Question

What will be the pH of the solution obtained by mixing 800mL of 0.05M NaOH and 200 mL of 0.1 M of HCl (assuming complete ionisation of the acid and the base)?

Answer 1

hey
answer is in the picture.

Answer 2

Initial concentration of OH⁻ = 800 x 0.05 = 40 milimoles

Initial concentration of H⁺ = 200 x 0.1 = 20 milimoles

Thus, same amount of H⁺ will neutralise OH⁻

Remaining OH⁻ concentration = (40-20) = 20 milimoles = 20/1000 moles = 0.02 moles

pOH = - log [OH⁻]

or, pOH = - log [0.02]

or, pOH = 1.7

Now, pH + pOH = 14

or, pH = 14 - 1.7

or, pH = 12.3

pH of the solution will be 12.3.