0.08 g of CaF_2 is dissolved in 2.90 L of w | vedclass

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

Question

(A) Step $1$: Calculate the molarity $(S)$ of the saturated solution.$S = \frac{	ext{mass}}{	ext{molar mass} 	imes 	ext{volume in L}} = \frac{0.08

Vedclass · Accepted Answer

$1.767 	imes 10^{-10}$

Vedclass · Answer

(A) Step $1$: Calculate the molarity $(S)$ of the saturated solution.$S = \frac{	ext{mass}}{	ext{molar mass} 	imes 	ext{volume in L}} = \frac{0.08 \ g}{78.08 \ g/mol 	imes 2.90 \ L} \approx 3.53 	imes 10^{-4} \ mol/L$.Step $2$: Write the dissociation equilibrium for $CaF_2$.$CaF_2(s) ightleftharpoons Ca^{2+}(aq) + 2F^-(aq)$.Step $3$: Express $K_{sp}$ in terms of solubility $S$.$K_{sp} = [Ca^{2+}][F^-]^2 = (S)(2S)^2 = 4S^3$.Step $4$: Calculate $K_{sp}$.$K_{sp} = 4 	imes (3.53 	imes 10^{-4})^3 \approx 4 	imes 4.417 	imes 10^{-11} \approx 1.767 	imes 10^{-10}$.

$0.08 \ g$ of $CaF_2$ is dissolved in $2.90 \ L$ of water to form a saturated solution at $298 \ K$. Calculate the solubility product constant $(K_{sp})$ of $CaF_2$. (Molar mass of $CaF_2 = 78.08 \ g/mol$)

Similar Questions

Solid $Ba(NO_3)_2$ is gradually added to a $1.0 \times 10^{-4} \, M \, Na_2CO_3$ solution. At what concentration of $Ba^{2+}$ will precipitation start? (Given $K_{sp}$ for $BaCO_3 = 5.1 \times 10^{-9}$)

The solubility product expression for $Ca_3(PO_4)_2$ is represented as

The precipitation occurs if the ionic product is

At $T$ $K$,the solubility product of $AX$ is $10^{-10}$. What is the molar solubility of $AX$ in $0.1$ $M$ $HX$ solution?

Assign $A, B, C, D$ from the given type of reaction.
$Ca(OH)_2 + 2HF \longrightarrow CaF_2 \downarrow + 2H_2O$

Vedclass Test Series

Exam Paper Generator

Online Exam Module