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What happens when an acid reacts with a base, and how can we use that reaction to find an unknown concentration?

Write neutralization reactions producing a salt and water, and use titration data with solution stoichiometry to find an unknown concentration (MA STE supporting content, neutralization and titration).

A standard-level answer on neutralization and titration for Massachusetts high school chemistry: the acid-plus-base reaction that forms a salt and water, the titration procedure and endpoint, and using titration data with solution stoichiometry to find an unknown concentration, grounded in the framework's acid-base content.

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  1. What this topic is asking
  2. Neutralization
  3. What a titration is
  4. The titration calculation
  5. Try this

What this topic is asking

When an acid meets a base they react in a neutralization, the most important acid-base reaction. A Massachusetts high school chemistry course expects you to write a neutralization equation, name its products, and use a titration (a controlled neutralization) with solution stoichiometry to find an unknown concentration. This is where the molarity skills of this module are put to work.

Neutralization

A general neutralization is acid + base gives salt + water. For example, HCl+NaOH→NaCl+H2O\text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O}, where sodium chloride is the salt (an ionic compound made of the metal from the base and the non-hydrogen part of the acid). As base is added to an acid, the pH rises from acidic toward 7, reaching neutral when the moles of acid and base exactly match. Neutralization is everywhere: antacids neutralize excess stomach acid, and lime neutralizes acidic soil.

What a titration is

In a typical setup, a known volume of acid is placed in a flask with a few drops of indicator, and base of known concentration is run in from a burette. The endpoint is the point at which the indicator changes color, signalling that the acid has just been neutralized. The volume of base added is read from the burette. Repeating the titration to get consistent volumes improves accuracy.

The titration calculation

A titration calculation is solution stoichiometry, using the mole bridge from molarity and solution stoichiometry:

  1. Find the moles of the solution you know fully (molarity times volume in liters).
  2. Use the mole ratio from the balanced equation to find the moles of the unknown.
  3. Divide those moles by the unknown's volume (in liters) to get its molarity.

Try this

Q1. Write the salt formed when sulfuric acid neutralizes potassium hydroxide. [1]

  • Cue. Potassium sulfate, K2SO4\text{K}_2\text{SO}_4 (plus water).

Q2. In a 1 to 1 titration, 10.0 mL of 0.50 M acid neutralizes a base. How many moles of base reacted? [1]

  • Cue. Moles of acid =0.50×0.0100=0.0050= 0.50 \times 0.0100 = 0.0050 mol, so 0.0050 mol of base (1 to 1 ratio).

Exam-style practice questions

Practice questions written in the style of MA DESE exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

MA Chemistry (style)3 marksHydrochloric acid reacts with sodium hydroxide. (a) Write the balanced equation. (b) Name the type of reaction and its products. (c) State what happens to the pH as base is added to the acid.
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A 3-point neutralization item.

(a) 1 point: HCl+NaOH→NaCl+H2O\text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O}.
(b) 1 point: a neutralization reaction; the products are a salt (sodium chloride) and water.
(c) 1 point: as base is added the pH rises from acidic toward 7, reaching neutral when the acid is exactly used up. Markers reward the salt-plus-water products and the rising pH toward neutral.

MA Chemistry (style)3 marksIn a titration, 25.0 mL of hydrochloric acid is neutralized by 20.0 mL of 0.100 M sodium hydroxide. The reaction is 1 to 1. Find the concentration of the acid.
Show worked answer →

A 3-point titration calculation.

1 point: moles of NaOH =M×V=0.100×0.0200=0.00200= M \times V = 0.100 \times 0.0200 = 0.00200 mol.
1 point: the 1 to 1 ratio means moles of HCl =0.00200= 0.00200 mol.
1 point: concentration of acid =nV=0.002000.0250=0.0800= \dfrac{n}{V} = \dfrac{0.00200}{0.0250} = 0.0800 M. Markers reward the moles-ratio-concentration path with volumes in liters.

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