Chemistry - Ionic Equilibrium Question with Solution | TestHub

The equilibrium reaction is given by:

$\text{BiO(OH)(s)}\rightleftharpoons {\text{BiO}}^{+}\text{(aq)}+{\text{OH}}^{-}\text{(aq)}$

The equilibrium constant expression is:

$K=[{\text{BiO}}^{+}][{\text{OH}}^{-}]$

Let the solubility of $\text{BiO(OH)}$ be $s$. Then $[{\text{BiO}}^{+}]=s$ and $[{\text{OH}}^{-}]=s$.

Substituting the values into the equilibrium expression:

$s^2=4\times 10^{-10}$

$s=2\times 10^{-5}\text{ M}$

Therefore, the concentration of hydroxide ions is:

$[{\text{OH}}^{-}]=2\times 10^{-5}\text{ M}$

Calculating the pOH:

$\text{pOH}=-\log [{\text{OH}}^{-}]=-\log (2\times 10^{-5})=5-\log 2$

Given $\log 2=0.3010$:

$\text{pOH}=5-0.3010=4.699$

The pH of the solution at $298\text{ K}$ is:

$\text{pH}=14-\text{pOH}=14-4.699=9.301$

Answer: $9.301$