The values of K_{sp} of CaCO_{3} and CaC_{2}O | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Let the concentration of $Ca^{2+}$ from $CaCO_{3}$ be $x$ and from $CaC_{2}O_{4}$ be $y$. Total $[Ca^{2+}] = x + y$. For $CaCO_{3}$: $K_{sp} = [Ca

Vedclass · Accepted Answer

$7.746$

Vedclass · Answer

(D) Let the concentration of $Ca^{2+}$ from $CaCO_{3}$ be $x$ and from $CaC_{2}O_{4}$ be $y$. Total $[Ca^{2+}] = x + y$. For $CaCO_{3}$: $K_{sp} = [Ca^{2+}][CO_{3}^{2-}] = (x + y)x = 4.7 	imes 10^{-9} \quad (i)$ For $CaC_{2}O_{4}$: $K_{sp} = [Ca^{2+}][C_{2}O_{4}^{2-}] = (x + y)y = 1.3 	imes 10^{-9} \quad (ii)$ Dividing $(i)$ by $(ii)$: $\frac{x}{y} = \frac{4.7}{1.3} \approx 3.615$. So,$x = 3.615y$. Substitute into $(ii)$: $(3.615y + y)y = 1.3 	imes 10^{-9} \implies 4.615y^{2} = 1.3 	imes 10^{-9}$. $y^{2} = 0.2817 	imes 10^{-9} = 28.17 	imes 10^{-11} \implies y \approx 5.308 	imes 10^{-6} \, M$. $x = 3.615 	imes 5.308 	imes 10^{-6} \approx 1.919 	imes 10^{-5} \, M$. Total $[Ca^{2+}] = x + y = 1.919 	imes 10^{-5} + 0.531 	imes 10^{-5} = 2.45 	imes 10^{-5} \, M$. Re-evaluating the calculation: $x+y = \sqrt{\frac{K_{sp1} + K_{sp2}}{1}} = \sqrt{6.0 	imes 10^{-9}} \approx 7.746 	imes 10^{-5} \, M$. Thus,the concentration is $7.746 	imes 10^{-5} \, M$.

The values of $K_{sp}$ of $CaCO_{3}$ and $CaC_{2}O_{4}$ are $4.7 \times 10^{-9}$ and $1.3 \times 10^{-9}$ respectively at $25 \, ^oC$. If the mixture of these two is washed with water,what is the concentration of $Ca^{2+}$ ions in water $\dots \times 10^{-5} \, M$?

Similar Questions

The solubility of a salt of weak acid $(AB)$ at $pH = 3$ is $Y \times 10^{-3} \ mol \ L^{-1}$. The value of $Y$ is (Given that the value of solubility product of $AB$ $(K_{sp}) = 2 \times 10^{-10}$ and the value of ionization constant of $HB$ $(K_{a}) = 1 \times 10^{-8}$)

The solubility product of $Mg(OH)_2$ is $9 \times 10^{-12}$. If the concentration of $Mg^{+2}$ is $0.01 \ M$,then the concentration of $OH^{-}$ is = .....

The solubility product of Lead sulphate,$PbSO_4$ is $1.3 \times 10^{-8}$. Calculate its solubility in pure water. The molecular mass of $PbSO_4 = 303 \ g \ mol^{-1}$.

The solubility product of the sparingly soluble salt $AB_2$ is $2.56 \times 10^{-10}$ at $298 \ K$. Calculate its solubility in $mol \ dm^{-3}$ at the same temperature?

The $K_{sp}$ for bismuth sulphide $(Bi_{2}S_{3})$ is $1.08 \times 10^{-73}$. The solubility of $Bi_{2}S_{3}$ in $mol \ L^{-1}$ at $298 \ K$ is ......

Vedclass Test Series

Exam Paper Generator

Online Exam Module