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

The correct option is A $7.746\times {10}^{-5}M$

$K_{sp}\left(CaC{O}_3\right)=4.7{\times 10}^{-9}{K}_{sp}\left(Ca{C}_2{O}_4\right)=1.3{\times 10}^{-9}CaC{O}_3\to {Ca}^{2+}+{CO}_3^{2-}4.7{\times 10}^{-9}---\left(1\right)S+S_{1}SCa{C}_2{O}_4\to {Ca}^{2+}+{C_{2}O}_4^{2-}1.3{\times 10}^{-9}---\left(2\right)S+S_1{S}_{1}Ondividing\left(1\right)\&\left(2\right),\frac{(S+S_1)\times S}{(S+S_1)\times S_1}=\frac{4.7{\times 10}^{-9}}{1.3{\times 10}^{-9}}\frac{S}{S_1}=\frac{47}{13}=3.61S=3.61S_1------\left(3\right)From\left(1\right),(S+S_1)\times S=4.7{\times 10}^{-9}From\left(2\right),(S+S_1)\times S_1=1.3{\times 10}^{-9}(3.61S_1+S_1)\times S_1=1.3{\times 10}^{-9}4.61{S_1}^2=1.3{\times 10}^{-9}{S_1}^2=0.281{\times 10}^{-9}{S}_1=\sqrt{0.281{\times 10}^{-9}}=1.67{\times 10}^{-5}From\left(3\right),S=3.61S_1=3.61\times 1.67{\times 10}^{-5}=6.02{\times 10}^{-5}\left[{Ca}^{2+}\right]=S+S_1=(6.02+1.67){\times 10}^{-5}=7.69{\times 10}^{-5}$