Question

In a titration of a 100.0mL 1.00M HCl strong acid solution with 1.00M NaOH, what is the pH of the solution after the addition of 139 mL of NaOH? (3 significant figures)

Answer 1

first write the balanced equation

NaOH + HCl ---NaCl + H2O

no fo moles of HCl = 1.0M x 0.1L = 0.1moles

no of moles of NaOH = 1.0M x 0.139L = 0.139 moles

0.1 mole of HCl will react with 0.1 moles of NaOH

no of moles of NaOH remaining = 0.139 - 0.1 = 0.039 moles of NaOH

total volume = 100 +139 = 239 mL = 0.239L

Concemtration of remaining NaOH = 0.039 moles / 0.239 L = 0.1631 moles/L

since NaOH is strong base

pOH = -log(0.1631)

pOH = 0.787

pH = 14-pOH = 14-0.787

= 13.213