# Exercise 6 | Buffers and Titration of Acids and Bases

Calculate the pH of the resulting solution when 20.0 mL of 0.20 M HCl is added to 20.0 mL of 0.15 M NaOH.

HCl → H+ + Cl-
NaOH → Na+ + HO-

Reaction: H+ + HO- → H2O

nH+,0 = [H+] x VH+ = 0.20 x 20.0 x 10-3 = 4.0 x 10-3 mol

nHO-,0 = [HO-] x VHO- = 0.15 x 20.0 x 10-3 = 3.0 x 10-3 mol

HO- is the limiting reagent and at the end nH+ = nH+,0 - nHO-,0 = 1.0 x 10-3 mol

[H+] = $\frac{{\mathrm{n}}_{\mathrm{H}+}}{\mathrm{V}}$ =  = 2.5 x 10-2 M

pH = - log [H+] = 1.60