Question

Consider the titration of 40mL of 0.120M NaOH with 0.0763M HCl. Calculate the pH after the addition of 0.20L of acid.

Answer 1

The chemical reaction involved here is

NaOH + HCl ----------> NaCl + H2O

Moles of NaOH = Molarity x Volume(in L) = 0.120 M 0.04 L = 4.8 x 10^-3 moles

Moles of HCl = Molarity x Volume(in L) = 0.0763 M x 0.20 L = 0.0153 mol

From the chemical reaction, it is clear that 1 mol of NaOH reacts with 1 mol of HCl

Thus, Moles of HCl left over = 0.0153 - (4.8 x 10^-3) = 0.0105 mol

Total volume = 0.04 + 0.20 = 0.240 L

Thus, [H+] = 0.0105 mol / 0.240 L = 0.044 M

pH = -log[H+]

pH = -log[0.044]

pH = 1.35