At $298 \ K$ temperature,the $K_{sp}$ of $CaF_2$ is $1.7 \times 10^{-10}$. If a person drinks $2.5 \ L$ of $CaF_2$ saturated water daily,how many grams of $CaF_2$ are present in the water consumed? (Molecular mass of $CaF_2$ is $78 \ g \ mol^{-1}$)
(A) The dissociation of $CaF_2$ is: $CaF_2(s) \rightleftharpoons Ca^{2+}(aq) + 2F^-(aq)$.
Let the solubility be $S \ mol \ L^{-1}$. Then $K_{sp} = [Ca^{2+}][F^-]^2 = (S)(2S)^2 = 4S^3$.
Given $K_{sp} = 1.7 \times 10^{-10}$,so $4S^3 = 1.7 \times 10^{-10}$.
$S^3 = 0.425 \times 10^{-10} = 42.5 \times 10^{-12}$.
$S = \sqrt[3]{42.5} \times 10^{-4} \approx 3.49 \times 10^{-4} \ mol \ L^{-1}$.
Amount in $2.5 \ L$ in moles $= S \times V = 3.49 \times 10^{-4} \ mol \ L^{-1} \times 2.5 \ L = 8.725 \times 10^{-4} \ mol$.
Mass of $CaF_2 = \text{moles} \times \text{molar mass} = 8.725 \times 10^{-4} \ mol \times 78 \ g \ mol^{-1} \approx 0.068 \ g$.
Let the solubility be $S \ mol \ L^{-1}$. Then $K_{sp} = [Ca^{2+}][F^-]^2 = (S)(2S)^2 = 4S^3$.
Given $K_{sp} = 1.7 \times 10^{-10}$,so $4S^3 = 1.7 \times 10^{-10}$.
$S^3 = 0.425 \times 10^{-10} = 42.5 \times 10^{-12}$.
$S = \sqrt[3]{42.5} \times 10^{-4} \approx 3.49 \times 10^{-4} \ mol \ L^{-1}$.
Amount in $2.5 \ L$ in moles $= S \times V = 3.49 \times 10^{-4} \ mol \ L^{-1} \times 2.5 \ L = 8.725 \times 10^{-4} \ mol$.
Mass of $CaF_2 = \text{moles} \times \text{molar mass} = 8.725 \times 10^{-4} \ mol \times 78 \ g \ mol^{-1} \approx 0.068 \ g$.
Explore More
Similar Questions
Equal volumes of $0.002 \, M$ solutions of sodium iodate and cupric chlorate are mixed together. Will it lead to precipitation of copper iodate?
(For cupric iodate $K_{sp} = 7.4 \times 10^{-8}$)
(For cupric iodate $K_{sp} = 7.4 \times 10^{-8}$)
Medium
View SolutionFormula Solubility product $(K_{sp})$ $AgBrO_3$ $2.5 \times 10^{-10}$ $MgSO_4$ $2.7 \times 10^{-27}$ $CaCO_3$ $2.7 \times 10^{-18}$ $FeC_2O_4$ $2.4 \times 10^{-15}$
Which of the following salts has the maximum solubility?
| Formula | Solubility product $(K_{sp})$ |
|---|---|
| $AgBrO_3$ | $2.5 \times 10^{-10}$ |
| $MgSO_4$ | $2.7 \times 10^{-27}$ |
| $CaCO_3$ | $2.7 \times 10^{-18}$ |
| $FeC_2O_4$ | $2.4 \times 10^{-15}$ |
Medium
View SolutionThe sulphide ion concentration $[S^{2-}]$ in a saturated $H_2S$ solution is $1 \times 10^{-22} \ M$. Which of the following sulphides should be quantitatively precipitated by $H_2S$ in the presence of dilute $HCl$?
Product Sulphide Solubility $(K_{sp})$ $I$ $1.74 \times 10^{-16}$ $II$ $1.2 \times 10^{-22}$ $III$ $8.2 \times 10^{-46}$ $IV$ $5.0 \times 10^{-34}$
| Product | Sulphide Solubility $(K_{sp})$ |
|---|---|
| $I$ | $1.74 \times 10^{-16}$ |
| $II$ | $1.2 \times 10^{-22}$ |
| $III$ | $8.2 \times 10^{-46}$ |
| $IV$ | $5.0 \times 10^{-34}$ |
Difficult
View SolutionWhat is the molar solubility of $Fe(OH)_2$ $(K_{sp} = 8.0 \times 10^{-16})$ at $pH$ $13.0$ ?
Medium
View SolutionIf $K_{sp}$ of $CaF_2$ at $25\,^{\circ}C$ is $1.7 \times 10^{-10}$,the combination amongst the following which gives a precipitate of $CaF_2$ is
DifficultAIEEE 2012
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo