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QJEEM21Chc39c: JEE Main 2021 Chemistry - Class 11 JEEM21Chc39c Solution: The solubility of AgCN in a buffer solution of pH = 3 is x. The …

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]

Multiple Choice Options

  • A 1.9 ×\times 10 −- 5
  • B 1.6 ×\times 10 −- 6
  • C 2.2 ×\times 10 −- 16
  • D 0.625 ×\times 10 −- 6

Step-by-Step Verified Solution

Let solubility is x

JEE Main 2021 (Online) 25th February Morning Shift Chemistry - Ionic Equilibrium Question 63 English Explanation

Ksp×1Ka=[Ag+][CN−]×[HCN][H+][CN−]{K_{sp}} \times {1 \over {{K_a}}} = [A{g^ + }][C{N^ - }] \times {{[HCN]} \over {[{H^ + }][C{N^ - }]}}

2.2×10−16×16.2×10−10=[S][S]10−32.2 \times {10^{ - 16}} \times {1 \over {6.2 \times {{10}^{ - 10}}}} = {{[S][S]} \over {{{10}^{ - 3}}}}

S2=2.26.2×10−9{S^2} = {{2.2} \over {6.2}} \times {10^{ - 9}}

S2=3.55×10−10{S^2} = 3.55 \times {10^{ - 10}}

S=3.55×10−10S = \sqrt {3.55 \times {{10}^{ - 10}}}

S=1.88×10−5=1.9×10−5S = 1.88 \times {10^{ - 5}} = 1.9 \times {10^{ - 5}}

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