One litre of a buffer solution containing `0.02` mol of propanoic acid and some sodium propanate has `pH = 4.75`. What will be the `pH` if `0.01` mol of hydrogen chloride is dissolved in the above buffer solution ?
[Dissociation constant of propanoic acid at `25^(@)C` is `1.34 xx 10^(-5)`].

Options

A. `4.11`
B. `2.11`
C. `0.11`
D. `3.11`

Answer

A

Explanation

Using the expression
`pH = pKa + "log" (["salt"])/(["acid"])`
We get
`4.75 = - log(1.34 xx 10^(-5)) + "log" (["salt"])/(["acid"])`
which gives `4.75 = 4.87 + "log" (["salt"])/(0.02M)`
or `(["Salt"])/(0.02M) = 0.76` or `["salt"] = 1.52 xx 10^(-2)M`
Hence, Amount of sodium propanoate to be added
`= 1.52 xx 10^(-2) "mol"`
The addition of `0.01` hydrogen chloride convert the equivalent amount of sodium propanoate into propanoic acid.