Question

Calculate the ph of a solution that is 0.115m benzoic acid and 0.230m sodium benzoate, a salt whose anion is the conjugate base of benzoic acid.

Answer 1

Step-by-step explanation:

According to the Handerson equation,

pH =
`pK_(a) + log (C_(6)H_(5)COONa)/(C_(6)H_(5)COOH)`

Also, as standard value of benzoic acid's
`K_(a) = 6.46 * 10^(-5)`

So,
`pK_(a) = -log K_(a)`

=
`-log(6.46 * 10^(-5))`

= 4.19

Therefore, pH =
`-log K_(a) + log (C_(6)H_(5)COONa)/(C_(6)H_(5)COOH)`

=
`4.19 + log (0.230)/(0.115)`

= 4.19 + 0.30

= 4.49

Thus, we can conclude that the pH of given solution is 4.49.

Answer 2

Ka of benzoic acid = 6.5 x 10⁻⁵
pKa = - log Ka = - log (6.5 x 10⁻⁵) = 4.2
Benzoic acid (C₆H₅COOH) and its conjugate base Benzoate (C₆H₅COO⁻) are considered as acidic buffer.
Using Henderson- Hasselbalch equation for calculation of pH of buffer
pH = pKa + log
`([C6H5COO^(-) ])/([C6H5COOH])`
= 4.2 + log (0.23) / (0.115) = 4.5