Determine [Zn2 ], [CN Write the pertinent reactions. Zn(CN)2(s) Zn2+(aq) + 2CN-(

Question

Determine [Zn2 ], [CN

Write the pertinent reactions. Zn(CN)2(s) Zn2+(aq) + 2CN-(aq) ksp = 3.0 times 10-16 CN-(aq) + H2O(l) HCN(aq) + OH-(aq) kb = 1.0 times 10-14/6.2 times 10-10 = 1.6 times 10-5 H2O(l) H+(aq) + OH-(aq) Kw = 1.0 times 10-14 Write the charge balance equation. Because the pH is fixed, there is no charge balance equation for this problem Write the mass balance equations. [CN-] + [HCN] = 2[Zn2+] Write the equilibrium constant expression for each reaction. Ksp = 3.0 times 10-16=[Zn2+][CN-] Kb = 1.6 times 10-5 = [HCN][OH-]/[CN-] Kw = 1.0 times 10-14 = [H+][OH-]

Explanation / Answer

Given that pH = 3.93, [H+] = 10-3.93

Kw = [H+][OH-] = 10-14  

=> [OH-] = 10-14/10-3.93 = 10-10.07

Now, Kb = ([HCN][OH-])/[CN-] = 1.6 x 10-5

Substituting [OH-],

[HCN] = 1.6 x 10-5 / 10-10.07 x [CN-] = 1.6 x 105.07 [CN-]

Similarly, Ksp = [Zn2+][CN-]2 = 3 x 10-16

=>[Zn2+] = 3 x 10-16 / [CN-]2

Writing the mass balance equation,

[CN-] + [HCN] = 2 [Zn2+]

Substituting the values of [Zn2+] and [HCN], we get

[CN-] + 1.6 x 105.07 [CN-] = 2 x 3 x 10-16 / [CN-]2

=>[CN-]3+ 1.6 x 105.07 [CN-]3= 6 x 10-16

=> [CN-]3 = 6 x 10-16/ (1.6 x 105.07 + 1)

Since, we can say 1.6 x 105.07 + 1 = 1.6 x 105.07

[CN-]3 = 3.75 x 10-21.07

=> [CN-] = 1.47 x 10-7

[HCN] = 1.6 x 105.07 x [CN-] = 1.6 x 105.07 x 1.47 x 10-7 = 2.76 x 10-2

[Zn2+] = 3 x 10-16 / [CN-]2 = 3 x 10-16 / (1.47 x 10-7)2 = 1.39 x 10-2

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