What is the pH at equivalence point in the titration of 0.1 M CH 3 COOH and 0.1 M NaOH ?( K a. for acetic acid is 1.8 × 10-5 )

Question

What is the pH at equivalence point in the titration of 0.1 M CH 3 COOH and 0.1 M NaOH ?( K a. for acetic acid is 1.8 × 10-5 ) (A) 7 (B) Between 7 and 8 (C) Between 8 and 9 (D) Between 6 and 7

Tardigrade Admin · Accepted Answer

For weak acid and strong base, p H=7+(1/2) p Ka+(1/2) log C =7+(4.74/2)+(1/2) log 10-1

Q. What is the $p H$ at equivalence point in the titration of $0.1 M C H_{3} COO H$ and $0.1 M N a O H ? (K_{a}$ for acetic acid is $1.8 \times 1 0^{- 5}$ )

7

Between 7 and 8

Between 8 and 9

Between 6 and 7

For weak acid and strong base,
$p H = 7 + \frac{1}{2} p K_{a} + \frac{1}{2} lo g C$
$= 7 + \frac{4.74}{2} + \frac{1}{2} lo g 1 0^{- 1}$

Q. What is the pH at equivalence point in the titration of 0.1MCH3​COOH and 0.1MNaOH?(Ka​ for acetic acid is 1.8×10−5 )

7

Between 7 and 8

Between 8 and 9

Between 6 and 7

For weak acid and strong base, pH=7+21​pKa​+21​logC =7+24.74​+21​log10−1

Q. What is the $p H$ at equivalence point in the titration of $0.1 M C H_{3} COO H$ and $0.1 M N a O H ? (K_{a}$ for acetic acid is $1.8 \times 1 0^{- 5}$ )

For weak acid and strong base,
$p H = 7 + \frac{1}{2} p K_{a} + \frac{1}{2} lo g C$
$= 7 + \frac{4.74}{2} + \frac{1}{2} lo g 1 0^{- 1}$