Calculate the molar solubility of calcium hyd | vedclass

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

Question

(C) The dissociation of $Ca(OH)_2$ is given by: $Ca(OH)_2(s) ightleftharpoons Ca^{2+}(aq) + 2OH^-(aq)$.$K_{sp} = [Ca^{2+}][OH^-]^2 = 5.5 	imes 10^{

Vedclass · Accepted Answer

$5.5 	imes 10^{-4}$

Vedclass · Answer

(C) The dissociation of $Ca(OH)_2$ is given by: $Ca(OH)_2(s) ightleftharpoons Ca^{2+}(aq) + 2OH^-(aq)$.$K_{sp} = [Ca^{2+}][OH^-]^2 = 5.5 	imes 10^{-6}$.In $0.10 \ M \ NaOH$ solution, the concentration of $OH^-$ ions is $0.10 \ M$ due to complete dissociation of $NaOH$.Let the molar solubility of $Ca(OH)_2$ be $S$.Then $[Ca^{2+}] = S$ and $[OH^-] = (0.10 + 2S) \approx 0.10 \ M$ (since $S$ is very small).Substituting these values into the $K_{sp}$ expression:$5.5 	imes 10^{-6} = S 	imes (0.10)^2$.$5.5 	imes 10^{-6} = S 	imes 0.01$.$S = \frac{5.5 	imes 10^{-6}}{0.01} = 5.5 	imes 10^{-4} \ M$.

Calculate the molar solubility of calcium hydroxide $Ca(OH)_2$ in $0.10 \ M \ NaOH$ solution. The solubility product constant $(K_{sp})$ of calcium hydroxide is $5.5 \times 10^{-6}$.

Similar Questions

What is the correct representation of the solubility product of $Ag_2CrO_4$?

Calculate the solubility product of sparingly soluble salt $BA$ at $300 \ K$ if its solubility is $9.1 \times 10^{-3} \ mol \ dm^{-3}$ at same temperature.

Assign $A, B, C, D$ from the given type of reaction: $PbCl_2(s) + \text{Hot water} \longrightarrow Pb^{2+}(aq) + 2Cl^{-}(aq)$

Calculate solubility $(mol \ dm^{-3})$ of a sparingly soluble electrolyte $AB$ at $298 \ K$ if its solubility product is $1.6 \times 10^{-5}$?

The molar solubility $(mol \ L^{-1})$ of a sparingly soluble salt $MX_4$ is $s$. The corresponding solubility product is $K_{sp}$. $s$ is given in terms of $K_{sp}$ by the relation:

Vedclass Test Series

Exam Paper Generator

Online Exam Module