Question

Solid $Ba(N{O}_3{)}_2$ is gradually dissolved in a $1\times {10}^{-4}$M $N{a}_{2}C{O}_3$ solution. At which concentration of ${Ba}^{2+}$, precipitate of $BaC{O}_3$ begins to form?$(K_{sp}$ for $BaC{O}_3=5.1\times {10}^{-9})$

• A. $5.1\times {10}^{-5}$M
• B. $7.1\times {10}^{-8}$M
• C. $4.1\times {10}^{-5}$M
• D. $8.1\times {10}^{-7}$M

Answer 1

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

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

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

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

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

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

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