Question

Give reason why $BaS{O}_4$ will precipitate out when equal volumes of $2\times {10}^{-3}MBaC{l}_2$ solution and $2\times {10}^{-4}MN{a}_{2}S{O}_4$ solution are mixed. Given that the solubility product of $BaS{O}_4$ is $1\times {10}^{-10}.$

• A. Ionic product of $BaS{O}_4$ is less than solubility product of $BaS{O}_4$
• B. Ionic product of $BaS{O}_4$ is greater than solubility product of $BaS{O}_4$
• C. Ionic product of $BaS{O}_4$ is equal to solubility product of $BaS{O}_4$
• D. None of the above

Answer 1

The correct option is B Ionic product of $BaS{O}_4$ is greater than solubility product of $BaS{O}_4$

$BaC{l}_2$ and $N{a}_{2}S{O}_4$ are ionised completely in the solution:

$BaC{l}_2\left(s\right)\rightleftharpoons B{a}^{+2}\left(aq\right)+2C{l}^{-}\left(aq\right)$
$\left[B{a}^{+2}\right]=\left[BaC{l}_2\right]=2\times {10}^{-3}M$
$N{a}_{2}S{O}_4\left(s\right)\rightleftharpoons 2N{a}^{+}\left(aq\right)+S{O}_4^{-2}\left(aq\right)$
$\left[N{a}_{2}S{O}_4\right]=\left[S{O}_4^{-2}\right]=2\times {10}^{-4}M$

Equal volumes of the two solutions are mixed together. Therefore, the concentration of $B{a}^{+2}$ ions and $S{O}_4^{-2}$ ions after mixing will be:

$\left[B{a}^{+2}\right]=\frac{2\times {10}^{-3}}{2}={10}^{-3}M$
and $\left[S{O}_4^{-2}\right]=\frac{2\times {10}^{-4}}{2}={10}^{-4}M$

$\therefore$ Ionic product of $BaS{O}_4=\left[B{a}^{+2}\right]\left[S{O}_4^{-2}\right]={10}^{-3}\times {10}^{-4}={10}^{-7}M$

Here, ionic product of $BaS{O}_4$ is greater than solubility product value. Hence, a precipitate of $BaS{O}_4$ will be formed.