50 mL of 0.1 M CH3COOH is being titrated against 0.1 M NaOH. When 25 mL of NaOH has been added, the pH of the solution will be_________* 10-2. (Nearest integer) (Given : pKa (CH3-COOH) = 4.76) log 2 = 0.30 log 3 = 0.48 log 5 = 0.69 log 7 = 0.84 log 11 = 1.04

**50 mL of 0.1 M CH₃COOH is being titrated against 0.1 M NaOH. When 25 mL of NaOH has been added, the pH of the solution will be_________* 10^-². (Nearest integer)

(Given : pKa (CH₃-COOH) = 4.76)

log 2 = 0.30

log 3 = 0.48

log 5 = 0.69

log 7 = 0.84

log 11 = 1.04**

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Payal Gupta | Contributor-Level 10

6 months ago

Here, total meq of acetic acid = 50 * 0.1 = 5

And total meq of NaOH = 25 * 0.1 = 2.5

After neutralization process

Meq of left acetic acid = 2.5

And meq of formed CH₃COONa = 2.5

$p H = p K a + l o g_{10} \frac{[S]}{[A]}$

$p H = 4.76 + l o g_{10} \frac{2.5}{2.5} = 4.76 = 476 * 1 0^{- 2}$

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Chemistry NCERT Exemplar Solutions Class 12th Chapter Six 2025

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