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AMU Medical AMU Solved Paper-2015

  • question_answer
    The pH of a solution obtained by mixing of 100 mL of a \[HCl\] solution of pH = 2 with 400 mL of another \[HCl\] solution of pH = 3 will be

    A)  2                            

    B)  3        

    C)  2.5                        

    D)  2.8

    Correct Answer: C

    Solution :

                     100 mL of a \[HCl\] solution of \[pH=2\Rightarrow [{{H}^{+}}]={{10}^{-2}}\] 400 mL of a \[HCl\] solution of \[pH=3\Rightarrow [{{H}^{+}}]={{10}^{-3}}\] Resulting \[[{{H}^{+}}]=\frac{100\times {{10}^{-2}}+400\times {{10}^{-3}}}{500}\]\[=\frac{1+0.4}{500}=\frac{1.4}{500}=0.0028\] \[\because \]     \[pH=-\log [{{H}^{+}}]\] \[\therefore \]  \[pH=-\log (0.0028)=2.55\]


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