Question

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

• 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

According to the problem

$N{a}_{2}C{O}_3\to 2N{a}^{+}+C{O_3}^{2-}$

$1\times {10}^{-4}M$

For,

$BaC{O}_{3}B{a}^{2+}+C{O_3}^{2-}$

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

Hence,

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

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

$5.1\times {10}^{-5}M$