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  4. The values of Ksp of CaCO3 and CaC2O4 are 4.7 × 10-9 and 1.3 × 10-9 respectively at 25°C. If the mixture of these two is washed with water, what is the concentration of Ca2+ ions in water?

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Q. The values of $K_{sp}$ of $CaCO_3$ and $CaC_2O_4$ 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 $Ca^{2+}$ ions in water?

NEETNEET 2013Equilibrium

Solution:

$K _{ sp }\left( CaCO _{3}\right)=4.7 \times 10^{-9}$
$K _{ sp }\left( CaC _{2} O _{4}\right)=1.3 \times 10^{-9}$
$\begin{array}{cccc} CaCO _{3} \rightarrow Ca ^{2+} & + CO _{3}^{2-} & 4.7 \times 10^{-9}---(1) \\ & S + S _{1} & S \\ CaC _{2} O _{4} \rightarrow Ca ^{2+} & + C _{2} O _{4}^{2-} & 1.3 \times 10^{-9}---(2) \\ & & & S + S _{1} & S _{1}\end{array}$
On dividing (1) &(2),
$\frac{\left(S+S_{1}\right) \times S}{\left(S+S_{1}\right) \times S_{1}}=\frac{4.7 \times 10^{-9}}{1.3 \times 10^{-9}}$
$\frac{S}{S_{1}}=\frac{47}{13}=3.61$
$S =3.61 S _{1}...(3)$
From (1),
$\left( S + S _{1}\right) \times S =4.7 \times 10^{-9}$
From (2),
$\left( S + S _{1}\right) \times S _{1}=1.3 \times 10^{-9}$
$\left(3.61 S _{1}+ S _{1}\right) \times S _{1}=1.3 \times 10^{-9}$
$4.61 S _{1}{ }^{2}=1.3 \times 10^{-9}$
$S _{1}{ }^{2}=0.281 \times 10^{-9}$
$S _{1}=\sqrt{0.281 \times 10^{-9}}=1.67 \times 10^{-5}$
From (3),
$S =3.61 S _{1}=3.61 \times 1.67 \times 10^{-5}=6.02 \times 10^{-5}$
$\left[ Ca ^{2+}\right]= S + S _{1}=(6.02+1.67) \times 10^{-5}=7.69 \times 10^{-5}$