Question

The values of ${K_{sp}}$ of $CaC{O_3}$ and $Ca{C_2}{O_4}$ are $4.7 \times {10^{ - 9}}$ and $1.3 \times {10^{ - 9}}$ respectively at ${25^ \circ }C$ . If the mixture of these two is washed with water, what is the concentration of $C{a^{ + 2}}$ ions in water?
A) $7.746 \times {10^{ - 5}}M$
B) $5.831 \times {10^{ - 5}}M$
C) $6.856 \times {10^{ - 5}}M$
D) $3.606 \times {10^{ - 5}}M$

Answer 1

Hint: In this problem we are given the solubility constant ${K_{sp}}$ . The value of ${K_{sp}}$ is given as the product of the solubilities of the respective ions present in the salt, raised to their respective stoichiometries. The salts given to us are Calcium Carbonate and Calcium Oxalate.

Complete answer:
The given salts are calcium carbonate ( $CaC{O_3}$ ) and Calcium Oxalate ( $Ca{C_2}{O_4}$ ). We can see that in both salts Calcium ion is common. In this question we are given that $CaC{O_3}$ and $Ca{C_2}{O_4}$ are washed together in water. Here the common ion effect will come into play. The common ion in both the solutions is $C{a^{ + 2}}$ . Hence the concentration of $C{a^{ + 2}}$ will increase in the resultant solution.
We are given the value of ${K_{sp}}$ of $CaC{O_3}$ as $4.7 \times {10^{ - 9}}$ and that of $Ca{C_2}{O_4}$ is given as $1.3 \times {10^{ - 9}}$ . Let us consider the solubility of $CaC{O_3}$ to be ‘x’ and that of $Ca{C_2}{O_4}$ be ‘y’. The dissociation of both the salts can be given as:
$CaC{O_3}{\text{ }} \rightleftharpoons {\text{ C}}{{\text{a}}^{ + 2}}{\text{ }} + {\text{ }}CO_3^{2 - }$

T=0	a	-	-
T=equilibrium	$a - x$	$x + y$	$x$

${\text{Ca}}{{\text{C}}_2}{{\text{O}}_4}{\text{ }} \rightleftharpoons {\text{ C}}{{\text{a}}^{ + 2}}{\text{ }} + {\text{ }}{{\text{C}}_2}O_4^{2 - }$

T=0	$b$	-	-
T=equilibrium	$b - y$	$x + y$	$y$

Now, the ${K_{sp}}$ of $CaC{O_3}$ as $4.7 \times {10^{ - 9}}$. Mathematically can be given as: ${K_{sp}}(CaC{O_3}) = [C{a^{ + 2}}][CO_3^{2 - }] = 4.7 \times {10^{ - 9}}$
$[C{a^{ + 2}}][CO_3^{2 - }] = (x + y)(x) = 4.7 \times {10^{ - 9}}$ -- (1)
Now, the ${K_{sp}}$ of $Ca{C_2}{O_4}$ as $1.3 \times {10^{ - 9}}$ . Mathematically can be given as: ${K_{sp}}(Ca{C_2}{O_4}) = [C{a^{ + 2}}][{C_2}{O_4}^{ - 2}] = 1.3 \times {10^{ - 9}}$
$[C{a^{ + 2}}][{C_2}O_4^{2 - }] = (x + y)(y) = 1.3 \times {10^{ - 9}}$ -- (2)
Dividing equation (1) by (2) we get: $\dfrac{x}{y} = 3.6$
Cross multiplying, we get, $x = 3.6y$
Substitute this value for x in equation (2)
$y(3.6y + y) = 1.3 \times {10^{ - 9}}$
On solving and simplifying we get the value of y as: $y = 1.68 \times {10^{ - 5}}$
Calculating the value of x from $x = 3.6y$ we get, $x = 3.6(1.68 \times {10^{ - 5}}) \to x = 6.048 \times {10^{ - 5}}$
The concentration of calcium ions, $[C{a^{ + 2}}] = (x + y) = (1.68 \times {10^{ - 5}}) + (6.048 \times {10^{ - 5}})$
$[C{a^{ + 2}}] = 7.728 \times {10^{ - 5}}M$ which is nearby to $7.746 \times {10^{ - 5}}M$

Option A is the correct option.

Note:
If we are given two or more solutions having at least one common ion, always the common ion effect will come into play, and an increase in the concentration of that common ion is observed. If concentration of one ion is increased, the system will try to decrease the concentration and the equilibrium will move forward.