Question

If 20mL $3.0\times {10}^{-4}MB{a}^{2+}$ and 30mL $2.0\times {10}^{-4}MC{O}_3^{2-}$ mixed, Which of following is correct ?
Given:$K_{sp}\left(BaC{O}_3\right)=5.1\times {10}^{-9}{M}^2$.

• A. Precipitate is formed
• B. No precipitate is formed
• C. Can't be determined.
• D. None of the above

Answer 1

The correct option is A Precipitate is formed

After mixing,
Total volume= 50 ml.

using,
$M_1{V}_1=M_1{V}_2$
$(3.0\times {10}^{-4}M)\left(20mL\right)=M_2\left(50mL\right)$
$[B{a}^{+2}=1.2\times {10}^{-4}M]$

$M_1{V}_1=M_1{V}_2$
$(2.0\times {10}^{-4}M)\left(30mL\right)=M_2\left(50mL\right)$ $[C{O}_3^{+2}=1.2\times {10}^{-4}M]$

$\text{Ionic Product}=\left[B{a}^{+2}\right]\left[C{O}_3^{-2}\right]=[1.2\times {10}^{-4}][1.2\times {10}^{-4}]=1.4\times {10}^{-8}$

Since $\text{Ionic Product}>K_{sp}$,
$\Rightarrow$Preicipitate is formed.