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JEE Main 202125 FebruaryMorning Shift
Exam
JEE Main
Year
2021
Date
25 February
Shift
Morning Shift
Subject
Chemistry
Chapter
Ionic Equilibrium

JEE Main 2021 · 25 February · Morning Shift · Chemistry · Ionic Equilibrium

The solubility of AgCN in a buffer solution of pH = 3 is x . The...

The solubility of AgCNAgCN in a buffer solution of pH=3pH = 3 is xx. The value of xx is: [Assume : No cyano complex is formed ; Ksp(AgCN)=2.2×10−16{K}_{sp}\left(AgCN\right) = 2 . 2 \times {10}^{- 16} and Ka(HCN)=6.2×10−10{K}_{a}\left(HCN\right) = 6 . 2 \times {10}^{- 10}]

Options

  1. A1.9×10−51 . 9 \times {10}^{- 5}
  2. B2.2×10−162 . 2 \times {10}^{- 16}
  3. C0.625×10−60 . 625 \times {10}^{- 6}
  4. D1.6×10−61 . 6 \times {10}^{- 6}

Correct answer

A. 1.9×10−51 . 9 \times {10}^{- 5}

Step-by-step solution

Let, the solubility be xx

Given that,

Ksp(AgCN)=2.2×10−16{K}_{sp}\left(AgCN\right) = 2 . 2 \times {10}^{- 16} and Ka(HCN)=6.2×10−10{K}_{a}\left(HCN\right) = 6 . 2 \times {10}^{- 10}

AgCN⇌Ag+x+CN−xKsp=2.2×10−16AgCN ⇌ \mathop{{Ag}^{+}}_{x} + \mathop{{CN}^{-}}_{x}{K}_{sp} = 2 . 2 \times {10}^{- 16}

H++CN−⇌HCNK=1Ka=16.2×10−10{H}^{+} + {CN}^{-} ⇌ HCNK = \dfrac{1}{{K}_{a}} = \dfrac{1}{6 . 2 \times {10}^{- 10}}

Ksp×1Ka=[Ag+][CN−]×[HCN][H+][CN−]{K}_{sp} \times \dfrac{1}{{K}_{a}} = \left[{Ag}^{+}\right]\left[{CN}^{-}\right] \times \dfrac{[ HCN ]}{\left[{H}^{+}\right]\left[{CN}^{-}\right]}

2.2×10−16×16.2×10−10=[S][S]10−32 . 2 \times {10}^{- 16} \times \dfrac{1}{6 . 2 \times {10}^{- 10}} = \dfrac{[ S ] [ S ]}{{10}^{- 3}}

S2=2.26.2×10−9{S}^{2} = \dfrac{2 . 2}{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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