𝐴, with pKa = 9.50

25.0 mL of 𝐾𝑂𝐻 solution The measured voltage was -0.440 V. Calculate the molarity of the KOH solution. Assume that essentially all of the copper(I) is in the form 𝐶𝑢(𝐶𝑁) . For the right side (the cathode), the half cell reaction is:

𝐶𝑢(𝐶𝑁) + 𝑒 ⇌ 𝐶𝑢(𝑠) + 2𝐶𝑁                                                              𝐸° = −0.429  𝑉

7. (20 pts.) A 3.67 g sample of insecticide was decomposed in acid, any As5+ was reduced to As3+ and diluted to 250.0 mL in a volumetric flask. A 5.00 mL aliquot was added to 125.0 mL of 0.0500 M KI buffered to pH 7. A coulometric titration was carried out with electrically-generated I3-, which oxidized As3+ to As5+ according to the reaction.

As3+ + I3- 3I- + As5+ The titration required 287 s at a constant current of 24.25 mA to reach the endpoint. Calculate the percentage of As2O3 (197.84 g/mol) in the insecticide.

8. (15 pts.) An ion selective electrode used to measure 𝑀𝑛 is also sensitive to 𝐿𝑖 and obeys the equation

𝐸 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 + 57.1  𝑚𝑉

2 log{[𝑀𝑛 ] + 0.002[𝐿𝑖 ] }

When the electrode was immersed in 100.0 mL of unknown containing 𝑀𝑛 in 0.200  𝑀  𝐿𝑖𝐶𝑙, the reading was 194.6  𝑚𝑉. When 1.00  𝑚𝐿 of 1.07 × 10  𝑀  𝑀𝑛 (in 0.200  𝑀  𝐿𝑖𝐶𝑙 ) was added to the unknown, the reading increased to 200.7  𝑚𝑉. Find the concentration of 𝑀𝑛 in the original unknown.