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pH when 50 mL of 0.1 M acetic acid is partly neutralised by 0.1 M NaOH (pKa 4.76)

The same question, asked 2 times: 26 Jun 2022*, 13 Apr 2023. Every wording below, oldest first.

Try each one first, then open its answer.

  1. Question 1

    26 Jun 2022

    53
    50 mL50\text{ }mL of 0.1MCH3COOH0.1MCH_{3}COOH is being titrated against 0.1MNaOH0.1MNaOH. When 25 mL25\text{ }mL of NaOHNaOH has been added, the pHpH of the solution will be ×10−2\times10^{- 2}. (Nearest integer) (Given:  pKa(CH3COOH)=4.76)\left. \ pK_{a}\left( CH_{3}COOH \right)= 4.76 \right) log⁡2=0.30\log2 = 0.30 log⁡3=0.48\log3 = 0.48
    log⁡5=0.69\log5 = 0.69
    log⁡7=0.84\log7 = 0.84
    log⁡11=1.04\log11 = 1.04

  2. Question 2

    13 Apr 2023

    146
    20 mL20\text{ }mL of 0.1MNaOH0.1MNaOH is added to 50 mL50\text{ }mL of 0.1M0.1M acetic acid solution. The pHpH of the resulting solution is____ ×10−2\times10^{- 2} (Nearest integer) Given: pKa(CH3COOH)=4.76pKa\left( CH_{3}COOH \right)= 4.76
    log⁡2=0.30\log2 = 0.30
    log⁡3=0.48\log3 = 0.48