Chemistry - Ionic Equilibrium Question with Solution | TestHub

Precipitation occurs when the ionic product exceeds the $K_{sp}$ value. When equal volumes of two solutions are mixed, the concentration of each is reduced by half.

Therefore,

in first case,

ionic product : $I.P.={\left(\frac{1}{2}\times 10^{-4}\right)}^2\times \left(\frac{1}{2}\times 10^{-4}\right)=\frac{1}{8}\times 10^{-12}=1.25\times 10^{-13}$

As $I.P.<K_{sp}$,

No precipitation occurs.

In second case, $I.P.={\left(\frac{1}{2}\times 10^{-2}\right)}^2\times \left(\frac{1}{2}\times 10^{-3}\right)=\frac{1}{8}\times 10^{-7}=1.25\times 10^{-8}$

As $I.P.>K_{sp}$,

precipitation occurs.

In the third case, $I.P.={\left(\frac{1}{2}\times 10^{-5}\right)}^2\times \left(\frac{1}{2}\times 10^{-3}\right)=\frac{1}{8}\times 10^{-13}=1.25\times 10^{-14}$

As $I.P.<K_{sp}$,

no precipitation occurs.

In the fourth case,

$I.P.={\left(\frac{1}{2}\times 10^{-4}\right)}^2\times \left(\frac{1}{2}\times 10^{-5}\right)=\frac{1}{8}\times 10^{-13}=1.25\times 10^{-14}$

As $I.P.<K_{sp}$,

no precipitation occurs.