Question

What is the solubility ( in mol/L ) of Fe(OH)${}_3$ in a solution of pH=$8.0$ ? $\left[K_{sp}forFe\right(OH{)}_3=1.0\times {10}^{-36}]$ Calculate the solubility of A${}_2$ X${}_3$ in pure water, assuming that neither kind of ion reacts with water.For A${}_2$X${}_3$ $[K_{sp}=1.1\times {10}^{-23}]$

Answer 1

$pH=8$

$\therefore pOH=14-8=6$

$\therefore -\log \left[O{H}^{-}\right]=6$

$\therefore \left[O{H}^{-}\right]={10}^{-6}$ $M$

$Fe(OH{)}_3\leftrightharpoons \stackrel{3+}{Fe}+3O{H}^{-}$

At $t=0$- $s$ $0$ $0$

At $t=$ equilibrium- $(s-x)$ $s$ $(3s+{10}^{-6})$

$\therefore K_{sp}=s\times (3s+{10}^{-6}{)}^3$

$⟹{10}^{-36}=s\times (3s{)}^3$

$⟹{10}^{-36}=27s^4$

$⟹s^4-0.037\times {10}^{-36}$

$\therefore s=4.386\times {10}^{-10}$ $mol/L$

Therefore, solubility of $Fe(OH{)}_3$ is $4.386\times {10}^{-10}$ $mol/L$

$A_2{X}_3\leftrightharpoons 2A^{x+}+3X^{y-}$

At $t=0$- $s$ $0$ $0$

At $t=$ equilibrium- $(s-x)$ $\left(2s\right)$ $\left(3s\right)$

$K_{sp}=\left(2s{)}^2\right(3s{)}^3=(4\times 27)s^5$

$⟹1.1\times {10}^{-23}=108s^5$

$⟹s={\left(\frac{1.1\times {10}^{-23}}{108}\right)}^{1/5}$

$⟹s=1\times {10}^{-5}$ $mol/L$

Therefore solubility of $A_2{X}_3$ is ${10}^{-5}$ $mol/L$