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

Suppose moles of sodium propionate is added to one line of an aqueous solution containing 0.02 mole of proponic acid
Using `pH =pK+log ""(["salt"])/(["acid"])`
`4.75=-log (1.34 xx 10^(-5)) +log""(x)/(0.02)`
`x=1.5 xx 10^(-2)`
=0.0015 M
Further, addition of 0.01 mole of HCI will produce an extra 0-01 mole of propionic acid and remove 0:01 mole of sodium propionate in a one-litre solution
`[C_(2)H_(5)COOH]=0.02+0.01=0.03M`
`[C_2H_5COONa]=0.015 -0.01=0.005M`
`pH=-log (1.34 xx 10^(-5))+log""(0.005)/(0.03)=4.06`
Now, since the pH of 0.01 M, i.e., `10^(-2)` M HCl is 2, therefore, the pH of the buffer solution is about two times greater than that of 0.01 M HCI.