Question

What is the [H+] in a solution with pOH of 0.253? 5.58 × 10−15 M 1.79 × 10−14 M 3.21 × 10−2 M 5.58 × 10−1 M

Answer 1

Okay so we know that pOH + pH = 14, so if pOH is 0.253 the pH would be 13.747.

And pH = -log [H+], so [H+] = 10^(-pH) -->This is just the antillog...

so [H+] = 1.791e-14, which would make sense for the strongly basic solution (so B is the correct answer)

Hope this helps...

Answer 2

Answer:
`3.6* 10^(-14)M`

Explanation:-

First we have to calculate the pOH.

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

`0.253=-\log[OH^-]`

`[OH^-]=0.558M`

Now we have to calculate the pH.

`pH+pOH=14\\\\pH=14-0.558\\\\pH=13.442`

Now we have to calculate the
`OH^-` concentration.

`pH=-\log [H^+]`

`13.442=-\log [H^+]`

`[H^+]=3.6* 10^(-14)M`

Therefore, the
`H^+` concentration is,
`3.6* 10^(-14)M`