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)`.

Answer

B, C, D

Explanation

Let the moles of sodium propinote added be `a mol L^(-1)` For an acidic buffer.
`pH = pK_(a) +"log" (["Salt"])/(["Acid"])`
or `4.75 =- log (1.34 xx 10^(-5)) +"log" (a)/(0.020)`
or `a = 1.5 xx 10^(-2)`
`CH_(3)CH_(2)COOH hArr CH_(3)CH_(2)COO^(Theta) +H^(o+)`
`CH_(3)CH_(2)COONa hArr CH_(3)CH_(2)COO^(Theta) +Na^(o+)`
Rule A A A: When `HCI` is added, the amount of acid increases because propanote ions recat with hydrogen ions to form undissociated salt.
Therefore, the amount of propanoate decreases.
New `["Acid"] = 0.020 + 0.010 = 0.030`
New `["Salt"] = 0.015 - 0.010 = 0.005`
`pH = pK_(a) +"log" (["Salt"])/(["ACid"])`
`=- log (1.34 xx 10^(-4)) +"log" (0.005)/(0.030) = 4.09`
`pH` of `0.010 M HCI =- log (1xx 10^(-2)) = 2`
pH of `HCI` is less than that of the buffer.