JEE MAIN - Chemistry (2019 - 12th April Evening Slot - No. 3)
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 = 12 is :
2.49 × 10–10 M
1.84 × 10–9
M
6.23 × 10–11 M
$${{2.49} \over {1.84}} \times {10^{ - 9}}M$$
Explanation
| Cd(OH)2(s) | ⇌ | Cd2+(aq) | + | 2OH$$-$$(aq) |
|---|---|---|---|---|
| S | 2S |
$$ \therefore $$ Ksp[Cd(OH)2] = 4(S)3 = 4(1.84 × 10–5)3 [Given S = 1.84 × 10–5]
For buffer solution of pH = 12 :
As pH = 12 so pOH = 2
$$ \Rightarrow $$ [OH$$-$$] = 10-2
Let the solubilty = S1
| Cd(OH)2(s) | ⇌ | Cd2+(aq) | + | 2OH$$-$$(aq) |
|---|---|---|---|---|
| S1 | 2S1 + 10-2 |
$$ \therefore $$ Ksp[Cd(OH)2] = S1(10-2)2 = 4(1.84 × 10–5)3
$$ \Rightarrow $$ S1 = 2.49 × 10–10 M
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