Question:
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.
Solution:
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 + 10-4/2
=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.