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?

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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?

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Answer 1 · 2016-04-11T16:30:57

$p H = 4.43$ $pH=4.43$ .

Explanation:

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

$p H = p K_{a} + log (\frac{[B a s e]}{[A c i d]})$ $pH=pK_a+log(([Base])/([Acid]))$

where, $p K_{a}$ $pK_a$ for benzoic acid is $4.20$ $4.20$ .

and $[B e n z o a t e] = \frac{n}{V} and n = \frac{m}{M M} = \frac{20.0 g}{144 \frac{g}{m o l}} = 0.139 m o l$ $[Benzoate]=n/V and n=m/(MM)=(20.0cancel(g))/(144cancel(g)/(mol))=0.139mol$

$[B e n z o a t e] = \frac{n}{V} = \frac{0.139 m o l}{0.500 L} = 0.278 M$ $[Benzoate]=n/V=(0.139mol)/(0.500L)=0.278M$

and $[Benzoic acid] = \frac{n}{V} and n = \frac{m}{M M} = \frac{10.0 g}{122 \frac{g}{m o l}} = 0.0820 m o l$ $["Benzoic acid"]=n/V and n=m/(MM)=(10.0cancel(g))/(122cancel(g)/(mol))=0.0820mol$

$[B e n z o a t e] = \frac{n}{V} = \frac{0.0820 m o l}{0.500 L} = 0.164 M$ $[Benzoate]=n/V=(0.0820mol)/(0.500L)=0.164M$

Therefore, $p H = 4.20 + log (\frac{0.278}{0.164}) = 4.43$ $pH=4.20+log((0.278)/(0.164))=4.43$

Acid - Base Equilibria | Buffer Solution.

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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

Explanation:

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