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CGP EDU Academic Team
Published on: August 13, 2026
20 mL of 0.1 MNaOH is added to 50 mL of 0.1M acetic acid solution. The pH of the resulting solution is x 10 -2 . (Nearest integer) Given : pKa(CH 3 COOH)=4.76
log 2 = 0.30
log 3 = 0.48
Text Solution
Verified by ExpertsThe correct answer is:
4.58
(4.58)
When a strong base is added to a weak acid solution, it results in the formation of a salt. Here, acid is present in a limiting reagent and base is present in excess amounts. So, by using the pH formula:
CH 3 COOH + NaOH 
CH 3 COONa + H 2 O
5 2 - -
3 0 2
pH = pK a + log 
pH = 4. 76 + 0.30 - 0.48
= 4. 76-.18
= 4.58
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