2. Questions Bank
  3. JEE Main & Advanced
  4. Chemistry
  5. Equilibrium / साम्यावस्था

JEE Main & Advanced Chemistry Equilibrium / साम्यावस्था Question Bank Self Evaluation Test - Equilibrium

  • question_answer
    The solubility (in \[mol\text{ }{{L}^{-1}}\]) of \[AgCl\] \[({{K}_{sp}}=1.0\times {{10}^{-10}})\] in a 0.1 M \[KCl\] solution will be

    A) \[1.0\times {{10}^{-9}}\]         

    B) \[1.0\times {{10}^{-10}}\]

    C) \[1.0\times {{10}^{-5}}\]         

    D) \[1.0\times {{10}^{-11}}\]

    Correct Answer: A

    Solution :

    [a] Let solubility of \[AgCl=x\,mole/L\] \[AgCl\rightleftharpoons A{{g}^{+}}+C{{l}^{-}}\] i.e. \[{{K}_{sp(Agcl)}}=x\times x\] \[KCl\xrightarrow{{}}{{K}^{+}}+\underset{0.1}{\mathop{C{{l}^{-}}}}\,\] \[\left[ C{{l}^{-}} \right]\] from \[KCl=0.1m\] Total \[\left[ C{{l}^{-}} \right]\] in solution \[=x+0.1\] \[{{K}_{sp}}(AgCl)=[A{{g}^{+}}][C{{l}^{-}}]=x(x+0.1)\] \[1.0\times {{10}^{-10}}=x(x+0.1)\] \[1.0\times {{10}^{-10}}={{x}^{2}}+0.1x\] \[1.0\times {{10}^{-10}}=0.1x\left( as\,{{x}^{2}}<<1 \right)\] \[x=1.0\times {{10}^{-9}}mol/L\]


You need to login to perform this action.
You will be redirected in 3 sec spinner