JEE Main & Advanced Chemistry Equilibrium Question Bank Common ion effect, Isohydric solutions, Solubility product, Ionic product of water and salt hydrolysis

  • question_answer The solubility product of \[BaS{{O}_{4}}\] at \[25{}^\circ C\] is \[1.0\times {{10}^{-9}}\]. What would be the concentration of \[{{H}_{2}}S{{O}_{4}}\] necessary to precipitate \[BaS{{O}_{4}}\] from a solution of \[0.01\,M\,B{{a}^{2+}}\]ions [RPMT 1999]

    A)                 \[{{10}^{-9}}\]  

    B)                 \[{{10}^{-8}}\]

    C)                 \[{{10}^{-7}}\]  

    D)                 \[{{10}^{-6}}\]

    Correct Answer: C

    Solution :

    \[BaS{{O}_{4}}\]⇌\[\underset{0.01\,\,\,\,}{\mathop{\underset{(S)\,\,\,\,\,\,\,}{\mathop{B{{a}^{++}}}}\,}}\,+\underset{5\,\,\,\,}{\mathop{\underset{(S)\,\,\,\,}{\mathop{SO_{4}^{-\,-}}}\,}}\,\]                    \[{{K}_{sp}}={{S}^{2}}=S\times S=0.01\times S\]                                 \[{{S}_{(SO_{4}^{2-})}}=\frac{{{K}_{sp}}}{{{S}_{(B{{a}^{++}})}}}=\frac{1\times {{10}^{-9}}}{0.01}={{10}^{-7}}\]mole/litre


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