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NEET 2026
Exam
NEET
Year
2026
Subject
Chemistry
Chapter
Chapter

NEET 2026 · Chemistry · Chapter

The solubility product ( K_ sp ) of zinc hydroxide, Zn(OH)_2 , is...

The solubility product (KspK_{sp}) of zinc hydroxide, Zn(OH)2\text{Zn(OH)}_2, is 3.2×10−143.2 \times 10^{-14} at 298 K298\text{ K}. What is the pH of its saturated aqueous solution at this temperature?
(Given : log⁡2=0.3010\log 2 = 0.3010)

Options

  1. A9.3019.301
  2. B9.6029.602
  3. C4.3984.398
  4. D4.6994.699

Correct answer

B. 9.6029.602

Step-by-step solution

Let the solubility of Zn(OH)2\text{Zn(OH)}_2 be ss.

Zn(OH)2(s)⇌Zn2+(aq)+2OH−(aq)\text{Zn(OH)}_2(s) \rightleftharpoons \text{Zn}^{2+}(aq) + 2\text{OH}^-(aq)

Ksp=[Zn2+][OH−]2=(s)(2s)2=4s3K_{sp} = [\text{Zn}^{2+}][\text{OH}^-]^2 = (s)(2s)^2 = 4s^3

4s3=3.2×10−144s^3 = 3.2 \times 10^{-14}

s3=0.8×10−14=8×10−15s^3 = 0.8 \times 10^{-14} = 8 \times 10^{-15}

s=2×10−5 Ms = 2 \times 10^{-5} \text{ M}

[OH−]=2s=4×10−5 M[\text{OH}^-] = 2s = 4 \times 10^{-5} \text{ M}

pOH=−log⁡[OH−]=−log⁡(4×10−5)=5−2log⁡2\text{pOH} = -\log[\text{OH}^-] = -\log(4 \times 10^{-5}) = 5 - 2\log 2

pOH=5−2(0.3010)=5−0.6020=4.398\text{pOH} = 5 - 2(0.3010) = 5 - 0.6020 = 4.398

pH=14−pOH=14−4.398=9.602\text{pH} = 14 - \text{pOH} = 14 - 4.398 = 9.602

Answer: 9.6029.602
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