At the point in a titration where 5.00 mL of 0.225 mol/L hydrochloric acid, HCl(aq), is added to 40.00 mL of 0.175 mol/L potassium hydroxide, KOH(aq), what is the pH of the solution?

Answer 1

HCl and KOH react in a 1:1 mol ratio as follows:

HCl + KOH ==> KCl + H₂O

moles HCl present = 5.00 ml x 1 L/1000 ml x 0.225 mol/L = 0.001125 moles HCl

moles KOH present = 40.00 ml x 1 L/1000 ml x 0.175 mol/L = 0.0070 moles KOH

KOH is present in excess. How many moles of KOH are left after reaction?

0.007 mol KOH - 0.001125 moles = 0.005875 moles KOH

Total volume = 5 ml + 40 ml = 45 ml = 0.045 L

[KOH] = 0.005875 moles / 0.045 L = 0.131 M

pOH = -log [OH-] = -log 0.131 = 0.883

pH = 14 - pOH

pH = 14 - 0.883

pH = 13.1