If [H_3O^+] = 1.005 * 10^-5 M, what is the pH?

Dec 24, 2015

$4.9978$

Explanation:

Right from the start, you can say that the pH of this solution will be very close to $5$.

Here's why that is so.

As you know, the pH of pure water at room temperature is equal to $7$. This value is derived from the self-ionization of water, in which two water molecules exchange a proton, ${\text{H}}^{+}$, to form hydronium and hydroxide ions

$2 {\text{H"_2"O"_text((l]) rightleftharpoons "H"_3"O"_text((aq])^(+) + "OH}}_{\textrm{\left(a q\right]}}^{-}$

At room temperature, the ion product constant---which you may see as ${K}_{w}$---for this reaction is equal to ${10}^{- 14}$, which is why

${10}^{- 14} = \left[{\text{H"_3"O"^(+)] * ["OH}}^{-}\right]$

["H"_3"O"^(+)] = ["OH"^(-)] = sqrt(10^(-14)) = 10^(-7)"M"

As you know, pH is defined as

"pH" = -log( ["H"_3"O"^(+)])

For pure water, this is equal to

$\text{pH} = - \log \left({10}^{- 7}\right) = 7$

In your case, assuming that you're at room temperature as well, the concentration of hydronium ions is higher than ${10}^{- 7}$. This tells you that

• the solution is acidic
• its pH must be lower than $7$

Since pH is defined as the negative log base $10$ of the concentration of hydronium ions, a hydronium concentration that is approximately two times higher will result in a pH that is approximately two units lower.

$\text{pH} = - \log \left(1.005 \cdot {10}^{- 5}\right) = \textcolor{b l u e}{4.9978}$

Indeed, the resulting pH is very close to $5$, since $1.005 \cdot {10}^{- 5}$ is very close to ${10}^{- 5}$.