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 which concentration of $$B{a^{2 + }},$$ precipitate of $$BaC{O_3}$$ begins to form ? $$\left( {{K_{sp}}\,{\text{for}}\,BaC{O_3} = 5.1 \times {{10}^{ - 9}}} \right)$$
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 :
$$5.1 \times {10^{ - 5}}M$$
Solution :
$$\eqalign{
& {\text{Given}}\,\,{\text{N}}{{\text{a}}_2}C{O_3} = 1.0 \times {10^{ - 4}}M \cr
& \therefore \,\,\left[ {CO_3^{2 - }} \right] = 1.0 \times {10^{ - 4}}M \cr
& i.e.\,\,\,s = 1.0 \times {10^{ - 4}}M \cr
& {\text{At equilibrium}} \cr
& \left[ {B{a^{2 + }}} \right]\left[ {CO_3^{2 - }} \right] = {K_{sp}}\,{\text{of}}\,BaC{O_3} \cr
& \left[ {B{a^{2 + }}} \right] = \frac{{{K_{sp}}}}{{\left[ {CO_3^{2 - }} \right]}} \cr
& = \frac{{5.1 \times {{10}^{ - 9}}}}{{1.0 \times {{10}^{ - 4}}}} \cr
& = 5.1 \times {10^{ - 5}}M \cr} $$