# Figuring out the pH

So, I’m refreshing my college chemistry a bit. Could someone give me a pointer on the following practice exercise?

What is the pH at 25 degrees Celsius in a bucket with 10 liter pure water in which a monoprotic strong acid with molecular mass of 100 is being dissolved?

Edit: So forgot to add the total mass of the acid which is 10 mg.

I think I figured it out:

total mass: $m = 10\ \mathrm{mg} = 0.01\ \mathrm{g}$
molecular mass: $M = 100\ \mathrm{g\ mol^{-1}}$
volume: $V=10\ \mathrm{l}$
amount of substance: $n = \frac{0.01\ \mathrm{g}}{100\ \mathrm{g\ mol^{-1}}} = 10^{−4}\ \mathrm{mol} =$ moles of dissociated protons
$\mathrm{pH}=-\log\left(\frac{10^{-4}\ \mathrm{mol}}{10\ \mathrm{l}}\right) = -\log\left(10^{-5}\right) = 5$

• I'm fairly sure this problem is unsolvable without additional information. – Dissenter Jun 26 '14 at 20:13

$\mathrm{p}\ce{H} = -\log([\ce{H+}])$
where $[\ce{H+}]$ is the molar concentration of protons. It's a monoprotic strong acid, which means that for every mole of the acid, one mole of protons will completely dissociate.
Assuming that you can find the mass of the acid, you would then use the molar mass to find moles of the acid. You know it's a $1:1$ ratio of protons to acid, so that also tells you the number of moles of protons. Then divide by the volume in liters, take the negative logarithm base 10, and you will have $\mathrm{p}\ce{H}$.
This works as long as the concentration of acid is relatively large. At small concentrations the autoionization of water becomes important. It also completely ignores the absorption of $\ce{CO2}$ from the atmosphere, which would have a noticeable effect on $\mathrm{p}\ce{H}$.