Question

Question 6 The mineral barite, (BaSO.) has a ke of 1.1 x 10" at 25°C. Calculate the solubility of barium sulfate in water, in: 6.1. moles per liter 6.2. grams per liter

Answer 1

Explanation:

(6.1).    The reaction equation will be as follows.

             $BaSO_{4}(s) \rightleftharpoons Ba^{2+}(aq) + SO^{2-}_{4}(aq)$

Assuming the value of $K_{sp}$ as $1.1 \times 10^{-10}$ and let the solubility of each specie involved in this reaction is "s". The expression for $K_{sp}$ will be as follows.

            $K_{sp} = [Ba^{2+}][SO^{-}_{2}]$    (Solids are nor considered)

                        = $s \times s$

                   s = $\sqrt{K_{sp}}$

                      = $\sqrt{1.1 \times 10^{-10}}$

                      = $1.05 \times 10^{-5}$

Therefore, solubility of barium sulfate in water is $1.05 \times 10^{-5}$.

(6.2).   As the molar mass of $BaSO_{4}$ is 233.38 g/mol

Therefore, the solubility is g/L will be calculated as follows.

                $233.38 g/mol \times 1.05 \times 10^{-5}$

                  = $2.45 \times 10^{-3} g/L$

Therefore, solubility of barium sulfate in grams per liter is $2.45 \times 10^{-3} g/L$.