Question

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

Answer 1

\[\ce{pH}\]  of Solution A = 6

Therefore, concentration of \[\ce{[H+]}\] ion in solution A = 10^–6 mol L^–1

\[\ce{pH}\] of Solution B = 4

Therefore, Concentration of \[\ce{[H+]}\] ion concentration of solution B = 10^–4 mol L^–1

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

Amount of \[\ce{H+}\] ions in 1L of Solution A= Concentration × volume V = 10^–6 mol × 1L

Amount of \[\ce{H+}\] ions in 1L of solution B = 10^–4 mol × 1L

∴ Total amount of \[\ce{H+}\] ions in the solution formed by mixing solutions A and B is (10^–6 mol + 10^–4 mol)

This amount is present in 2L solution.

∴ Total H⁺ = `(10^(-4) (1 + 0.01))/2`

= `(1.01 xx 10^(-4))/2` mol L^–1

= `(1.01 xx 10^(-4))/2` mol L^–1

= 0.5 × 10^–4 mol L^–1

= 5 × 10^–5 mol L^–1 

\[\ce{pH = – log [H+] = – log (5 × 10^{–5})}\]

= – [log 5 + (– 5 log 10)]

= – log 5 + 5

= 5 – log 5

= 5 – 0.6990

= 4.3010 = 4.3