How much moles of sodium propionate should be added to `1L` of an aqueous solution containing `0.020 mol` of propionic acid to obtain a buffer solution of `pH 4.75 `? What will be the `pH` if `0.010 mol` of `HCI` is dissolved in the above buffer solution. Compare the last `pH` value with the `pH of 0.010 M HCI` solution. Dissociation constant of propionic acid, `K_(a)`, at `25^(@)C` is `1.34 xx 10^(-5)`.

Explanation

For acidic buffer, the Henderson’s equation is
`pH = pK_(a) + log(mole of air)/(mol e of acid)`
`4.75 = -log(1.34 xx 10^(-5)) + log x/0.02`
`implies x = 0.015` mol e of sodium propionate.
Addition of 0.01 mole HCl increase moles of propinic acid by 0.01 and moles of sodium propionate will decrease by same amount.
New moles of acid = 0.02 + 0.01 = 0.03
`pH = -log(1.34 xx 10^(-5)) + log (0.005/0.030) = 4.09`
pH of 0.001 HCl = 2, just half of the pH of final buffer solution