Question

Give reason why BaSO4 will precipitate out when equal volumes of 2×10-3M BaCl2 solution and 2×10-4M Na2SO4 solution are mixed. given that the solubility product of BaSO4 is 1×10-10

Answer 1

$\sf{BaCl_{2}}$ and $\sf{Na_{2}SO_{4}}$ ionises completely in the solution.

$\sf{BaCl_{2}}$ $\longrightarrow$ $\sf{Ba^{+2}}$ + $\sf{2Cl^{-}}$

$\sf{[Ba^{+2}]}$ = $\sf{[BaCl_{2}]}$ $=\boxed{2 \times {10}^{ - 3} \: M}$


$\sf{Na_{2}SO_{4}}$ $\longrightarrow$ $\sf{2Na^{+}}$ + $\sf{SO_{4}^{-2}}$

$\sf{[Na_{2}SO_{4}]}$ = $\sf{[SO_{4}^{-2}]}$ $=\boxed{2 \times {10}^{ - 4} \: M}$


Equal volumes of the two solutions are mixed together, therefore, the concentration of $\sf{Ba^{+2}}$ ions and $\sf{SO_{4}^{-2}}$ ions after mixing will be:


$\sf{[Ba^{+2}]}$ $= \frac{2 \times {10}^{ - 3} }{2} = \boxed{{10}^{ - 3} \: M}$


$\sf{[SO_{4}^{-2}]}$ $= \frac{2 \times {10}^{ - 4} }{2} =\boxed{{10}^{ - 4} \: M}$


$\text{Therefore,}$

$\text{Ionic\:product\:of:}$

$\sf{BaSO_{4}}$ = $\sf{[Ba^{+2}]}$ $\sf{[SO_{4}^{-2}]}$ = $\boxed{{10}^{ - 7} \: M}$


$\text{Here,}$

Ionic product of $\sf{BaSO_{4}}$ is greater than solubility product value. Hence, a precipitate of $\sf{BaSO_{4}}$ will be formed.