Question

If pH of a saturated solution of $\text{Ba}\left(\text{OH}\right)_{2}$ is $12,$ the value of its $K_{\left(\right. \text{sp} \left.\right)}$ is found to be $n\times 10^{- 7}M^{3}.$ find the value of $‘n’$ here?

Answer 1

As $\text{pH}=\text{12}$

$\text{pOH}=\text{14}-\text{12}=2$

$\left[\right.\text{OH}\left]\right.=10^{- 2}M=0.0\text{1M}/L$

In case of $Ba(OH)_2 \rightleftharpoons Ba^{+2} + 2OH^{-2}$

As $\text{2s}=0.0\text{1}m / L$

$s=\frac{0.01}{2}=5\times 10^{- 3}m / L$

$\mathrm{Ksp}$ for $\mathrm{Ba}(\mathrm{OH})_2=4 \mathrm{~s}^3=4\left(5 \times 10^{-3}\right)^3=5 \times 10^{-7} \mathrm{M}^3$

Hence $'n'$ = 5