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?

1 Answer
Apr 11, 2016

Answer:

#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.