A 2.901 L saturated solution is formed by 0. | vedclass

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

Question

(A) $1$. Calculate the moles of $CaF_2$: $n = \frac{	ext{mass}}{	ext{molar mass}} = \frac{0.08 \ g}{78 \ g \ mol^{-1}} \approx 1.0256 	imes 10^{-3}

Vedclass · Accepted Answer

$1.715 	imes 10^{-10}$

Vedclass · Answer

(A) $1$. Calculate the moles of $CaF_2$: $n = \frac{	ext{mass}}{	ext{molar mass}} = \frac{0.08 \ g}{78 \ g \ mol^{-1}} \approx 1.0256 	imes 10^{-3} \ mol$. $2$. Calculate the molar solubility $(s)$ in $mol \ L^{-1}$: $s = \frac{n}{V} = \frac{1.0256 	imes 10^{-3} \ mol}{2.901 \ L} \approx 3.535 	imes 10^{-4} \ mol \ L^{-1}$. $3$. The dissociation equilibrium is $CaF_2(s) ightleftharpoons Ca^{2+}(aq) + 2F^-(aq)$. $4$. The solubility product expression is $K_{sp} = [Ca^{2+}][F^-]^2 = (s)(2s)^2 = 4s^3$. $5$. Substitute the value of $s$: $K_{sp} = 4 	imes (3.535 	imes 10^{-4})^3 \approx 4 	imes 4.416 	imes 10^{-11} \approx 1.766 	imes 10^{-10}$.

$A$ $2.901 \ L$ saturated solution is formed by $0.08 \ g$ of $CaF_2$ at $298 \ K$. Calculate the $K_{sp}$ of $CaF_2$. (Molar mass of $CaF_2$ is $78 \ g \ mol^{-1}$)

Similar Questions

In an aqueous solution containing $AgCl_{(s)}$,$AgBr_{(s)}$,and $AgI_{(s)}$,find the ratio $[Cl^{-}] : [Br^{-}] : [I^{-}]$ at equilibrium if $K_{sp}(AgCl) = 10^{-10}$,$K_{sp}(AgBr) = 10^{-13}$,and $K_{sp}(AgI) = 10^{-16}$.

In $1 \ L$ saturated solution of $AgCl$ $[K_{sp}(AgCl)=1.6 \times 10^{-10}]$,$0.1 \ mol$ of $CuCl$ $[K_{sp}(CuCl)=1.0 \times 10^{-6}]$ is added. The resultant concentration of $Ag^{+}$ in the solution is $1.6 \times 10^{-x}$. The value of $x$ is

The solubility of $Ag_2C_2O_4$ is $2 \times 10^{-4} \ mol \ L^{-1}$ at $298 \ K$. What is its solubility product?

The solubility of $Ca(OH)_2$ in water is ....... .
[Given : The solubility product of $Ca(OH)_2$ in water $= 5.5 \times 10^{-6}$]

The $K_{sp}$ of $Mg(OH)_2$ is $1.2 \times 10^{-11}$. Calculate its solubility in pure water.

Vedclass Test Series

Exam Paper Generator

Online Exam Module