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.

Answers (1)

pH of solution A = 6

Therefore, concentration of [H+ ] ion in solution A = 10-6 mol L-1

pH of solution B = 4

Therefore, concentration of [H+ ] ion in solution B = 10-4 mol L-1 .

On mixing one litre of each solution, total volume = 1L + 1L = 2L.

Amount of H+ ions in 1L of solution A= concentration x volume V= 10-6 mol X 1L

Amount of H+ ions in 1L of solution B= 10-4 mol X 1L

The total amount of H+ ions in the solution formed by mixing solutions A and B is (10-6 + 10-4 ) mol X 1L

This amount is present in 2L solution.

Total\; [H^{+}]in\; 1L\; solution=\frac{10^{-6}+10^{-4}}{2}

=10^{-4}(1+\frac{0.01}{2})

=10^{-4}\times \frac{0.01}{2}

=5\times 10^{-5}mol/L

Therefore, pH=-log[H^{+}]=-log[5\times 10^{-5}]

=-log(5)+(-5\; log10)

=-log(5)+5=4.3

Hence, the pH will be 4.3

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