Question

The solubility of barium sulphate at 298 K is 1.1 ├Ч10^-5 mol L-1. Calculate the solubility product of barium sulphate at the same temperature.

Answer 1

Answer : The solubility product of barium sulphate is $1.21\times 10^{-10}mole^2/L^2$.

Solution : Given,

The solubility of barium sulphate is $1.1\times 10^{-5}mole/L$.

The equilibrium reaction is,

$BaSO_4\rightleftharpoons Ba^{2+}+SO^{2-}_4$

From the reaction, we conclude that the 1 mole of barium sulphate gives 1 mole of barium ion and 1 mole of sulphate ion.

That means,

$[Ba^{2+}]$ = $1.1\times 10^{-5}mole/L$

$[SO^{2-}_4]$ = $1.1\times 10^{-5}mole/L$

The solubility product expression is,

$K_{sp}=[Ba^{2+}][SO^{2-}_4]$ = $(1.1\times 10^{-5}mole/L)\times (1.1\times 10^{-5}mole/L)$ = $1.21\times 10^{-10}mole^2/L^2$

Answer 2

Answer:

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