question_answer 1)

  A sparingly soluble salt gets precipitated only when the product of concentration of its ions in the solution \[({{Q}_{sp}})\]becomes greater than its solubility product. If the solubility of \[BaS{{O}_{4}}\] in water is\[8\times {{10}^{-4}}mol\,d{{m}^{-3}}\]. Calculate its solubility in \[0.01\,mol\,d{{m}^{-3}}\] of\[{{H}_{2}}S{{O}_{4}}\].

Answer:

  Solubility product of\[BaS{{O}_{4}}\]. \[{{K}_{sp}}=[B{{a}^{2+}}][SO_{4}^{2-}]\] \[={{s}^{2}}\] \[=(8\times {{10}^{-4}})=64\times {{10}^{-8}}\] Let solubility of \[BaS{{O}_{4}}\] in \[0.01M{{H}_{2}}S{{O}_{4}}\] is\[{{s}_{1}}\]. \[\underset{0.01M}{\mathop{{{H}_{2}}S{{O}_{4}}}}\,\to 2{{H}^{+}}+\underset{0.01M}{\mathop{SO_{4}^{2-}}}\,\] \[\underset{{{s}_{1}}}{\mathop{BaS{{O}_{4}}}}\,\to \underset{{{s}_{1}}}{\mathop{B{{a}^{2+}}}}\,+\underset{({{s}_{1}}+0.01)}{\mathop{SO_{4}^{2-}}}\,\] \[{{K}_{sp}}=[B{{a}^{2+}}][SO_{4}^{2-}]\] \[64\times {{10}^{-8}}={{s}_{1}}({{s}_{1}}+0.01)\] \[=s_{1}^{2}+0.01{{s}_{1}}\] Neglecting higher power of\[{{s}_{1}}\], we get:                        \[64\times {{10}^{-8}}=0.01{{s}_{1}}\] \[{{s}_{1}}=64\times {{10}^{-6}}M\]