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 $ B{a_2}C{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 $

Hint: The solubility product constant is the equilibrium constant for the dissolution of a solid substance into an aqueous solution. It is denoted by the symbol $ {K_{sp}} $ . It represents the level at which a solute dissolves in solution; the more soluble a substance is, the higher the $ {K_{sp}} $ value it has.
 $ (d)5.1 \times {10^{ - 5}}M $

Complete answer:
First, we need to write a balanced equation for the reaction. Remember the polyatomic ions remain together as a unit and do not break apart into separate elements.
 $ N{a_2}C{O_3} \rightleftharpoons 2N{a^{2 + }} + C{O_3}^{2 - } $
We are given with, $ [N{a_2}C{O_3}] = [C{O_3}^{2 - }] = 1 \times {10^{ - 4}}M $
 $ BaC{O_3} \rightleftharpoons B{a^{2 + }} + C{O_3}^{2 - } $
Next, write the $ {K_{sp}} $ expression:
 $ {K_{SP}} = [B{a^{2 + }}][C{O_3}^{2 - }] $
As we need to find the concentration of barium ion we will calculate the concentration through the $ {K_{sp}} $ .
 $ [B{a^{2 + }}] = \dfrac{{{K_{Sp}}(BaC{O_3})}}{{[C{O_3}^{2 - }]}} $
 $ [B{a^{2 + }}] = \dfrac{{5.1 \times {{10}^{ - 9}}}}{{1 \times {{10}^{ - 4}}}} $
 $ [B{a^{2 + }}] = 5.1 \times {10^{ - 5}}M $
Hence, at $ 5.1 \times {10^{ - 5}}M $ concentration of $ B{a^{2 + }} $ , precipitation will begin to form.

Note:
Solubility is a kind of equilibrium constant and its value depends on temperature. $ {K_{sp}} $ usually increases with an increase in temperature due to increased solubility. It also says that when solid barium carbonates in equilibrium with its saturated solution, the product of concentrations of ions of both barium and sulphate is equal to the solubility product constant.