The ionization constant of formic acid is `1.8xx10^(-4)`. Around what pH will its mixture with sodium formed give buffer solution of higher capacity. Calculate the ratio of sodium formate and formic acid in a buffer of `pH 4.25`.

Explanation

For acidic buffer mixtures.
`pH= – log K_a + log (["Salt"])/(["Acid"])`
The buffer capacity is `(dpH)/(dn_("acid or base"))`
Thus highest buffer capacity of this is `(dpH)/(dn_("acid"))`. This will be maximum when pH is near to `pK_a`. Also the best results are obtained by buffer when `(["Salt"])/(["Acid"])=10 or 1/10" i.e, within the range "pK_(a) pm 1`
Also `pH =-log K_(a)+log (["Salt"])/(["Acid"])`
`4.25=-1.8 xx 10^(-4)+log (["Salt"])/(["Acid"])`
`or 4.25 =3.74 +log (["Salt"])/(["Acid"])`
`therefore (["Salt"])/(["Acid"])=3.24`