At 298 , K,the solubility product (K_{sp}) of | vedclass

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

Question

(B) The dissociation of $M_2SO_4$ is given by: $M_2SO_4(s) \rightleftharpoons 2M^+(aq) + SO_4^{2-}(aq)$.Let the solubility of $M_2SO_4$ be $s \, mol/L

Vedclass · Accepted Answer

$2.89 	imes 10^{-2} \, M$

Vedclass · Answer

(B) The dissociation of $M_2SO_4$ is given by: $M_2SO_4(s) ightleftharpoons 2M^+(aq) + SO_4^{2-}(aq)$.Let the solubility of $M_2SO_4$ be $s \, mol/L$.Then,$[M^+] = 2s$ and $[SO_4^{2-}] = s$.The solubility product expression is $K_{sp} = [M^+]^2 [SO_4^{2-}] = (2s)^2(s) = 4s^3$.Given $K_{sp} = 1.2 	imes 10^{-5}$,we have $4s^3 = 1.2 	imes 10^{-5}$.$s^3 = 0.3 	imes 10^{-5} = 3.0 	imes 10^{-6}$.$s = (3.0 	imes 10^{-6})^{1/3} \approx 1.442 	imes 10^{-2} \, M$.The concentration of $M^+$ is $[M^+] = 2s = 2 	imes 1.442 	imes 10^{-2} = 2.884 	imes 10^{-2} \, M \approx 2.89 	imes 10^{-2} \, M$.

At $298 \, K$,the solubility product $(K_{sp})$ of $M_2SO_4$ (where $M^+$ is a monovalent metal ion) is $1.2 \times 10^{-5}$. What is the maximum concentration of $M^+$ ions in a saturated solution of this salt at $298 \, K$?

Similar Questions

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.

The solubility of $AgCN$ in a buffer solution of $pH = 3$ is $x$. The value of $x$ is [Assume : No cyano complex is formed; $K_{sp}(AgCN) = 2.2 \times 10^{-16}$ and $K_{a}(HCN) = 6.2 \times 10^{-10}$]

Which of the following has the maximum solubility? (Given $K_{sp}$ values)

If the solubility product $(K_{sp})$ of $Ni(OH)_2$ is $1.9 \times 10^{-15}$,the molar solubility of $Ni(OH)_2$ in $1.0 \ M \ NaOH$ is:

Which of the following substances is insoluble in acetic acid $(CH_3COOH)$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module