JEE MAIN - Chemistry (2021 - 1st September Evening Shift - No. 15)
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)
(Given : The solubility product of Zn(OH)2 is 2 $$\times$$ 10$$-$$20)
Answer
2
Explanation
_1st_September_Evening_Shift_en_15_1.png)
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,
$${K_{sp}} = {(0.1)^2} \times S$$
$$2 \times {10^{ - 20}} = 0.01 \times S$$
$$ \Rightarrow S = {{2 \times {{10}^{ - 20}}} \over {0.01}} = 2 \times {10^{ - 18}}$$
$$\therefore$$ x = 2
Comments (0)
