K_{sp} for BaSO_4 at 25 , ^oC is 1.1 times 10 | vedclass

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

Question

(B) The solubility product constant $(K_{sp})$ for $BaSO_4$ is given by: $K_{sp} = [Ba^{2+}][SO_4^{2-}] = 1.1 	imes 10^{-10}$.Let the solubility of $

Vedclass · Accepted Answer

$0.01$

Vedclass · Answer

(B) The solubility product constant $(K_{sp})$ for $BaSO_4$ is given by: $K_{sp} = [Ba^{2+}][SO_4^{2-}] = 1.1 	imes 10^{-10}$.Let the solubility of $BaSO_4$ in the presence of $Na_2SO_4$ be $S = 1.1 	imes 10^{-8} \, M$.In the solution,$BaSO_4$ dissociates as $BaSO_4 ightleftharpoons Ba^{2+} + SO_4^{2-}$,so $[Ba^{2+}] = S = 1.1 	imes 10^{-8} \, M$.$Na_2SO_4$ is a strong electrolyte and dissociates completely as $Na_2SO_4 ightarrow 2Na^+ + SO_4^{2-}$. Let the concentration of $Na_2SO_4$ be $C$.Then,$[SO_4^{2-}] = S + C \approx C$ (since $S$ is very small compared to $C$).Substituting these into the $K_{sp}$ expression: $1.1 	imes 10^{-10} = (1.1 	imes 10^{-8}) 	imes C$.Solving for $C$: $C = \frac{1.1 	imes 10^{-10}}{1.1 	imes 10^{-8}} = 10^{-2} = 0.01 \, M$.

$K_{sp}$ for $BaSO_4$ at $25 \, ^oC$ is $1.1 \times 10^{-10}$. To make the new solubility equal to $1.1 \times 10^{-8} \, M$,it is necessary to use a solution of $Na_2SO_4$ of the following concentration $....... \ M$

Similar Questions

The saturated solution of $Ag_2SO_4$ is $2.5 \times 10^{-2} \ M$. Its solubility product $(K_{sp})$ is

Which of the following is least soluble in water?

Which of the following is insoluble in water?

An ionic compound is dissolved simultaneously in heavy water and simple water. Its solubility is

The solubility of $AgCl$ in water is $1.43 \times 10^{-3} \ g \ L^{-1}$. What is the value of $K_{sp}$ for $AgCl$ at $25 \ ^oC$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module