I came across the following question in Chapter 15 (Acids and Bases) of Holt High School Chemistry:

Neutralizing $\pu{5.00 L}$ of an acid rain sample required $\pu{11.3 mL}$ of $\pu{0.0102 M}\ \ce{KOH}$. Calculate the hydronium ion $[\ce{H3O+}]$ concentration in the rain sample.

This is a reasonably straightforward question requiring substitution into the equation
$$c_1V_1 = c_2V_2,$$
which lets us derive the concentration of $\ce{H3O+}$.

I initially mis-read the question (which does not mention $\ce{H3O+}$) to read it as wanting the concentration of $\ce{OH-}$. However, had I not mis-read the question, would it be a valid question?

For example, can we (after getting the concentration of $\ce{H3O+}$) via the equation above, use the equation with the ionization constant of water
$$K_\mathrm{w} = [\ce{H3O+}][\ce{OH-}]$$ to also get the $\ce{OH-}$ concentration?

  • $\begingroup$ Of course we can. $\endgroup$ – Mithoron Sep 10 '17 at 16:19

Yes, that works. In fact, some textbooks and teachers ask this type of question (asking for the lesser ion of the autodeprotonation of water) to test which students actually read the question and which gloss over details. I fell into the trap of calculating the wrong one and stopping too early, too.

So if your question had asked for the hydroxide concentration, you would indeed calculate the hydronium concentration first (because that is actually accessible) and then invert and multiply $10^{-14}$ to get the desired one as you suggest.


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