Equilibrium - Result Question 74
####78. The values of $K _{s p}$ of $CaCO_3$ and $CaC_2 O_4$ 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 $Ca^{2+}$ ions in water?
[NEET Kar. 2013]
(a) $7.746 \times 10^{-5} M$
(b) $5.831 \times 10^{-5} M$
(c) $6.856 \times 10^{-5} M$
(d) $3.606 \times 10^{-5} M$
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Solution:
- (a) $CaCO_3 \longrightarrow \underset{x}{Ca^{2+}}+CO_3^{2-}$
$CaC_2 O_4 \longrightarrow \underset{y}{Ca^{2+}}+\underset{y}{C_2 O_4^{2-}}$
$\therefore[Ca^{2+}]=x+y$
Now, $K _{s p}(CaCO_3)=[Ca^{2+}][CO_3^{2-}]$
or $4.7 \times 10^{-9}=(x+y) x$
similarly, $K _{s p}(CaC_2 O_4)=[Ca^{2+}][C_2 O_4^{2-}]$
or $1.3 \times 10^{-9}=(x+y) y$
On solving, we get
$[Ca^{2+}]=7.746 \times 10^{-5} M$
