Question

1.75 g of solid NaOH is added to 0.25 $d{m}^3$ of 0.1 M $NiC{l}_2$ solution. Calculate pH of final solution. (write your answer to nearest integer)

Given_ $K_{sp}$ of $Ni(OH{)}_2=1.6\times {10}^{-14}$.

Answer 1

Also for $Ni(OH{)}_2$ solution,

$Ni(OH{)}_2\rightleftharpoons N{i}^{2+}+2O{H}^{⊝}$

$\left[N{i}^{2+}\right][O{H}^{⊝}{]}^2=K_{sp}$
$\left[\frac{0.025-\frac{0.0438}{2}}{0.25}\right][O{H}^{⊝}{]}^2=1.6\times {10}^{-14}$

$\therefore \left[\stackrel{⊝}{O}H\right]=11.35\times {10}^{-7}\therefore pOH=5.94\therefore pH=8.06$

Note : $Ni(OH{)}_2$ formed will be precipitated out since maximum solubility of $Ni(OH{)}_2$ is $1.58\times {10}^{-5}M$