JEE Main · 2019 · Shift-IIhardIEQ-066

The molar solubility of Cd(OH)2 is 1.84 10-5 M in water. The expected solubility of Cd(OH)2 in a buffer solution of pH…

Ionic Equilibrium · Class 11 · JEE Main Previous Year Question

Question

The molar solubility of Cd(OH)2\mathrm{Cd(OH)_2} is 1.84×10−51.84 \times 10^{-5} M in water. The expected solubility of Cd(OH)2\mathrm{Cd(OH)_2} in a buffer solution of pH = 12 is:

Options
  1. a

    2.49×10−102.49 \times 10^{-10} M

    ✓
  2. b

    2.491.84×10−9\frac{2.49}{1.84} \times 10^{-9} M

  3. c

    1.84×10−91.84 \times 10^{-9} M

  4. d

    6.23×10−116.23 \times 10^{-11} M

Correct Answera

2.49×10−102.49 \times 10^{-10} M

Detailed Solution

Step 1 — Find KspK_{sp} from water solubility: Cd(OH)2⇌Cd2++2OH−s=1.84×10−5 MKsp=s(2s)2=4s3=4(1.84×10−5)3=4×6.23×10−15=2.49×10−14\mathrm{Cd(OH)_2 \rightleftharpoons Cd^{2+} + 2OH^-} s = 1.84 \times 10^{-5}\ \mathrm{M} K_{sp} = s(2s)^2 = 4s^3 = 4(1.84 \times 10^{-5})^3 = 4 \times 6.23 \times 10^{-15} = 2.49 \times 10^{-14}

Step 2 — Solubility at pH = 12: pOH=14−12=2⇒[OH−]=0.01 M\mathrm{pOH} = 14 - 12 = 2 \Rightarrow [\mathrm{OH^-}] = 0.01\ \mathrm{M}

Let solubility = s′s' (assume s′≪0.01s' \ll 0.01): Ksp=s′(0.01)2=s′×10−4s′=2.49×10−1410−4=2.49×10−10 MK_{sp} = s'(0.01)^2 = s' \times 10^{-4} s' = \frac{2.49 \times 10^{-14}}{10^{-4}} = 2.49 \times 10^{-10}\ \mathrm{M}

Answer: Option (1) — 2.49×10−102.49 \times 10^{-10} M

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