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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1 Answer
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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
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