Question

Solid $Be(N{O}_3{)}_2$ 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 percipitate begin to form?
$(K_{SP}$ for $BaC{O}_3=5\times {10}^{-9})$.

• 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

Answer: (D)

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

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

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

$BaC{O}_3⟶B{a}^{2+}+C{O}_3^{2-}$

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

$5\times {10}^{-9}=\left[B{a}^{2+}\right][1\times {10}^{-4}]$

$\left[B{a}^{2+}\right]=\frac{5\times {10}^{-9}}{1\times {10}^{-4}}$

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