question_answer 134)

  Calculate the pH of a solution formed by mixing equal volumes of two solutions A and B of a strong acid having pH = 6 and pH = 4 respectively.

Answer:

  Solution (A):\[pH=6\] \[\therefore \]  \[[{{H}^{+}}]={{10}^{-6}}M\] Solution (B): \[pH=4\] \[\therefore \]  \[[{{H}^{+}}]={{10}^{-4}}M\] Let 1L of both these solution are mixed then resultant concentration of \[[{{H}^{+}}]\] ion can be calculated as. \[{{M}_{1}}{{V}_{1}}+{{M}_{2}}{{V}_{2}}={{M}_{R}}({{V}_{1}}+{{V}_{2}})\]                 \[{{10}^{-6}}\times 1+10\times 1={{M}_{R}}\times 2\]                                 \[{{M}_{R}}=\frac{101\times {{10}^{-6}}}{2}=50.5\times {{10}^{-6}}M\]                                 \[pH=-\log [{{H}^{+}}]\]                                                 \[=-\log [50.5\times {{10}^{-6}}]\]                                                 \[=4.296\approx 4.3\]