At 310 , K,the solubility of CaF_{2} in water | vedclass

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

Question

(A) First,calculate the solubility $S$ in $mol / L$:$S = \frac{2.34 	imes 10^{-3} \, g}{78 \, g/mol} 	imes \frac{1}{0.1 \, L} = \frac{0.03 	imes 10

Vedclass · Accepted Answer

$0.0108$

Vedclass · Answer

(A) First,calculate the solubility $S$ in $mol / L$:$S = \frac{2.34 	imes 10^{-3} \, g}{78 \, g/mol} 	imes \frac{1}{0.1 \, L} = \frac{0.03 	imes 10^{-3}}{0.1} = 3 	imes 10^{-4} \, mol / L$.The dissociation of $CaF_{2}$ is $CaF_{2} ightleftharpoons Ca^{2+} + 2F^-$.$K_{sp} = [Ca^{2+}][F^-]^{2} = (S)(2S)^{2} = 4S^{3}$.$K_{sp} = 4 	imes (3 	imes 10^{-4})^{3} = 4 	imes 27 	imes 10^{-12} = 108 	imes 10^{-12} = 0.0108 	imes 10^{-8} \, (mol / L)^{3}$.Thus,$x = 0.0108$.

At $310 \, K$,the solubility of $CaF_{2}$ in water is $2.34 \times 10^{-3} \, g / 100 \, mL$. The solubility product of $CaF_{2}$ is $x \times 10^{-8} \, (mol / L)^{3}$. Find the value of $x$. (Given molar mass: $CaF_{2} = 78 \, g \, mol^{-1}$)

Similar Questions

If the solubility products of $AgBrO_3$ and $Ag_2SO_4$ are $5.5 \times 10^{-5}$ and $2 \times 10^{-5}$ respectively,the relationship between the solubilities of these can be correctly represented as

The solubility product of $AgCl$ at $25 \, ^\circ C$ is $5 \times 10^{-13}$. Its solubility is = .......

At $298 \, K$,the solubility of $PbCl_2$ is $2 \times 10^{-2} \, mol/L$,then $K_{sp} = ?$

The solubility of $CaF_2$ is $s$ moles/litre. Then its solubility product is ....

In which of the following is the solubility of $AgCl$ maximum?

Vedclass Test Series

Exam Paper Generator

Online Exam Module