Question

A solution contains `[OH^-] = 4.0 times 10^-5` `M`, what is the concentration of `[H_3O^+]`?

Answer 1

`[H_3O^+]=2.50xx10^-10*mol*L^-1`

Explanation:

In aqueous solution, it is a fact that the following equilibrium operates under standard conditions.........

`2H_2O(l) rightleftharpoons H_3O^(+) +HO^(-)`

Under standard conditions............

`K_"rxn"=K_w=[H_3O^+][HO^-]=10^-14`. And typically we would use logarithms to reduce this expression to.......

`pH+pOH=14`, where the `pH` function means `-log_10[H_3O^+]` etc.

So `pOH=-log_10[4.0xx10^-5]=4.40`, and `pH=9.60`.

And thus `[H_3O^+]=10^(-9.60)=2.50xx10^-10*mol*L^-1.`

Answer 2

`[H_3O^+]=2.5*10^(-10) M`

Explanation:

Given that no hydrolysis happened in the solution (inferred from question) Calculate `[H_3O^+]` from `k_w` directly.
`[H_3O^+]`
`=k_w/([OH^-])`
`=10^-14/(4.0*10^-5)`
`=2.5*10^-10M`

Answer 3

`2.5 xx 10 ^-10` `"M"`

Explanation: