Question

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`.

Answer 1

For acidic 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)+-1`
Also `pH= -log K_(a)+log (["Salt"])/(["Acid"])`
`4.25= -1.8xx10^(-4)+ log((["Salt"])/(["Acid"]))`
or `4.25=3.74+log ((["Salt"])/(["Acid"]))`
`:. ((["Salt"])/(["Acid"]))=3.24`