Question

The solubility of $Mg(OH{)}_2$ in pure water is $9.57\times {10}^{-3}g{L}^{-1}$, its solubility (in $g{L}^{-1}$) in $0.02MMg(N{O}_3{)}_2$ will be ?

• A. $8.7\times {10}^{-4}$
• B. $8.7\times {10}^{-3}$
• C. $8.7\times {10}^{-2}$
• D. $8.7\times {10}^{-5}$

Answer 1

The correct option is A $8.7\times {10}^{-4}$

solution:

solubility of $Mg(OH{)}_2$ in pure water is 9.57 x 10${}^{-3}$ g/L

= $\frac{9.57X10-3}{58}$ mole/L = 1.65 x 10${}^{-4}$mol/L

$K_s$ for $Mg(OH{)}_2$ = s x $(2s{)}^2$ = $4s^3$

= 4 x (1.65 x 10${}^{-4}$ mol/L)${}^3$

= 17.96 x 10${}^{-12}$

Suppose, the solubility of Mg(OH)${}_2$ in the presence of Mg(NO3)${}_2$ = s'

So, Mg${}^{2+}$ = s'+c = s' + 0.02

[OH'] = 2s'

Therefore, $K_s$ = (s' + 0.02) (2s')${}^2$

17.96 x 10${}^{-12}$ = 4 (s' + 0.02) (s')${}^2$

or, 17.96 x 10${}^{-12}$/4 = s'${}^3$+ 0.02 (s' )${}^2$

, on neglecting s'${}^3$

4.4921 x 10${}^{-12}$

= 0.02 (s' )${}^2$

(s' )${}^2$= 14.98 x 10${}^{-6}$ mol/l

= 14.98 x 10${}^{-6}$ x58 = 8.69 x 10${}^{-4}$ g/l

hence the correct opt : A