50 mL of 0.1 M is being titrated against 0.1 M NaOH. When 25 mL of NaOH has been added, the pH of the solution will be . (Nearest integer)

Question

50 mL of 0.1 M  is being titrated against 0.1 M NaOH. When 25 mL of NaOH has been added, the pH of the solution will be . (Nearest integer)

Canvas Classes · Accepted Answer

Step 1 — Moles at half-equivalence point:
Moles of  =  mmol
Moles of NaOH added =  mmol

After reaction:
-  remaining = 2.5 mmol
-  formed = 2.5 mmol

Step 2 — Henderson-Hasselbalch:

At half-neutralisation:

Expressed as : , so

Answer: 476

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