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.

Given, pH of solution A = 6

[H⁺] of solution A = 10^-6 mol/lit.

pH of solution B = 4

[H⁺] of solution B = 10^-4 mol/lit.

On mixing 1L of each solution we will get total 2L of Solution.

Amount of [H⁺] in 1L: solution A = 10^-6× 1L = 10^-6

: Solution B = 10^-4 × 1L = 10^-4

Total [H⁺] in Solution = 10^-6 + 10^-4

=10^-4 (1+0.01/2)

=10^-4× 1.01/2

=5×10^-5 mol/L

pH = -log[H⁺]= -log[5×10^-5]

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

=-log5 + 5 =4.3

The pH of the Solution formed by mixing will be 4.3