What is the pH at equivalence point in the titration of 0.1 M CH 3 COOH and 0.1 M NaOH ? (ka 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 ? (ka 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;
pH =7+(1/2) pK a +(1/2) log c
=7+(4.74/2)+(1/2) log 10-1=8.87

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;

$p H = 7 + \frac{1}{2} p K_{a} + \frac{1}{2} lo g c$

$= 7 + \frac{4.74}{2} + \frac{1}{2} lo g 1 0^{- 1} = 8.87$

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=8.87

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;

$p H = 7 + \frac{1}{2} p K_{a} + \frac{1}{2} lo g c$

$= 7 + \frac{4.74}{2} + \frac{1}{2} lo g 1 0^{- 1} = 8.87$