When 10 mL of 0.1 M acetic acid (pKa = 5.0) is titrated against 10 mL of 0.1 M ammonia solution (pKb = 5.0), the equivalence point occurs at pH

Question

When 10 mL of 0.1 M acetic acid (pKa = 5.0) is titrated against 10 mL of 0.1 M ammonia solution (pKb = 5.0), the equivalence point occurs at pH (A) 5.0 (B) 6.0 (C) 7.0 (D) 9.0

Tardigrade Admin · Accepted Answer

p Ka =- log Ka text and p Kb=- log Kb pH =-(1/2)[ log Ka+ log Kw- log Kb] =-(1/2)[-5+ log 10-14-(-5)] =-(1/2)[-5-14+5] =7 Thus, the end point or equivalence point is obtained at pH =7 in neutral medium.

Q. When $10 m L$ of $0.1 M$ acetic acid $(p K_{a} = 5.0)$ is titrated against $10 m L$ of $0.1 M$ ammonia solution $(p K_{b} = 5.0)$ , the equivalence point occurs at $p H$

5.0

6.0

7.0

9.0

$p K_{a} = - lo g K_{a} and p K_{b} = - lo g K_{b}$
$p H = - \frac{1}{2} [lo g K_{a} + lo g K_{w} - lo g K_{b}]$
$= - \frac{1}{2} [- 5 + lo g 1 0^{- 14} - (- 5)]$
$= - \frac{1}{2} [- 5 - 14 + 5]$
$= 7$
Thus, the end point or equivalence point is obtained at $p H = 7$ in neutral medium.

Q. When 10mL of 0.1M acetic acid (pKa​=5.0) is titrated against 10mL of 0.1M ammonia solution (pKb​=5.0), the equivalence point occurs at pH

5.0

6.0

7.0

9.0

pKa​=−logKa​ and pKb​=−logKb​ pH=−21​[logKa​+logKw​−logKb​] =−21​[−5+log10−14−(−5)] =−21​[−5−14+5] =7 Thus, the end point or equivalence point is obtained at pH=7 in neutral medium.

Q. When $10 m L$ of $0.1 M$ acetic acid $(p K_{a} = 5.0)$ is titrated against $10 m L$ of $0.1 M$ ammonia solution $(p K_{b} = 5.0)$ , the equivalence point occurs at $p H$

$p K_{a} = - lo g K_{a} and p K_{b} = - lo g K_{b}$
$p H = - \frac{1}{2} [lo g K_{a} + lo g K_{w} - lo g K_{b}]$
$= - \frac{1}{2} [- 5 + lo g 1 0^{- 14} - (- 5)]$
$= - \frac{1}{2} [- 5 - 14 + 5]$
$= 7$
Thus, the end point or equivalence point is obtained at $p H = 7$ in neutral medium.