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. –11
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pH of Solution A = 6 + –6 –1 Therefore, concentration of [H ] ion in solution A = 10 mol L pH of Solution B = 4 + –4 –1 Therefore, Concentration of [H ] ion concentration of solution B = 10 mol L On mixing one litre of each solution, total volume = 1L + 1L = 2L + Amount of H ions in 1L of Solution A= Concentration × volume V –6 = 10 mol × 1L + –4 Amount of H ions in 1L of solution B = 10 mol × 1L + ∴ Total amount of H ions in the solution formed by mixing solutions A –6 –4 and B is (10 mol + 10 mol) This amount is present in 2L solution. −4 –4 –4 10 (1 + 0.01) 1.01 × 10 –1 1.01 × 10 –1 ∴ Total [H+] = = mol L = mol L 2 2 2 –4 –1 = 0.5 × 10 mol L –5 –1 = 5 × 10 mol L + pH = – log [H ] = – log (5 × 10–5) = – [log 5 + (– 5 log 10)] = – log 5 + 5 = 5 – log 5 = 5 – 0.6990 = 4.3010 = 4.3