Question

The solubility of barium sulphate at 298 k is 1.1 *10 ^-5 mol/l calculate the solubility products of barium sulphate at the same temperature

Answer 1

BaSO4 ↔ Ba2+ +SO42−Let the Solubility of Ba2+  and SO42− is SKsp = [Ba2+] [SO42−] = S × S = S2     = (1.1 × 10−5)2    = 1.21 × 10−10

Answer 2

Answer: The solubility product of barium sulfate is $1.21\times 10^{-10}$

Explanation:

We are given:

Molar concentration of the ions of barium sulfate which are $SO_4^{2-}\text{ and }Ba^{2+}$ is $1.1\times 10^{-5}mol/L$

The chemical equation that represents the ionization of barium sulfate is:

$BaSO_4\rightleftharpoons Ba^{2+}+SO_4^{2-}$

Solubility product is defined as the product of the concentration of the ions each raised to the power their stoichiometric coefficients. Equation for the solubility product of barium sulfate follows:

$K_{sp}=[Ba^{2+}][SO_4^{2+}]$

$K_{sp}=(1.1\times 10^{-5})(1.1\times 10^{-5})=1.21\times 10^{-10}$

Thus, the solubility product of barium sulfate is $1.21\times 10^{-10}$