Question

An aqueous solution at 25°C has a OH− concentration of 1.9x10−8M . Calculate the H3O+ concentration. Be sure your answer has the correct number of significant digits.

Answer 1

Answer : The $H_3O^+$ concentration is, $5.0\times 10^{-7}M$

Explanation:

First we have to calculate the pOH.

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

Given :

Concentration of $OH^-$ = $1.9\times 10^{-8}M$

$pOH=-\log (1.9\times 10^{-8})$

$pOH=7.7$

Now we have to calculate the pH.

$pH+pOH=14\\\\pH=14-pOH\\\\pH=14-7.7=6.3$

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

pH : It is defined as the negative logarithm of hydrogen ion or hydronium ion concentration.

Formula used :

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

$6.3=-\log [H_3O^+]$

$[H_3O^+]=5.0\times 10^{-7}M$

Therefore, the $H_3O^+$ concentration is, $5.0\times 10^{-7}M$

Answer 2

Answer:

= 5.26 × 10^-7 M

Explanation:

We know that;

[H3O+][OH-] = 1 x 10^-14

Therefore;

Given; OH− concentration = 1.9x10^−8 M

Then; [H3O+] = (1 x 10^-14)/[OH-]

                      = (1 x 10^-14)/(1.9x10^−8)

                      = 5.26 × 10^-7 M