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 $25°C$. If the mixture of these two is washed with water, what is the concentration of $C{a}^{2+}$ ions in water?

• A. $5.831\times 10^{-5}M$
• B. $6.856\times 10^{-5}M$
• C. $3.606\times 10^{-5}M$
• D. $7.746\times 10^{-5}M$

Answer 1

Answer: (D)

$K_{sp}\left(CaC{O}_3\right)=4.7\times 10^{-9}$
$K_{sp}\left(Ca{C}_2{O}_4\right)=1.3\times 10^{-9}$
$\begin{array}{ccccc}CaC{O}_3\to C{a}^{2+}& +C{O}_3^{2-}& 4.7\times 10^{-9}---(1)& & \\ & S+S_1& S& & \\ Ca{C}_2{O}_4\to C{a}^{2+}& +C_2{O}_4^{2-}& 1.3\times 10^{-9}---(2)& & \\ & & & S+S_1& S_1\end{array}$
On dividing (1) &(2),
$\frac{\left(S+S_1\right)\times S}{\left(S+S_1\right)\times S_1}=\frac{4.7\times 10^{-9}}{1.3\times 10^{-9}}$
$\frac{S}{S_1}=\frac{47}{13}=3.61$
$S=3.61S_1...(3)$
From (1),
$\left(S+S_1\right)\times S=4.7\times 10^{-9}$
From (2),
$\left(S+S_1\right)\times S_1=1.3\times 10^{-9}$
$\left(3.61S_1+S_1\right)\times S_1=1.3\times 10^{-9}$
$4.61S_1{}^2=1.3\times 10^{-9}$
$S_1{}^2=0.281\times 10^{-9}$
$S_1=\sqrt{0.281\times 10^{-9}}=1.67\times 10^{-5}$
From (3),
$S=3.61S_1=3.61\times 1.67\times 10^{-5}=6.02\times 10^{-5}$
$\left[C{a}^{2+}\right]=S+S_1=(6.02+1.67)\times 10^{-5}=7.69\times 10^{-5}$