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

5 Views|Posted 4 months ago
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4 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 1 0 [ S ] [ A ]

p H = 4 . 7 6 + l o g 1 0 2 . 5 2 . 5 = 4 . 7 6 = 4 7 6 * 1 0 2  

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Chemistry Ncert Solutions Class 11th 2023

Chemistry Ncert Solutions Class 11th 2023

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