Question

In the laboratory, a general chemistry student measured the pH of a 0.341 M aqueous solution of benzoic acid, C6H5COOH to be 2.351. Use the information she obtained to determine the Ka for this acid.

Answer 1

Answer:
`K_a` for the acid is
`5.75* 10^(-5)`

Step-by-step explanation:

`C_6H_5COOH\rightarrow H^+C_6H_5COO^-`

cM 0 0

`c-c\alpha`
`c\alpha`

So dissociation constant will be:

`K_a=((c\alpha)^(2))/(c-c\alpha)`

Give c= 0.341 M and = 2.351

`K_a=?`

`pH=-log[H^+]`

`[H^+]=10^(-2.351)=0.0044`

`[H^]=c\alpha=0.0044`

`K_a=((0.0044)^(2))/(0.341-0.0044)`

`K_a=5.75* 10^(-5)`

Thus
`K_a` for the acid is
`5.75* 10^(-5)`