Question

If a buffer made by dissolving 20.0 g of sodium benzoate (144g/mol) and 10 grams of benzoic acid (122g/mol) in water to a final volume of 500.0 ml, what is the pH of the buffer?

Answer 1

`pH=4.43`.

Explanation:

To find the pH of the buffered solution in question, we can simply use the Henderson-Hasselbach equation:

`pH=pK_a+log(([Base])/([Acid]))`

where, `pK_a` for benzoic acid is `4.20`.

and `[Benzoate]=n/V and n=m/(MM)=(20.0cancel(g))/(144cancel(g)/(mol))=0.139mol`

`[Benzoate]=n/V=(0.139mol)/(0.500L)=0.278M`

and `["Benzoic acid"]=n/V and n=m/(MM)=(10.0cancel(g))/(122cancel(g)/(mol))=0.0820mol`

`[Benzoate]=n/V=(0.0820mol)/(0.500L)=0.164M`

Therefore, `pH=4.20+log((0.278)/(0.164))=4.43`

Acid - Base Equilibria | Buffer Solution.