Question

Calculate the solubility of solid zinc hydroxide at a $pH$ of $5$.

Given:
$Zn{\left(OH\right)}_2\left(s\right)\rightleftharpoons Zn{\left(OH\right)}_2\left(aq\right)$ $k_1={10}^{-6}M.........\left(1\right)$

$Zn{\left(OH\right)}_2\left(aq\right)\rightleftharpoons Zn{\left(OH\right)}^{+}+{OH}^{-}$ $k_2={10}^{-7}M.........\left(2\right)$

$Zn{\left(OH\right)}^{+}\rightleftharpoons {Zn}^{2+}+{OH}^{-}$ $k_3={10}^{-4}......\left(3\right)$

$Zn{\left(OH\right)}_2\left(aq\right)+{OH}^{-}\rightleftharpoons Zn{\left(OH\right)}_3^{-}$ $k_4={10}^3{M}^{-1}.......\left(4\right)$

$Zn{\left(OH\right)}_3^{-}\left(aq\right)+{OH}^{-}\rightleftharpoons Zn{\left(OH\right)}_4^{2-}$ $k_5=10M^{-1}..........\left(5\right)$

Answer 1

$s=\left[Zn{\left(OH\right)}_2\right(aq\left)\right]+Zn{\left(OH\right)}^{+}+{Zn}^{+2}+Zn{\left(OH\right)}_3^{-}+Zn{\left(OH\right)}_4^{-2}$

$s=k_1+\frac{k_1{k}_2}{\left[{OH}^{-}\right]}+\frac{k_1{k}_2{k}_3}{{\left[{OH}^{-}\right]}^2}+k_1{k}_4\left[{OH}^{-}\right]+k_1{k}_4{k}_5{\left[{OH}^{-}\right]}^2$

$s={10}^{-6}+\frac{{10}^{-13}}{\left[{OH}^{-}\right]}+\frac{{10}^{-17}}{{\left[{OH}^{-}\right]}^2}+{10}^{-3}\left[{OH}^{-}\right]+{10}^{-2}{\left[{OH}^{-}\right]}^2$

(a) $pH=5,pOH=9,\left[{OH}^{-}\right]={10}^{-9}$

$s={10}^{-6}+{10}^{-4}+10+{10}^{-12}+{10}^{-20}=10M$

(b) $pH=9$, $pOH=5$, $\left[{OH}^{-}\right]={10}^{-5}$

$s={10}^{-6}+{10}^{-8}+{10}^{-7}+{10}^{-8}+{10}^{-12}=1.12\times {10}^{-6}M$

(c) $pH=13,pOH=1,\left[{OH}^{-}\right]={10}^1$

$s={10}^{-6}+{10}^{-12}+{10}^{-15}+{10}^4+{10}^{-4}=2\times {10}^{-4}M$