Question

The ka of benzoic acid is 6.4 x 10–5. what is the approximate ph of a 1.5 m solution of benzoic acid?

Answer 1

I don't know if you need the answer explain to you but here's the answer 2.00

Answer 2

Answer: 2.01

Step-by-step explanation:

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

initial : c 0 0

eqm:
`c(1-\alpha)`
`c\alpha`
`c\alpha`

`K_a=(c\alpha* c\alpha)/(c(1-\alpha))`

when
`\alpha` is very very small the, the expression will be,

`k_a=(c^2\alpha^2)/(c)=c\alpha^2\\\\\alpha=\sqrt{(k_a)/(c)}`

And,

`[H^+]=c\alpha`

Thus the expression will be,

`[H^+]=√(k_a* c)`

Now put all the given values in this expression, we get

`[H^+]=\sqrt{(6.4* 10^(-5))* 1.5}`

`[H^+]=9.7* 10^(-3)M`

`pH=-log[H^+]`

`pH=-log[9.7* 10^(-3)]=2.01`