Question

In an exactly.200M solution of benzoic acid, a monoprotic acid, the [H+] = 3.55 × 10–3M. What is the Ka for benzoic acid?

Answer 1

6.3 *10^-5 I hope this helps

Answer 2

Answer : The value of
`k_a` for benzoic acid is,
`6.4* 10^(-5)`

Solution :

The balanced equilibrium reaction will be,

`C_6H_5COOH\rightleftharpoons H^++C_6H_5COO^-`

initial conc. 0.2 M 0 0

at eqm.
`(0.2-0.00355)M`
`0.00355M`
`0.00355M`

The expression for dissociation constant for a benzoic acid will be,

`k_a=([H^+]* [C_6H_5COO^-])/([C_6H_5COOH])`

Now put all the given values in this formula, we get the value of
`k_a`

`k_a=((3.55* 10^(-3))* (3.55* 10^(-3)))/((0.2-3.55* 10^(-3)))=6.4* 10^(-5)`

Therefore, the value of
`k_a` for benzoic acid is,
`6.4* 10^(-5)`