Question
The molar solubility of Zn(OH)2 in 1 M ammonia solution at room temperature is (Ksp of Zn(OH)2 = 1.6 × 10−17; Kstab of Zn(NH3)42+ = 1.6 × 1010)
Options
Solution
Zn(OH)2(s) + 4NH3 $\rightleftharpoons$ $Zn\left( NH_{3} \right)_{4}^{2 +}$+ 2OH–
1 M 0 0
Equilibrium (1 – 4S)M SM 2SM
$K_{eq} = \frac{\left\lbrack Zn\left\lbrack NH_{3} \right\rbrack_{4}^{2 +}\left\lbrack OH^{-} \right\rbrack^{2} \right\rbrack}{\left\lbrack NH_{3} \right\rbrack^{4}} \times \frac{\left\lbrack Zn^{2 +} \right\rbrack}{\left\lbrack Zn^{2 +} \right\rbrack} $$${= K_{sp}.K_{stab} = 1.6 \times 10^{- 17} \times 1.6 \times 10^{10} = 2.56 \times 10^{- 7} }{\text{Now, }2.56 \times 10^{- 17} \times 1.6 \times 10^{10} = 2.56 \times 10^{- 7} }{\Rightarrow S = 4 \times 10^{- 3}M}$$
Create a free account to view solution
View Solution Free