Multistep problem that I\'m confused on the entire thing: A) The addition of 5E-
ID: 875685 • Letter: M
Question
Multistep problem that I'm confused on the entire thing:
A) The addition of 5E-3 total moles of Zn2+ to a 1.0L solution of NaCN gives a solution of the complex ion [Zn(CN)4-2](Kf=4.2E19). What is the concentration of uncomplexed Zn2+ ions if the concentration of cyanide ions in the final solution is 0.5M?
B) ZnCO3 is sparaingly soluble salt with a Ksp=1.0E-7. The addition of CN- (aq) to ZnCO3(s) yields the complex ion [Zn(Cn)42-] (aq) with the Kf mentioned before in Part A. Write the reaction.
C) What will be the Keq for this overall reaction?
D) What is the solubility of ZnCO3 (s) in a solution of NaCN (aq) where the equilibrium concentration of cyanide in the solution is 0.2 M? Use the calculated Keq from Part C.
Explanation / Answer
Kf is formation of complex
Ksp is dissolution constant.
A) Zn2+ + 4NaCN --------> [Zn(CN)4]2-
concentration of Zn2+ = 5 x 10^-3 moles / 1L = 5 x 10^-3
Zn2+ + 4NaCN --------> [Zn(CN)4]2-
5 x 10^-3 4(5 x 10^-3) 0
-x 0.5 5 x 10^-3
Feed the values,
Kf = 4.2 x 10^-3 = 5 x 10^-3 / [x](0.5)^4
[x] = 1.9 x 10^-21
Zn2+ uncomplexed in solution = 5 x 10^-3 - 1.9 x 10^-21 = 5 x 10^-3 M
B) The chemical equation would be,
ZnCO3(s) + 4CN-(aq) -------------> [Zn(CN)4]2-(aq) + CO32-(aq)
C) Keq = Kf x Ksp
= 4.2 x 10^19 x 1 x 10^-7
= 4.2 x 10^12
D) Keq. = [Zn(CN)4]2-/[Zn(CO3)][CN-]^4
4.2 x 10^12 = 5 x 10^-3/(ZnCO3)(0.2)^4
[Zn(CO3)] = 7.44 x 10^-20 is the solubility of ZnCO3(s)
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