Question

How would you make 500 mL of a 1.5M buffer of benzoic acid/sodium benzoate that effectively buffers to a pH of 4.40?

Answer 1

This is a relatively challenging buffer question that requires a clever use of the Henderson-Hasselbach equation.

Consider benzoic acid's equilibrium,

`C_6H_5COOH rightleftharpoons H^(+) + C_6H_5COO^(-)`

where, `"pK"_"a" = 4.20`.

Moreover, recall the Henderson-Hasselbach equation,

`"pH" = "pK"_"a" + log(([A^-])/([HA]))`

The ratio of the weak acid and conjugate base needed is,

`=> ([A^-])/([HA]) = 10^("pH" - "pK"_"a") approx 1.58`

I would take `500"mL"-x` of `1.5"M"` `C_6H_5COOH` solution, and assuming `"pH" approx "pK"_"a"` (a reasonable assumption on paper), add,

`(1.5+x)/(1.5-x) = 1.58`

`=> x = 0.337"L" = 33.7"mL"` `1"M"` `HCl`

or practically, `67.4"mL"` of `5"M"` `HCl`

to increase the `"pH"` by that amount.

To be sure, your reagents are,

`67.4"mL"` `5"M"` hydrochloric acid
`432.6"mL"` `1.5"M"` benzoic acid