Answer:
pH = 12.61
Explanation:
First of all, we determine, the milimoles of base:
0.120 M = mmoles / 300 mL
mmoles = 300 mL . 0120 M = 36 mmoles
Now, we determine the milimoles of acid:
0.200 M = mmoles / 100 mL
mmoles = 100 mL . 0.200M = 20 mmoles
This is the neutralization:
HCOOH + OH⁻ ⇄ HCOO⁻ + H₂O
20 mmol 36 mmol 20 mmol
16 mmol
We have an excess of OH⁻, the ones from the NaOH and the ones that formed the salt NaHCOO, because this salt has this hydrolisis:
NaHCOO → Na⁺ + HCOO⁻
HCOO⁻ + H₂O ⇄ HCOOH + OH⁻ Kb → Kw / Ka = 5.55×10⁻¹¹
These contribution of OH⁻ to the solution is insignificant because the Kb is very small
So: [OH⁻] = 16 mmol / 400 mL → 0.04 M
- log [OH⁻] = pOH → 1.39
pH = 14 - pOH → 12.61
