JEE MainChemistryModeratePhysical Chemistry

When pure water is saturated with both CaCO3 and CaC2O4, the
concentration of calcium ion in the solution under equilibrium is 8 ×
10-5 M. If the ratio of the solubility product of CaCO3 to that of
CaC2O4 is 3, what is the solubility product of CaCO3 in pure water ?
Report your answer after multiplying by 109 and rounding it off to
nearest whole number.

Show answer and solution

{CaCO}_{3} \rightleftharpoons {Ca}^{+ 2} + {CO}_{3}^{- 2}

{CaC}_{2}O_{4} \ \ \rightleftharpoons \ {Ca}^{+ 2} + \ C_{2}O_{4}^{- 2}

x x y y
(x+y) x (x+y) y

\frac{K_{SP}\left( {CaCO}_{3} \right)}{K_{SP}\left( {CaC}_{2}O_{4} \right)} = \frac{x}{y} = 3 \Rightarrow x = 3y

(x + y) = 8 \times 10^{- 5}

(3y + y) = 8 \times 10^{- 5}

\Rightarrow y = 2 \times 10^{- 5}x = 6 \times 10^{- 5}

K_{SP}\left( {CaCO}_{3} \right) = \left\lbrack {Ca}^{+ 2} \right\rbrack\left\lbrack {CO}_{3}^{- 2} \right\rbrack

= 8 \times 10^{- 5} \times 6 \times 10^{- 5}

= 48 \times 10^{- 10}

K_{SP}\left( {CaCO}_{3} \right) = 4.8 \times 10^{- 9}

Source: Grass-G / CH-22 Chemistry Paper 19 Dec. Eng.docx

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