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← JEE Main 2022 · 26 Jun · Shift 1
ChemistryInteger+4 / -1Question 83 of 90

JEE Main 2022 · 26 Jun · Shift 1, Chemistry question 83

50 mL of 0.1 M CH3COOH\mathrm{CH_3COOH} 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\times10^{-2}. (Nearest integer)

(Given : pKa(CH3COOH)=4.76\mathrm pK_a(\mathrm{CH_3COOH})=4.76, 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)

Answer and worked solution

Official answer: 476

Half neutralised. 2.5 mmol NaOH reacts with half of the 5 mmol acid, leaving equal acid and acetate.

Buffer. pH =pKa=4.76=476×10−2=\mathrm pK_a=4.76=476\times10^{-2}.

Answer: 476