If the solubility product of AB_2 is 3.20 tim | vedclass

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

Question

(C) The dissociation of $AB_2$ is given by: $AB_{2(s)} \rightleftharpoons A^{2+}_{(aq)} + 2B^{-}_{(aq)}$Let the solubility be $s \ mol \ L^{-1}$. Then

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(C) The dissociation of $AB_2$ is given by: $AB_{2(s)} ightleftharpoons A^{2+}_{(aq)} + 2B^{-}_{(aq)}$Let the solubility be $s \ mol \ L^{-1}$. Then $[A^{2+}] = s$ and $[B^{-}] = 2s$.The solubility product expression is: $K_{sp} = [A^{2+}][B^{-}]^2 = (s)(2s)^2 = 4s^3$.Given $K_{sp} = 3.20 	imes 10^{-11}$,we have $4s^3 = 3.20 	imes 10^{-11}$.$s^3 = \frac{3.20 	imes 10^{-11}}{4} = 0.80 	imes 10^{-11} = 8.0 	imes 10^{-12}$.Taking the cube root: $s = \sqrt[3]{8.0 	imes 10^{-12}} = 2 	imes 10^{-4} \ mol \ L^{-1}$.Thus,the value is $2 	imes 10^{-4} \ mol \ L^{-1}$.

If the solubility product of $AB_2$ is $3.20 \times 10^{-11} \ mol^3 L^{-3}$,then the solubility of $AB_2$ in pure water is......... $\times 10^{-4} \ mol \ L^{-1}$. [Assuming that neither kind of ion reacts with water]

Similar Questions

At $298 \ K$ temperature,the $K_{sp}$ of $Mg(OH)_2$ is $1.8 \times 10^{-11}$. If $0.1 \ M$ $NaOH$ solution is added to it,what is the concentration of $Mg^{2+}$ ions? Calculate its solubility in water.

The compound insoluble in water is

Which of the following would increase the solubility of $Pb(OH)_2$?

Solubility product of a salt $AB$ is $1 \times 10^{-8}$ in a solution in which the concentration of $A^{+}$ ions is $10^{-3} \ M$. The salt will precipitate when the concentration of $B^{-}$ ions is kept

Assign $A, B, C, D$ from the given type of reaction.
$2AgNO_3 + Na_2C_2O_4 \longrightarrow Ag_2C_2O_4 \downarrow + 2NaNO_3$

Vedclass Test Series

Exam Paper Generator

Online Exam Module