Equal volumes of a 0.020 M Zn²⁺ solution and a 2.0 M NH₃ solution are mixed. Kf for [Zn(NH₃)₄]²⁺ is 4.1 × 10⁸. If enough sodium oxalate is added to make the solution 0.10 M in oxalate, will ZnC₂O₄ precipitate? What is Q? Ksp ZnC₂O₄= 2.7 × 10^-8

Question

Quizplus · Accepted Answer

The balanced equation for the dissolution of ZnS in water is:

ZnS(s) ⇌ Zn2+(aq) + S2-(aq)

The solubility product expression for ZnS is:

Ksp = [Zn2+][S2-]

When the ZnS solution is mixed with the NHSB solution, the following reaction occurs:

ZnS + 2NHSB ⇌ [Zn(NHSB)2] + H2S

The equilibrium constant expression for this reaction is:

K = [Zn(NHSB)2]/[ZnS][NHSB]2

We can relate K to Ksp using the formation constant Kf for [Zn(NHSB)S]:

Kf = [Zn(NHSB)S]/[Zn2+][NHSB]

Rearranging the expression for Kf and substituting it into the equilibrium constant expression K, we get:

K = [Zn(NHSB)S]/Kf

Substituting the given values, we get:

K = (0.020 M)(2.0 M)2/4.1 × 10^-18 M = 1.91 × 10^13

When enough sodium oxalate is added to make the solution 0.10 M in oxalate, it reacts with the Zn2+ ions to form the complex ion [Zn(C2O4)4]2-. The reaction is:

Zn2+ + 4C2O42- ⇌ [Zn(C2O4)4]2-

The equilibrium constant expression for this reaction is:

K = [Zn(C2O4)4]2-/([Zn2+][C2O42-]^4)

The value of K can be calculated from the given Ksp for ZnC2O4 and the solubility product expression:

K = Ksp/([Zn2+][C2O42-]^4)

Substituting the given values, we get:

K = 2.7 × 10^-18/([Zn2+][C2O42-]^4)

At equilibrium, Q = K. Therefore,

K = Q = [Zn(NHSB)S][Zn2+][NHSB]/([ZnS][NHSB]2)

Substituting the given values, we get:

1.91 × 10^13 = [Zn(NHSB)S][Zn2+]/([ZnS][NHSB])

Assuming that the volumes of the solutions are equal, the initial concentration of Zn2+ is [Zn2+] = 0.020 M. The concentration of NHSB is 2.0 M, and the concentration of oxalate after addition of the sodium oxalate is 0.10 M. Therefore,

[Zn(NHSB)S] = [Zn(NHSB)S]eq = Kf[Zn2+][NHSB]/([NHSB] + [C2O42-])

Substituting the given values, we get:

[Zn(NHSB)S]eq = (4.1 × 10^-18 M)(0.020 M)(2.0 M)/0.10 M = 8.2 × 10^-17 M

Substituting these values into the equation for K, we get:

1.91 × 10^13 = (8.2 × 10^-17 M)[Zn2+]/(0.020 M)

Solving for [Zn2+], we get:

[Zn2+] = 7.8 × 10^-6 M

Substituting this value into the solubility product expression for ZnC2O4 and solving for Q, we get:

Q = [Zn2+][C2O42-]^4 = (7.8 × 10^-6 M)(0.10 M)^4 = 2.9 × 10^-12

Since Q is greater than Ksp, ZnC2O4 will precipitate. Therefore, the answer is C.

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