Addition of sodium hydroxide solution to a weak acid (HA) results in a buffer of pH 6. If ionisation constant of HA is 10-5, the ratio of salt to acid concentration in the buffer solution will be :

Question

Addition of sodium hydroxide solution to a weak acid (HA) results in a buffer of pH 6. If ionisation constant of HA is 10-5, the ratio of salt to acid concentration in the buffer solution will be : (A) 4: 5 (B) 1: 10 (C) 10: 1 (D) 5: 4

Tardigrade Admin · Accepted Answer

K a =10-5 pK a =- log K a =- log 10-5=5 pH = pKa + log ([ textsalt ]/[ textacid ]) 6=5+ log ([ textsalt ]/[ textacid ]) 1= log ([ textsalt ]/[ textacid ]) ([ text salt ]/[ text acid ])=(10/1)

Q. Addition of sodium hydroxide solution to a weak acid (HA) results in a buffer of pH 6. If ionisation constant of HA is $1 0^{- 5}$ , the ratio of salt to acid concentration in the buffer solution will be :

4 : 5

1 : 10

10 : 1

5 : 4

$K_{a} = 1 0^{- 5}$
$p K_{a} = - lo g K_{a} = - lo g 1 0^{- 5} = 5$
$p H = p K a + lo g \frac{[ salt ]}{[ acid ]}$
$6 = 5 + lo g \frac{[ salt ]}{[ acid ]}$
$1 = lo g \frac{[ salt ]}{[ acid ]}$
$\frac{[ salt ]}{[ acid ]} = \frac{10}{1}$

Q. Addition of sodium hydroxide solution to a weak acid (HA) results in a buffer of pH 6. If ionisation constant of HA is 10−5, the ratio of salt to acid concentration in the buffer solution will be :