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Ionic Equilibrium question

2021 · 25 Feb · Shift 1 · Q1
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  5. /2021 · 25 Feb · Shift 1 · Q1

Ionic Equilibrium question

2021 · 25 Feb · Shift 1 · Q1

JEE MainChemistryIonic EquilibriumMCQ+4 / −1
The solubility of AgCN in a buffer solution of pH = 3 is x. The value of x is : [Assume : No cyano complex is formed; Ksp(AgCN) = 2.2 ×\times× 10 −-− 16 and Ka(HCN) = 6.2 ×\times× 10 −-− 10]
  1. A
    1.9 ×\times× 10 −-− 5
  2. B
    1.6 ×\times× 10 −-− 6
  3. C
    2.2 ×\times× 10 −-− 16
  4. D
    0.625 ×\times× 10 −-− 6
View written solutionFree

Correct answer: A

  1. Write the dissolution and acid-base equilibria

Silver cyanide dissolves as: AgCN(s)⇌Ag++CN−\text{AgCN}(s) \rightleftharpoons \text{Ag}^+ + \text{CN}^-AgCN(s)⇌Ag++CN− with Ksp=[Ag+][CN−]=2.2×10−16K_{sp} = [\text{Ag}^+][\text{CN}^-] = 2.2 \times 10^{-16}Ksp​=[Ag+][CN−]=2.2×10−16

Since the solution is buffered at pH=3\text{pH} = 3pH=3, [H+]=10−3 M[\text{H}^+] = 10^{-3} \text{ M}[H+]=10−3 M

Cyanide ion is protonated: HCN⇌H++CN−\text{HCN} \rightleftharpoons \text{H}^+ + \text{CN}^-HCN⇌H++CN− with Ka=[H+][CN−][HCN]=6.2×10−10K_a = \frac{[\text{H}^+][\text{CN}^-]}{[\text{HCN}]} = 6.2 \times 10^{-10}Ka​=[HCN][H+][CN−]​=6.2×10−10

So, [CN−][HCN]=Ka[H+]=6.2×10−1010−3=6.2×10−7\frac{[\text{CN}^-]}{[\text{HCN}]} = \frac{K_a}{[\text{H}^+]} = \frac{6.2 \times 10^{-10}}{10^{-3}} = 6.2 \times 10^{-7}[HCN][CN−]​=[H+]Ka​​=10−36.2×10−10​=6.2×10−7

Thus most dissolved cyanide exists as HCN.

  1. Relate solubility to free CN−\text{CN}^-CN− concentration

Let the solubility of AgCN be xxx mol L−1^{-1}−1. Then:

  • Total silver dissolved =[Ag+]=x= [\text{Ag}^+] = x=[Ag+]=x
  • Total cyanide dissolved =x=[CN−]+[HCN]= x = [\text{CN}^-] + [\text{HCN}]=x=[CN−]+[HCN]

Using [CN−][HCN]=6.2×10−7\frac{[\text{CN}^-]}{[\text{HCN}]} = 6.2 \times 10^{-7}[HCN][CN−]​=6.2×10−7 we get [CN−]=(6.2×10−7)[HCN][\text{CN}^-] = (6.2 \times 10^{-7})[\text{HCN}][CN−]=(6.2×10−7)[HCN]

Since [HCN]≫[CN−][\text{HCN}] \gg [\text{CN}^-][HCN]≫[CN−], x≈[HCN]x \approx [\text{HCN}]x≈[HCN] Hence, [CN−]≈6.2×10−7x[\text{CN}^-] \approx 6.2 \times 10^{-7} x[CN−]≈6.2×10−7x

  1. Apply solubility product

Ksp=[Ag+][CN−]K_{sp} = [\text{Ag}^+][\text{CN}^-]Ksp​=[Ag+][CN−] 2.2×10−16=x(6.2×10−7x)2.2 \times 10^{-16} = x(6.2 \times 10^{-7}x)2.2×10−16=x(6.2×10−7x) 2.2×10−16=6.2×10−7x22.2 \times 10^{-16} = 6.2 \times 10^{-7} x^22.2×10−16=6.2×10−7x2

So, x2=2.2×10−166.2×10−7x^2 = \frac{2.2 \times 10^{-16}}{6.2 \times 10^{-7}}x2=6.2×10−72.2×10−16​ x2≈3.55×10−10x^2 \approx 3.55 \times 10^{-10}x2≈3.55×10−10

Therefore, x≈3.55×10−10x \approx \sqrt{3.55 \times 10^{-10}}x≈3.55×10−10​ x≈1.88×10−5x \approx 1.88 \times 10^{-5}x≈1.88×10−5

Thus, x≈1.9×10−5 Mx \approx 1.9 \times 10^{-5} \text{ M}x≈1.9×10−5 M

  1. Check options

The calculated solubility is: 1.9×10−51.9 \times 10^{-5}1.9×10−5 So the correct option is A.

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