III. Consider the following reaction between aqueous mercury ions and a complexi
ID: 512442 • Letter: I
Question
III. Consider the following reaction between aqueous mercury ions and a complexing agent “RWW” also known as a chelating agent (14 points).
Hg^2+ (aq) + 2RWW (aq) = Hg(RWW)2^2+ (aq) Kf = 7.91x10^10
Suppose a typical human (about 48.0 L of water by volume) ingests 510mg of mercuric ions. Consider then a subsequent chelating treatment consisting of injection of 25.0 mL of 0.35M RWW solution. Set up a reaction table (in mmol) for the complexation and calculate the concentration and mass in the body of the uncomplexed mercury. Neglect the volume of RWW solution compared to total body volume.
Rxn Table (in mmol):
[Hg^2+] = 1.15E-7
mass Hg^2+ = 1.104 mg
These are the right answers I just need to know how to get them.
Explanation / Answer
Hg2+ (aq) + 2RWW (aq) ----> Hg(RWW)22+ (aq) Kf = 7.91x1010
Kf = [Hg(RWW)22+] / [Hg2+] [RWW]2
Concentration of mercuric ions injested
510 x 10-3 g / Molar mass of Hg is 200.5900 g/mol = 0.00254 mol
= 0.00254 mol / 48 L
= 5.2916 x 10-5 M
Concentration of RWW
25 mL x 0.35 mmol / mL = 0.00875 mol
= 0.00875 mol / 48 L
= 0.0001822 M
1 mole of Hg2+ complexes with 2 moles of RWW
So amount of RWW required to completely react with 5.2916 x 10-5 M of mercuric ion is
= 5.2916 x 10-5 x 2
= 0.00010583 M
The concentration of RWW after complete reaction is
= 0.0001822 - 0.0001058
= 7.64 x 10-5M
ICE Table Hg2+ (aq) + 2RWW (aq) ----> Hg(RWW)22+ (aq)
Initial 5.2916 x 10-5 0.0001822
After
completing 0 7.64 x 10-5 5.2916 x 10-5
Reaction
Change x 2x -x
Final x 7.64 x 10-5 + 2x 5.2916 x 10-5 - x
Kf = [Hg(RWW)22+] / [Hg2+] [RWW]2
7.91x1010 = (5.2916 x 10-5 - x )/ x(7.64 x 10-5 + 2x)2
7.91x1010 = (5.2916 x 10-5 - x )/ x(5.8369 x 10-9 + 4x2)
assuming that x << 5.2916 x 10-5
7.91x1010 = (5.2916 x 10-5 )/ x(5.8369 x 10-9)
x(5.8369 x 10-9) = 5.2916 x 10-5 / 7.91x1010
x = 6.6897 x 10-16 /5.8369 x 10-9
x = 1.146 x 10-7 ~ 1.15 x 10-7 M = [Hg2+]
Mass of [Hg2+]
48 L x 1.15 x 10-7 mol / L = 5.52 x 10-6 mol
5.52 x 10-6 mol x 200.5900 g/mol = 1.107 mg
Mass of [Hg2+] = 1.107 mg