Question

60.0 mL of 0.100M KOH(aq) is mixed with 40.0 mL of 0.250M HCl(aq) at 25 oC. Assuming that volumes are additive, the pH of resulting solution is

Answer 1

mmol KOH = concentration   Volume = 0.100 M 60.0 ml = 6.00 mmol

mmol HCl = concentration   Volume = 0.250 M 40.0 ml = 10.0 mmol

Consider reaction, HCl + KOH KCl + H2O

According to reaction, 1 mmol KOH reacts with 1 mmol HCl.

6.00 mmol KOH reacts with 6.00 mmol HCl.

Excess mmol HCl = 10.0 - 6.00 = 4.00 mmol

Volume of solution = Volume of HCl + Volume of KOH = 40.0 + 60.0 = 100.0 ml

[ HCl ] = 4.00 mmol / 100.0 ml = 0.0400 M

HCl is a strong acid. It dissociates completely as shown below.

HCl (aq) + H2O (l) H3O + (aq) + Cl - (aq)

According to reaction, [ HCl ] = [ H3O + ] = [ Cl - ] = 0.0400 M

We have relation, pH = - log [ H3O + ]

pH = - log 0.0400

pH = 1.40

ANSWER : pH = 1.40