What is the H+ concentration (in molarity) of a solution with a pOH of 6.7?

Question

Mateo Aguilar · Accepted Answer

#[H^(+)] = 5.01xx10^(-8) M#

The relationship that you wish to use is this one:

#pOH + pH = 14#

Given the pOH, we can calculate the pH by deducting it from 14 in the following manner:

14 - pOH is equal to 7.3.

7.3 represents the pH of the solution and we can obtain the concentration of hydrogen ions in solution by taking the antilog of the pH. The antilog is #10^(-)# raised to some value, which is #10^(-7.3)# in our case.

#10^(-7.3)# = #5.01xx10^(-8) M#

Tips: #[H^(+)]# = antilog(-pH) pH = -log#[H^(+)]#

I hope you can understand this explanation!

Extra Assistance

Mateo Aguilar · Answer

undefined To find the $H^+$ concentration ($[H^+]$), you can use the relationship between $pOH$ and $[OH^−]$, and then use the fact that in a neutral solution, $[H^+] \times [OH^−] = 1.0 \times 10^{−14}$ at 25°C. First, calculate $[OH^−]$ using the formula $pOH = -\log[OH^−]$, then find $[H^+]$ by taking the reciprocal of $[OH^−]$.

Given:
$pOH = 6.7$

Calculate $[OH^−]$:
$$ [OH^−] = 10^{-pOH} = 10^{-6.7} $$

Calculate $[H^+]$:
$$ [H^+] = \frac{1.0 \times 10^{-14}}{[OH^−]} $$

$$ [H^+] = \frac{1.0 \times 10^{-14}}{10^{-6.7}} $$

$$ [H^+] = 10^{(-14 + 6.7)} $$

$$ [H^+] = 10^{-7.3} $$

So, the $H^+$ concentration of the solution is $10^{-7.3}$ M.