Question

When equal volumes of $2\times {10}^{-6}M$ $BaC{l}_2$ solution and $2\times {10}^{-5}M$ $N{a}_{2}S{O}_4$ solution are mixed - $BaS{O}_4$ is precipitated out - True or False?

The solubility product of barium sulphate is $1\times {10}^{-10}$.

• A. True
• B. False

Answer 1

The correct option is B

False

$BaC{l}_2\left(aq\right)\to B{a}^{2+}\left(aq\right)+2C{l}^{-}\left(aq\right)$

$N{a}_{2}S{O}_4\left(aq\right)\to 2N{a}^{+}\left(aq\right)+S{O}_4^{2-}\left(aq\right)$

Both these salts are soluble in water. They would be fully dissociated. Since equal volumes of two solutions were mixed, the resultant volume of the solution would have doubled after mixing.the concentration will reduce to half.

Before mixing, the $\left[B{a}^{2+}\right]$ = $2\times {10}^{-6}M$

After mixing, it will be $2\times {10}^{-6}\frac{M}{2}$ = $1\times {10}^{-6}M$

Similarly, the $\left[S{O}_4^{2-}\right]$ after mixing = $2\times {10}^{-5}\frac{M}{2}$ = $1\times {10}^{-5}M$

Now we can get the ionic product of $BaS{O}_4$ = $\left[B{a}^{2+}\right]\left[S{O}_4^{2-}\right]$ = ${10}^{-11}$

This is lesser than the solubility product of $BaS{O}_4$ = $1\times {10}^{-10}$

Hence, there won't be any precipitation