Question

Solid $Ba(N{O}_3{)}_2$ is gradually dissolved in a $1.0\times {10}^{-4}MN{a}_{2}C{O}_3$ solution. At what concentration of $B{a}^{2+}$ will a precipitate begin to form?
$(K_{SP}$ of $BaC{O}_3=5.1\times {10}^{-9})$.

• A. $8.1\times {10}^{-8}M$
• B. $8.1\times {10}^{-7}M$
• C. $4.1\times {10}^{-5}M$
• D. $5.1\times {10}^{-5}M$

Answer 1

The correct option is D $5.1\times {10}^{-5}M$

Given

Ksp=$5.1\ast {10}^{-9}$

Molarity of $N{a}_{2}C{O}_3^{2-}=1\ast {10}^{-4}$

Solution

$N{a}_{2}C{O}_3\to 2N{a}^{+}+C{O}_3^{2-}$

Therefore moles of $N{a}_{2}C{O}_3$=moles of $C{O}_3^{2-}$

$BaC{O}_3\to B{a}^{2+}+C{O}_3^{2-}$

$K_{sp}=\left[B{a}^{2+}\right]\ast \left[C{O}_3^{2-}\right]$

$K_{sp}=\left[B{a}^{2+}\right]\ast 1\ast {10}^{-4}$

$5.1\ast {10}^{-9}=\left[B{a}^{2+}\right]\ast 1\ast {10}^{-4}$

$\left[B{a}^{2+}\right]=5.1\ast {10}^{-5}$

the correct option is D