Question

A solution has a oh- = 1 x 10-5 m what are the h30 and the ph of the solution

Answer 1

Answer : The $H_3O^+$ concentration and the pH of the solution is, $1\times 10^{-9}M$ and 9 respectively.

Explanation:

First we have to calculate the pOH.

$pOH=-\log [OH^-]$

$pOH=-\log (1\times 10^{-5})$

$pOH=5$

Now we have to calculate the pH.

$pH+pOH=14\\\\pH=14-pOH\\\\pH=14-5=9$

Now we have to calculate the $H_3O^+$ concentration.

$pH=-\log [H_3O^+]$

$9=-\log [H_3O^+]$

$[H_3O^+]=1\times 10^{-9}M$

Therefore, the $H_3O^+$ concentration and the pH of the solution is, $1\times 10^{-9}M$ and 9 respectively.

Answer 2

I think it's easiest to find the pOH from the given [OH-] first.

-log(1x10^-5)

pOH=5

Then find the pH.

pOH+pH=14

5+pH=14

pH=9

Then find the [H+] using the pH.

antilog(-9) (if you dont have an antilog button use 10^-9)

[H+]=1x10^-9