Question

a.) Calculate the pH during the titration of 100.0 mL of 0.200 M HCl with 0.400 M NaOH. First what is the initial pH (before any NaOH is added)?

b.) What is the pH after 31.9 mL of NaOH are added?

c.) What is the pH after 50 mL of NaOH are added?

d.) What is the pH after 68.9 mL of NaOH are added?

Answer 1

a) moles of HCl = MXV = 0.2 X 0.1 = 0.02

pH = -log(0.02) = 1.69

b) moles of HCl = (100 X 0.2)/131.9 = 0.152, moles of base = (31.9 X0.4)/131.9 = 0.0967

moles of acid left = 0.152-.0967 = 0.0815

pH = -log(0.0815) = 1.089

c) moles of HCl = (100X0.2)/150 = 0.133, moles of base = (50 *0.4)/150 = 0.133

pH = 7 (solution is neutral)

d) moles of acid = 100*0.2/168.9 = 0.118, moles of base = 68.9*0.4/168.9 = 0.163

moles of base left = 0.163-0.118 = 0.045

pOH = -log(0.045) = 1.35

pH = 14-1.35 = 12.65