Question

What is the [H3O+] and the pH of a benzoic acid-benzoate buffer that consists of 0.17 M C6H5COOH and 0.42 M C6H5COONa? (Ka of benzoic acid = 6.3 × 10^−5). Be sure to report your answer to the correct number of significant figures.

Answer 1

`[H_(3)O^(+)]=x M = 2.5* 10^(-5)M` and pH = 4.6

Step-by-step explanation:

Construct an ICE table to calculate changes in concentration at equilibrium.

`C_(6)H_(5)COOH+H_(2)O\rightleftharpoons C_(6)H_(5)COO^(-)+H_(3)O^(+)`

I(M): 0.17 0.42 0

C(M): -x +x +x

E(M): 0.17-x 0.42+x x

So,
`([C_(6)H_(5)COO^(-)][H_(3)O^(+)])/([C_(6)H_(5)COOH])=K_(a)(C_(6)H_(5)COOH)`

or,
`((0.42+x)x)/((0.17-x))=6.3* 10^(-5)`

or,
`x^(2)+0.4201x-(1.071* 10^(-5))=0`

So,
`x=\frac{-0.4201+\sqrt{(0.4201)^(2)+(4* 1* 1.071* 10^(-5))}}{(2* 1)}M`

(
`ax^(2)+bx+c=0\Rightarrow x=\frac{-b\pm \sqrt{b^(2)-4ac}}{2a};x< 0.17M`)

So,
`x=2.5* 10^(-5)`M

Hence
`[H_(3)O^(+)]=x M = 2.5* 10^(-5)M`

`pH=-log[H_(3)O^(+)]=-logx=-log(2.5* 10^(-5))=4.6`