Solveeit Logo
Login

Question

Chemistry Question on Equilibrium

Addition of sodium hydroxide solution to a weak acid (HA) results in a buffer of pH 6. If ionisation constant of HA is 10−510^{-5}, the ratio of salt to acid concentration in the buffer solution will be :

A

4:05

B

1:10

C

10:01

D

5:04

Answer

10:01

Explanation

Solution

Ka=10−5K _{ a }=10^{-5}
pKa=−log⁡Ka=−log⁡10−5=5pK _{ a }=-\log K _{ a }=-\log 10^{-5}=5
pH=pKa+log⁡[salt][acid]pH = pKa +\log \frac{[\text{salt} ]}{[ \text{acid} ]}
6=5+log⁡[salt][acid]6=5+\log \frac{[ \text{salt} ]}{[ \text{acid} ]}
1=log⁡[salt][acid]1=\log \frac{[ \text{salt} ]}{[ \text{acid} ]}
[ salt][ acid]=101\frac{[\text{ salt} ]}{[\text{ acid} ]}=\frac{10}{1}