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QJEEM21Ch71da: JEE Main 2021 Chemistry - Class 11 JEEM21Ch71da Solution: The molar solubility of Zn(OH) 2 in 0.1 M NaOH solution is x 10 …

The molar solubility of Zn(OH)2 in 0.1 M NaOH solution is x ×\times 10 −- 18 M. The value of x is _________ (Nearest integer)

(Given : The solubility product of Zn(OH)2 is 2 ×\times 10 −- 20)

Step-by-Step Verified Solution

JEE Main 2021 (Online) 1st September Evening Shift Chemistry - Ionic Equilibrium Question 50 English Explanation

Due to common-ion effect (presence of NaOH) the concentration of OH −- will be (2S + 0.1) ≈\approx 0.1

( ∵\because 0.1 > > 2 S)

∴\therefore Solubility of product,

Ksp=(0.1)2×S{K_{sp}} = {(0.1)^2} \times S

2×10−20=0.01×S2 \times {10^{ - 20}} = 0.01 \times S

⇒S=2×10−200.01=2×10−18\Rightarrow S = {{2 \times {{10}^{ - 20}}} \over {0.01}} = 2 \times {10^{ - 18}}

∴\therefore x = 2

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