Question
Solid $$Ba{\left( {N{O_3}} \right)_2}$$ 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 for $$BaC{O_3} = 5.1 \times {10^{ - 9}}$$ )
A.
$$5.1 \times {10^{ - 5}}M$$
B.
$$8.1 \times {10^{ - 8}}M$$
C.
$$8.1 \times {10^{ - 7}}M$$
D.
$$4.1 \times {10^{ - 5}}M$$
Answer :
$$5.1 \times {10^{ - 5}}M$$
Solution :
$$\eqalign{
& \mathop {N{a_2}C{O_3}}\limits_{1 \times {{10}^{ - 4}}M} \to \mathop {2N{a^ + }}\limits_{1 \times {{10}^{ - 4}}M} + \mathop {C{O_3}^{2 - }}\limits_{1 \times {{10}^{ - 4}}M} \cr
& {K_{sp\left( {BaC{O_3}} \right)}} = \left[ {B{a^{2 + }}} \right]\left[ {CO_3^{2 - }} \right] \cr
& \left[ {B{a^{2 + }}} \right] = \frac{{5.1 \times {{10}^{ - 9}}}}{{1 \times {{10}^{ - 4}}}} = 5.1 \times {10^{ - 5}}M \cr} $$