Acids, bases and pH: how to calculate pH in Year 4

Late September 2026: in Year 4 of the liceo scientifico (Italian science-focused high school), natural sciences has started from solutions and concentration, and is heading towards thermochemistry and then kinetics and equilibrium — the topic I have already written about here. Right after equilibrium, typically between October and November, come acids, bases and pH, and the test on this block falls between the end of November and mid-December — nine to eleven weeks from now. If you start now, you have time to get there prepared, but the whole block rests on $K_c$: if equilibrium is still vague, it shows up straight away.

The minimum of logarithms you need

In many Year 4 classes, by the time pH arrives, logarithms have not yet been covered properly in maths. Chemistry does not wait, so the minimum you need here is mechanical, not conceptual:

• ${\log }_{10}$ of a power of ten is simply its exponent: $\log (10^n)=n$. So $\log (10^{-2})=-2$, $\log (10^{-5})=-5$.
• For any other number (not just an exact power of ten), you work out the logarithm with the calculator: nobody expects you to know it by heart.
• The inverse of the logarithm is raising ten to a power: if $\text{pH}=-\log [{\text{H}}^{+}]$, then $[{\text{H}}^{+}]=10^{-\text{pH}}$. You need it in both directions: from concentration to pH, and from pH to concentration.

With these three points, everything that follows is direct application: you need nothing more from logarithm theory to pass the test on this topic.

Arrhenius and Brønsted-Lowry: two definitions, one broader than the other

The Arrhenius definition is the simplest: an acid is a substance that, dissolved in water, releases ${\text{H}}^{+}$ ions; a base is a substance that releases ${\text{OH}}^{-}$ ions. It works well for the “classic” acids and bases (HCl, NaOH), but it does not explain why a substance like ammonia (${\text{NH}}_3$), which contains no ${\text{OH}}^{-}$ ions, still behaves as a base.

The Brønsted-Lowry definition is more general: an acid is a substance that donates a proton (${\text{H}}^{+}$), a base is a substance that accepts one. In this scheme, every acid that donates a proton produces its conjugate base, and every base that accepts a proton produces its conjugate acid. For acetic acid in water:

$\mathrm{C}{\mathrm{H}}_3\mathrm{COOH}$ is the acid, ${\mathrm{H}}_2\mathrm{O}$ behaves as a base (it accepts the proton and becomes ${\mathrm{H}}_3{\mathrm{O}}^{+}$), and $\mathrm{C}{\mathrm{H}}_3\mathrm{CO}{\mathrm{O}}^{-}$ is the conjugate base of the starting acid. From here on, when I write ${\text{H}}^{+}$ for simplicity, I always mean the hydrogen ion in water, which in fact exists as ${\mathrm{H}}_3{\mathrm{O}}^{+}$.

The self-ionisation of water and Kw

Even pure water, on its own, dissociates to a tiny extent into ${\text{H}}^{+}$ and ${\text{OH}}^{-}$ ions:

This equilibrium has its own constant, called the ionic product of water, which at $25{}^{\circ }\text{C}$ has a fixed value:

It holds in any aqueous solution at that temperature, not just in pure water: if a solution has a lot of ${\text{H}}^{+}$ (it is acidic), it must have little ${\text{OH}}^{-}$, and vice versa, because their product always stays at $1.0\times 10^{-14}$. It is the same mechanism as the $K_c$ you have already seen for chemical equilibrium in general: I won’t explain it again here, $K_w$ is just a special case applied to water.

pH and pOH

pH is defined as the base-10 logarithm, with its sign changed, of the concentration of ${\text{H}}^{+}$ ions:

In the same way, pOH is defined for ${\text{OH}}^{-}$ ions:

Taking the logarithm of both sides of $K_w=[{\text{H}}^{+}][{\text{OH}}^{-}]=1.0\times 10^{-14}$, you get a relation that at $25{}^{\circ }\text{C}$ always holds:

This lets you move from one to the other without recalculating everything from scratch: if you know the pOH of a solution, the pH is simply $14$ minus that value.

Strong acids and bases: the pH can be read directly from the concentration

An acid (or a base) is called strong when it dissociates completely in water: all the starting molecules become ions, none stays undissociated. For a strong monoprotic acid (one that donates a single proton, like $\text{HCl}$), the concentration of ${\text{H}}^{+}$ at equilibrium is exactly the initial concentration of the acid:

The same holds, mirrored, for a strong monoprotic base such as $\text{NaOH}$: $[{\text{OH}}^{-}]=C_{\text{strong base}}$, from which you first find the pOH and then the pH with $\text{pH}=14-\text{pOH}$.

Weak acids: the constant Ka and the approximation

A weak acid (like acetic acid, $\mathrm{C}{\mathrm{H}}_3\mathrm{COOH}$) dissociates only partly: at equilibrium, undissociated molecules and ions live side by side. Its acid dissociation constant $K_a$ is written with the same logic as $K_c$:

Starting from an initial concentration $C$ and calling $x$ the concentration of acid that dissociates, the ICE table gives $[{\text{H}}^{+}]=[{\text{A}}^{-}]=x$ and $[\text{HA}]=C-x$ at equilibrium, so:

When $K_a$ is much smaller than $C$ (the acid is weak and the concentration is not too dilute), $x$ is small compared with $C$, and you can approximate $C-x\approx C$: the quadratic equation simplifies into a direct one,

This approximation is not automatic: it has to be checked afterwards, by calculating the ratio $x/C$. If that ratio is under 5%, the approximation is considered acceptable; if it is above, you have to redo the calculation by solving the full quadratic equation, as in the ICE table for chemical equilibrium.

A preview: hydrolysis and titrations

Two topics that come right after this block, just to put them in context: salt hydrolysis is the fact that the salt of a weak acid (or of a weak base) does not always give a neutral solution when it dissolves in water — the ion that comes from the weak acid or base reacts in turn with water, shifting the pH. Titrations are the experimental method for finding the unknown concentration of an acid (or a base) by adding, drop by drop, a base (or an acid) of known concentration until the equivalence point. I don’t develop them here: they are a separate chapter, which comes later, with its own logic.

The typical mistake: getting lost in the logarithm, not in the chemistry

The most frequent mistake is not conceptual, it is in the numerical step. Three recurring cases:

1. Forgetting the minus sign. $\text{pH}=-\log [{\text{H}}^{+}]$ has a minus sign because $[{\text{H}}^{+}]$ is almost always less than 1, so its logarithm is negative: the minus sign makes it positive. If you calculate $\log [{\text{H}}^{+}]$ and leave it negative, you get a negative pH for an ordinary solution, a result that should set off an alarm bell straight away.

2. Applying $\sqrt{K_a\cdot C}$ without checking it. It is the fastest calculation, but only when you are allowed to use it. If $K_a$ is not much smaller than $C$ (a weak acid but a low concentration, or a relatively large $K_a$), the approximation gives a wrong pH: you must always check the ratio $x/C$ after finding $x$.

3. Confusing $[{\text{H}}^{+}]$ with pH. More ${\text{H}}^{+}$ means a lower pH, not a higher one: with $[{\text{H}}^{+}]=10^{-2}$ the pH is 2, with $[{\text{H}}^{+}]=10^{-5}$ the pH is 5. And every unit of pH lower means ten times more ${\text{H}}^{+}$, not “a bit more”.

Three worked examples

Exercise 1 — HCl, a strong acid.

Calculate the pH of a solution of $\text{HCl}$ with concentration $C=0.01\text{M}$.

$\text{HCl}$ is a strong monoprotic acid: it dissociates completely, so $[{\text{H}}^{+}]=C=0.01\text{M}=10^{-2}\text{M}$.

Exercise 2 — NaOH, a strong base.

Calculate the pOH and pH of a solution of $\text{NaOH}$ with concentration $C=0.005\text{M}$.

$\text{NaOH}$ is a strong monoprotic base: $[{\text{OH}}^{-}]=C=0.005\text{M}=5\times 10^{-3}\text{M}$.

The result makes sense: a base, even a dilute one, gives a pH above 7.

Exercise 3 — acetic acid, a weak acid.

Calculate the pH of a solution of acetic acid $\mathrm{C}{\mathrm{H}}_3\mathrm{COOH}$ with concentration $C=0.10\text{M}$, with $K_a=1.8\times 10^{-5}$.

I use the approximation, to be checked afterwards:

I check that the approximation is acceptable: $x/C=1.34\times 10^{-3}/0.10\approx 0.0134$, that is $1.3\%$, well under the 5% threshold. The approximation is valid, so $[{\text{H}}^{+}]\approx 1.34\times 10^{-3}\text{M}$:

Note the comparison with exercise 1: a weak acid at ten times the concentration ($0.10\text{M}$ against $0.01\text{M}$) still has a higher pH (is less acidic) than a more dilute strong acid, because only a small fraction of its molecules really dissociate.

Before the test

There is not yet a practice test on acids, bases and pH: the practice maths and physics tests are in the high school practice tests category. For this chapter, the revision is to redo the exercises above changing the numbers, plus the ones in your textbook.

For the chapter that comes before this one in the Year 4 programme, the concentration of solutions, you will find the method for mass percentage, molarity and dilutions in this post — it is written for the human sciences high school, but the calculation is identical. For chemical equilibrium and $K_c$, the chapter immediately before this one, there is the dedicated post: I don’t go over it again here, the logic of $K_a$ is the same. And if you need to revise how to write the formula of an acid or a salt correctly before getting to this point, there is this post on chemical nomenclature.

The full picture of the subject is on the sciences page. If you want to revise pH and acids/bases before the December test, start with the free introductory call.