A bottle in the lab says “sodium hydroxide, about 0.1 M”. About is not good enough for an analysis, and a car battery’s sulfuric acid or a factory’s waste water needs a number too. Chemists find such concentrations by titration: a solution of known molarity is added from a buret, drop by drop, until an indicator changes color, and the volume used tells you how much acid or base was in the sample.

In this lesson you will turn titration data into an unknown concentration, using the mole ratio of the balanced equation at the equivalence point (Figure 1).

1. Volume of titrantmL, then L
2. $\times C$molarity of the titrant
3. Moles of titrant
4. $\times \dfrac{c_\mathrm{s}}{c_\mathrm{t}}$mole ratio at the equivalence point
5. Moles of sample
6. $\dfrac{n}{V}$divide by the sample volume in L
7. Molarity of sampleM

$n = C \times V$ with V in L.$C = \dfrac{n}{V}$ for the sample.Volume of titrant: $V = \dfrac{n}{C}$.

Learning Objectives

After this lesson, you should be able to:

1. Use titration data to find an unknown concentration.
2. Apply the mole ratio at the equivalence point.

Key Ideas

1. What a Titration Measures

In a titration, a measured volume of one solution, the sample, sits in a flask with a few drops of an indicator. A second solution, the titrant, is a standard solution: its concentration is accurately known. The titrant is added from a buret, which reads the volume delivered.

The equivalence point is reached when the acid and the base have reacted in exactly the mole ratio of the balanced equation. The end point is where the indicator changes color; a good indicator changes color at, or very close to, the equivalence point. Phenolphthalein is colorless in acidic solution and pink in basic solution, so when a base is the titrant, the first faint pink that stays marks the end point (Figure 2).

Key idea: a titration gives you two measured volumes and one known molarity. At the equivalence point, the moles of acid and base are related by the coefficients of the balanced equation.

2. Amount of Titrant From Molarity and Volume

In Molarity you met $C = \frac{n}{V}$, the moles of solute in each L of solution. Rearranged, it gives the moles of titrant delivered by the buret:

The buret reads in mL, so convert first: $1\,\text{L} = 1000\,\text{mL}$. Of the three numbers you know, the molarity of the titrant and the volume of the titrant belong together: they give the moles of titrant. The volume of the sample is used at the very end.

Key idea: moles of titrant $= C \times V$, using the molarity and the volume of the titrant, with $V$ in L.

3. The Mole Ratio at the Equivalence Point

At the equivalence point, the moles of the sample follow from the moles of titrant through the mole ratio of the balanced equation, exactly as in the lesson Mole Ratios and Mole-to-Mole Conversions: wanted over given. Hydrochloric acid and sodium hydroxide react 1 : 1, but sulfuric acid needs 2 mol of NaOH for each mole of acid:

So at the equivalence point of Worked Example 2 there are half as many moles of sulfuric acid as moles of sodium hydroxide added (Figure 3).

Key idea: at the equivalence point, multiply the moles of titrant by the mole ratio from the balanced equation (wanted over given) to get the moles of the sample.

Try it: titrate to the equivalence point

Pick a reaction, set the sample and the titrant, predict the buret reading, then add titrant until the flask changes color.

4. From Moles to Molarity

The moles of the sample were all in the volume of sample you measured into the flask. Divide by that volume, in L, to get its molarity:

If the question asks instead for the volume of titrant needed, run the same chain from the other side: moles of the sample from its molarity and volume, the mole ratio, then divide the moles of titrant by its molarity (Worked Example 3).

Key idea: molarity of the sample $= \frac{n}{V}$ with the moles of the sample and the volume of the sample in L. Moles alone are not a concentration.

5. A Strategy for Titration Problems

Follow the roadmap in Figure 1:

1. Write the balanced equation for the acid and the base.
2. Convert the volumes to liters and find the moles of the solution whose molarity and volume you both know: $n = C \times V$.
3. Use the mole ratio at the equivalence point to get the moles of the other substance.
4. Divide: by the volume of the sample in L for a molarity, or by the molarity of the titrant for a volume (then convert L to mL). Keep 4 significant figures until the last step and round the answer to 3.

Exam tip: the shortcut $C_\mathrm{a}V_\mathrm{a} = C_\mathrm{b}V_\mathrm{b}$ works only when the acid and the base react 1 : 1. With sulfuric acid it gives the wrong answer (Common Mistakes). Going through moles and the mole ratio always works.

Lab note: the titrant is itself standardized first: its concentration is found by titrating a weighed amount of a pure solid called a primary standard. Read the buret at eye level, at the bottom of the curved surface.

Worked Example 1: The Molarity of Hydrochloric Acid

Problem: A 30.0 mL sample of hydrochloric acid is titrated with 0.124 M sodium hydroxide, with phenolphthalein as the indicator. The solution turns faint pink after 27.8 mL of NaOH has been added. What is the molarity of the hydrochloric acid?

Step 1. Convert both volumes to liters.

Step 2. Find the moles of sodium hydroxide added: $n = C \times V$.

Step 3. Use the mole ratio at the equivalence point. 1 mol of NaOH reacts with 1 mol of HCl.

Step 4. Divide the moles of HCl by the volume of the sample in L.

Final Answer: 0.115 M HCl

Check: less NaOH than sample was needed, and the reaction is 1 : 1, so the acid must be a little less concentrated than the base. It is.

1. Titrant volume in L$V = 0.0278\,\text{L}$Step 1
2. $\times 0.124\,\text{mol/L}$Step 2: molarity of NaOH
3. Moles of NaOH$n = 0.003447\,\text{mol}$
4. $\times \dfrac{1}{1}$Step 3: 1 HCl per 1 NaOH
5. Moles of HCl$n = 0.003447\,\text{mol}$
6. $\dfrac{0.003447\,\text{mol}}{0.0300\,\text{L}}$Step 4: divide by the sample volume
7. Molarity of HCl$C = 0.1149\,\text{mol/L} \approx 0.115\,\text{M}$

Your turn: a 1 : 1 titration

Same four steps as Worked Example 1: liters, moles of titrant from $n = C \times V$, the mole ratio, then divide by the volume of the sample.

Worked Example 2: Sulfuric Acid, a 1 : 2 Ratio

Problem: A 35.0 mL sample of sulfuric acid is titrated with 0.142 M sodium hydroxide. The end point is reached after 31.2 mL of the base is added. What is the molarity of the sulfuric acid?

Step 1. Convert both volumes to liters.

Step 2. Find the moles of sodium hydroxide added.

Step 3. Use the mole ratio. 2 mol of NaOH react with 1 mol of $\mathrm{H_2\!SO_4}$, so the moles are halved.

Step 4. Divide by the volume of the sample in L.

Final Answer: 0.0633 M H₂SO₄

The coefficient 2 in front of NaOH is the whole difference from Worked Example 1: skip it and the answer doubles.

1. Titrant volume in L$V = 0.0312\,\text{L}$Step 1
2. $\times 0.142\,\text{mol/L}$Step 2: molarity of NaOH
3. Moles of NaOH$n = 0.004430\,\text{mol}$
4. $\times \dfrac{1}{2}$Step 3: 1 H₂SO₄ per 2 NaOH
5. Moles of H₂SO₄$n = 0.002215\,\text{mol}$half the moles of NaOH
6. $\dfrac{0.002215\,\text{mol}}{0.0350\,\text{L}}$Step 4: divide by the sample volume
7. Molarity of H₂SO₄$C = 0.06329\,\text{mol/L} \approx 0.0633\,\text{M}$

Your turn: a ratio that is not 1 : 1

Same four steps as Worked Example 2: watch the coefficients in the mole ratio. The titrant can be the acid or the base.

Worked Example 3: The Volume of Titrant Needed

Problem: A 36.0 mL sample of 0.215 M potassium hydroxide is to be titrated with 0.164 M sulfuric acid. What volume of the acid, in mL, is needed to reach the equivalence point?

Step 1. Convert the sample volume to liters.

Step 2. Find the moles of potassium hydroxide in the sample.

Step 3. Use the mole ratio. 1 mol of $\mathrm{H_2\!SO_4}$ reacts with 2 mol of KOH.

Step 4. Find the volume of sulfuric acid: divide the moles by its molarity.

Step 5. Convert liters to milliliters.

Final Answer: 23.6 mL of H₂SO₄ solution

Without the mole ratio you would plan for 47.2 mL, twice the volume actually needed. The answer also fits in a single filling of a standard buret, as a well-planned titration should.

1. Sample volume in L$V = 0.0360\,\text{L}$Step 1
2. $\times 0.215\,\text{mol/L}$Step 2: molarity of KOH
3. Moles of KOH$n = 0.007740\,\text{mol}$
4. $\times \dfrac{1}{2}$Step 3: 1 H₂SO₄ per 2 KOH
5. Moles of H₂SO₄$n = 0.003870\,\text{mol}$
6. $\dfrac{0.003870\,\text{mol}}{0.164\,\text{mol/L}}$Step 4: divide by the molarity of H₂SO₄
7. Volume of H₂SO₄ in L$V = 0.02360\,\text{L}$
8. $\times \dfrac{1000\,\text{mL}}{1\,\text{L}}$Step 5: L to mL
9. Volume of H₂SO₄$V = 23.60\,\text{mL} \approx 23.6\,\text{mL}$

Your turn: volume to the equivalence point

Same five steps as Worked Example 3: moles of the sample, the mole ratio, divide by the molarity of the titrant, then L to mL.

Common Mistakes

Throughout, keep 4 significant figures in the moles and the unrounded molarity or volume, and round only the final answer to 3.

Common mistake: Leaving the volume in mL. In Worked Example 1, dividing 0.003447 mol by 30.0 (the sample volume in mL) gives 0.0001149 M, 1000 times too small. Molarity is in mol/L, so convert mL to L first.

Common mistake: Swapping the two volumes. In Worked Example 1, multiplying the molarity of NaOH by the volume of the sample and dividing by the volume of the titrant gives 0.134 M instead of 0.115 M. The titrant’s molarity goes with the titrant’s own volume.

Common mistake: Leaving out the coefficients. In Worked Example 2, using $C_\mathrm{a}V_\mathrm{a} = C_\mathrm{b}V_\mathrm{b}$, or taking the moles of $\mathrm{H_2\!SO_4}$ equal to the moles of NaOH, gives 0.127 M, twice the right answer of 0.0633 M.

Common mistake: Turning the mole ratio upside down. In Worked Example 2, multiplying by 2 instead of dividing gives 0.253 M, four times the right answer. Put the substance you want on top: $\frac{1\,\text{mol}\;\mathrm{H_2\!SO_4}}{2\,\text{mol}\;\mathrm{NaOH}}$.

Common mistake: Stopping at moles. In Worked Example 1, 0.003447 mol is the amount of HCl in the flask, not its concentration. Divide by the volume of the sample in L to get mol/L.

Key Takeaway

In an acid-base titration, find the moles of titrant from $n = C \times V$ (with $V$ in L), use the mole ratio of the balanced equation at the equivalence point to get the moles of the sample, and divide by the volume of the sample in L to get its molarity. To plan a titration, run the chain the other way and divide the moles of titrant by its molarity.

Practice: randomized problems

Each problem below uses fresh numbers every time and mixes every kind of problem in this lesson: work it out, check your answer, then open the worked solution to compare your steps.

Your results

This summary collects every practice problem you try on this page and suggests what to review next.

Figures: original diagrams by Dr. Ray’s Learning Center, licensed CC BY 4.0. Data: Whitten, Davis, Peck and Stanley, Chemistry, 9th ed. (Brooks/Cole, Cengage Learning, 2010), Section 11-1 (p. 375) for the reaction ratio of an acid and a base; Section 11-2 (pp. 379 to 380) for titrations, indicators and the end point; Section 11-3 (pp. 381 to 384) for acid-base titration calculations. Simulator: original work by Dr. Ray’s Learning Center, licensed CC BY 4.0; buret, drop size and phenolphthalein colors from Whitten 9th ed. (pp. 85, 379, 383, 757 to 759). The practice problems are generated and checked by an audited engine; see the lesson audit log.