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JEE Main & Advanced Chemistry Equilibrium / साम्यावस्था Question Bank Critical Thinking

  • question_answer
    If \[50\,ml\] of \[0.2\,M\,KOH\] is added to \[40\,ml\] of \[0.5\,M\,HCOOH,\] the \[pH\] of the resulting solution is \[({{K}_{a}}=1.8\times {{10}^{-4}})\] [MH CET 2000]

    A)                 3.4         

    B)                 7.5

    C)                 5.6         

    D)                 3.75

    Correct Answer: A

    Solution :

               \[pH=-\log {{K}_{a}}+\text{log}\frac{\text{ }\!\![\!\!\text{ Salt }\!\!]\!\!\text{ }}{\text{ }\!\![\!\!\text{ Acid }\!\!]\!\!\text{ }}\]                    \[\text{ }\!\![\!\!\text{ Salt }\!\!]\!\!\text{ }=\frac{0.2\times 50}{1000}=0.01\]; \[\text{ }\!\![\!\!\text{ Acid }\!\!]\!\!\text{ }=\frac{0.5\times 40}{1000}=0.02\]                    \[pH=-\text{log}\,\,(1.8\times {{10}^{-4}})+\text{log}\frac{0.01}{0.02}\]                    \[pH=4-\text{log}\,\,(1.8)+\text{log}\,\,\text{0}.5\]                    \[pH=4-\text{log}\,\,(1.8)-0.301\]                    \[pH=3.4\]


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