Mixing a strong acid with a strong base produces a fast, essentially complete neutralization. AP Chemistry focuses on using stoichiometry to find which reactant remains in excess and how that leftover sets pH.

Core idea: quantitative neutralization

Strong acid–strong base reactions go to completion because the reacting species are very strong: $H^+$ (or $H_3O^+$) and $OH^-$. As a result, equilibrium reasoning is not needed; treat the reaction as quantitative.

Plots of solution pH versus volume added for strong acid added to water and strong base added to water. Because strong acids/bases dissociate essentially completely, pH changes track the amount of $H^+$ or $OH^-$ introduced (a stoichiometric effect). The curves illustrate the rapid pH change at low added volumes and the approach to limiting pH at large added volumes. Source

This means the final pH depends only on which species is left over after reaction: excess $H_3O^+$ (acidic) or excess $OH^-$ (basic). If neither is in excess, the mixture is neutral at 25 °C.

Titration curve for a strong acid (e.g., HCl) titrated by a strong base (e.g., NaOH), highlighting the steep pH jump near the equivalence point. At the equivalence point, moles of acid and base are equal, so the solution is neutral (pH ≈ 7.00 at 25 °C). Past equivalence, the pH is controlled by excess $OH^-$. Source

Essential chemical equation

Use the net ionic equation to represent what actually changes in solution.

When writing full molecular equations, strong electrolytes are typically written as ions in solution; however, for mixture problems the net ionic form is the most direct.

Workflow for strong acid–strong base mixtures (no equilibrium)

1) Convert each reactant to moles of reacting ions

Track moles, not molarity, because mixing changes volume.

• Determine moles of $H^+$ provided by the strong acid.

• Determine moles of $OH^-$ provided by the strong base.

For typical monoprotic strong acids (e.g., HCl, HNO$_3$), moles acid = moles $H^+$. For hydroxide bases, moles base = moles $OH^-$. (If a base formula contains more than one $OH^-$ per formula unit, account for that stoichiometry when finding moles $OH^-$.)

2) Identify the limiting reagent and excess reagent

Because the stoichiometry is 1:1 between $H^+$ and $OH^-$:

• If $n(H^+) > n(OH^-)$, then acid is in excess and leftover $H^+$ remains.

• If $n(OH^-) > n(H^+)$, then base is in excess and leftover $OH^-$ remains.

• If $n(H^+) = n(OH^-)$, neither remains in excess.

3) Compute the concentration of the excess ion after mixing

First find leftover moles:

• $n_{\text{excess}} = |n(H^+) - n(OH^-)|$

Then convert to concentration using the total volume after mixing (sum of solution volumes, assuming additivity):

• $[H_3O^+] \approx \dfrac{n_{\text{excess }H^+}}{V_{\text{total}}}$ if acid is in excess

• $[OH^-] \approx \dfrac{n_{\text{excess }OH^-}}{V_{\text{total}}}$ if base is in excess

4) Translate the excess concentration to pH

• If acid is in excess, pH is controlled by $[H_3O^+]$.

• If base is in excess, pH is controlled by $[OH^-]$ (then relate to pH using pH and pOH relationships you have learned).

• If neither is in excess, the mixture is neutral (at 25 °C), so pH is 7.00.

Common AP Chemistry checkpoints

• Use stoichiometry first, then pH; do not start with pH formulas before determining what remains after reaction.

• Do not confuse concentration before mixing with concentration after mixing; dilution by added volume is often the dominant effect.

• The spectator ions (e.g., $Na^+$, $Cl^-$) do not affect the neutralization stoichiometry for this topic.