Question

The acid ionziation (hydrolysis) constant of $Z{n}^{2+}$ is $1.0\times {10}^{-9}$
(a) Calculate the $pH$ of a $0.001M$ solution of $ZnC{l}_2$
(b) What is the basic dissociation constant of $Zn(OH{)}^{+}$?

Answer 1

(Acid ionisation) hydrolysis constant of $Z{n}^{2+}=1.0\times {10}^{-9}$

Let it be $x⟹x={10}^{-9}$ also $c=0.001$

We know, $x=\sqrt{\frac{K_w}{K_b.C}}=\sqrt{\frac{{10}^{-14}}{{10}^{-3}.K_b}}={10}^{-9}$

$\frac{{10}^{-14}}{{10}^{-3}.K_b}={10}^{-18}=\frac{1}{{10}^{-7}}=K_b$

$K_b={10}^7$

$p{K}_b=-log{10}^7=-7$

$pH=\frac{1}{2}(14-p{K}_b-logC)$

$=\frac{1}{2}(14+7-log{10}^{-3})=\frac{24}{2}=12$

$pH=12$