E. Chemical Systems and Equilibrium · Acid-Base and Solubility EquilibriaTitration Curves I: Strong Acid and Strong Base
E2.5 E3.5 A1.6 A1.8About 70 minutes
Learning goals- I can describe how a pH titration curve is produced and identify its regions and the equivalence point.
- I can calculate the pH at any point in a strong acid-strong base titration.
- I can use the equivalence-point volume to calculate an unknown concentration, including for polyprotic acids.
- I can explain how the concentrations of acid and base affect the size of the pH jump.
Words to know
titration curve
A graph of pH (vertical axis) against the volume of titrant added (horizontal axis).
titrant
The solution of known concentration added from the burette.
analyte
The solution of unknown concentration in the flask.
equivalence point
The point where the moles of titrant added exactly react with the moles of analyte, according to the balanced equation.
endpoint
The point where an indicator changes colour. A good indicator makes the endpoint very close to the equivalence point.
Watching pH during a titration
In Grade 11 you titrated vinegar with sodium hydroxide and used an indicator to tell you when to stop. Now imagine a pH meter sitting in the flask the whole time, with a reading taken after every addition of base. Plot pH against volume and you get a titration curve. It has a surprising shape: almost flat for a long time, then a near-vertical cliff, then flat again. Reading that shape lets chemists find the equivalence point precisely, identify the type of acid, and even measure its Ka. Food, pharmaceutical and environmental labs collect curves like this with automatic titrators every day.
Remember from Grade 11
A titration adds a solution of known concentration (the titrant) from a burette to a measured volume of the analyte. At the equivalence point, moles of base = moles of acid × (mole ratio). You calculated concentrations with n = cV. The endpoint is where the indicator changes colour.
The four regions of a strong acid-strong base curve
Consider 25.00 mL of 0.100 mol/L HCl titrated with 0.100 mol/L NaOH. The net reaction is H₃O⁺ + OH⁻ → 2 H₂O, which goes to completion. To find the pH at any point, do stoichiometry first (which reactant is left over?) and then use the total volume:
- A. Before any base (0 mL): only HCl. [H₃O⁺] = 0.100 mol/L, pH = 1.00.
- B. Before equivalence: some H₃O⁺ is left over. [H₃O⁺] = (moles of acid − moles of base added) ÷ total volume. The pH rises slowly, because a lot of H₃O⁺ remains until the very end.
- C. At equivalence (25.00 mL): the flask contains only NaCl and water. Na⁺ and Cl⁻ are spectators, so pH = 7.00.
- D. After equivalence: extra OH⁻. [OH⁻] = (moles of base − moles of acid) ÷ total volume, then pH = 14.00 − pOH.
Near the equivalence point, a single drop makes a huge difference: at 24.95 mL the pH is 4.00, and at 25.05 mL it is 10.00.
| V(NaOH) added (mL) | Excess | pH |
|---|
| 0.00 | H₃O⁺ | 1.00 |
| 10.00 | H₃O⁺ | 1.37 |
| 20.00 | H₃O⁺ | 1.95 |
| 24.00 | H₃O⁺ | 2.69 |
| 24.90 | H₃O⁺ | 3.70 |
| 25.00 | none | 7.00 |
| 25.10 | OH⁻ | 10.30 |
| 26.00 | OH⁻ | 11.29 |
| 30.00 | OH⁻ | 11.96 |
| 40.00 | OH⁻ | 12.36 |
| 50.00 | OH⁻ | 12.52 |
Worked exampleExample 1: pH before the equivalence point
25.00 mL of 0.100 mol/L HCl is titrated with 0.100 mol/L NaOH. Find the pH after 10.00 mL of NaOH has been added.
- n(H₃O⁺) initially = 0.100 × 0.02500 = 2.500 × 10⁻³ mol.
- n(OH⁻) added = 0.100 × 0.01000 = 1.000 × 10⁻³ mol.
- H₃O⁺ left = 2.500 × 10⁻³ − 1.000 × 10⁻³ = 1.500 × 10⁻³ mol.
- Total volume = 25.00 + 10.00 = 35.00 mL = 0.03500 L. [H₃O⁺] = 1.500 × 10⁻³ ÷ 0.03500 = 0.04286 mol/L.
- pH = −log(0.04286) = 1.37.
Worked exampleExample 2: pH after the equivalence point
Same titration: find the pH after 30.00 mL of NaOH.
- n(OH⁻) added = 0.100 × 0.03000 = 3.000 × 10⁻³ mol; n(H₃O⁺) = 2.500 × 10⁻³ mol.
- OH⁻ left = 0.500 × 10⁻³ mol, in 55.00 mL = 0.05500 L.
- [OH⁻] = 5.00 × 10⁻⁴ ÷ 0.05500 = 9.09 × 10⁻³ mol/L. pOH = 2.04.
- pH = 14.00 − 2.04 = 11.96.
Worked exampleExample 3: Finding an unknown concentration
A pH curve shows that 20.00 mL of HCl(aq) reaches its equivalence point after 17.60 mL of 0.1250 mol/L NaOH. Find [HCl].
- n(NaOH) = 0.1250 × 0.01760 = 2.200 × 10⁻³ mol.
- HCl + NaOH → NaCl + H₂O is 1:1, so n(HCl) = 2.200 × 10⁻³ mol.
- c(HCl) = 2.200 × 10⁻³ ÷ 0.02000 = 0.1100 mol/L.
Worked exampleExample 4: A diprotic acid
10.00 mL of sulfuric acid needs 22.40 mL of 0.1000 mol/L NaOH to reach the final equivalence point. Find [H₂SO₄].
- Balanced equation: H₂SO₄ + 2 NaOH → Na₂SO₄ + 2 H₂O. The ratio is 1 : 2.
- n(NaOH) = 0.1000 × 0.02240 = 2.240 × 10⁻³ mol, so n(H₂SO₄) = 2.240 × 10⁻³ ÷ 2 = 1.120 × 10⁻³ mol.
- c = 1.120 × 10⁻³ ÷ 0.01000 = 0.1120 mol/L.
[H₂SO₄] = 0.1120 mol/L. Forgetting the 1 : 2 ratio would double the answer.
The reverse titration, and the effect of concentration
If you titrate a strong base with a strong acid, the curve is the mirror image: it starts high (pH 13.00 for 0.100 mol/L NaOH), stays high, plunges through pH 7.00 at equivalence, and levels off low.
The size of the jump depends on concentration. With 0.100 mol/L solutions, the pH goes from 3.70 at 0.10 mL before equivalence to 10.30 at 0.10 mL after: a jump of about 6.6 units. With 0.0100 mol/L solutions, the same two points are at pH 4.70 and 9.30: a smaller jump. Very dilute solutions give a jump that is hard to detect, which is one reason labs use concentrations around 0.1 mol/L.
Just for funNear the equivalence point of this titration, adding just two drops of base (about 0.10 mL) takes the pH from 4.00 to 10.00. That is a millionfold change in [H₃O⁺] from two drops.
Hands-on activityPlot a titration curve by hand
You need: graph paper (or a printed grid), a ruler, a pencil, a calculator, and the data table in this lesson.
- Draw axes: volume of NaOH from 0 to 50 mL across (2 squares per 5 mL works well), pH from 0 to 14 up.
- Plot every point from the table and join them with a smooth curve. Leave room near 25 mL; the curve is almost vertical there.
- Calculate three extra points yourself: 5.00 mL, 15.00 mL and 35.00 mL. Plot them and check that they fall on your curve. (Answers: 1.18, 1.60, 12.22.)
- On the same axes, sketch the curve for 0.0100 mol/L HCl titrated with 0.0100 mol/L NaOH. It starts at pH 2.00 and ends lower than 12.52, and its equivalence point is still at 25.00 mL and pH 7.00. Calculate pH at 24.90 mL and 25.10 mL for the dilute case.
- Mark the equivalence point on both curves and describe how they differ.
What to look for: both curves are flat-steep-flat with the steep part centred on 25.00 mL. The dilute curve starts higher, ends lower, and its vertical jump is shorter (about pH 4.7 to 9.3 over the same 0.20 mL). This matters when you choose an indicator in the next two lessons.
Your turn
1Same titration (25.00 mL of 0.100 mol/L HCl with 0.100 mol/L NaOH): find the pH after 20.00 mL of NaOH.
H₃O⁺ left = 2.500 × 10⁻³ − 2.000 × 10⁻³ = 5.00 × 10⁻⁴ mol in 0.04500 L = 0.0111 mol/L. pH = 1.95.
2Find the pH after 40.00 mL of NaOH.
OH⁻ left = 4.000 × 10⁻³ − 2.500 × 10⁻³ = 1.500 × 10⁻³ mol in 0.06500 L = 0.0231 mol/L. pOH = 1.64, pH = 12.36.
3Why is the pH exactly 7.00 at the equivalence point of a strong acid-strong base titration (at 25 °C)?
The flask contains only water and the salt NaCl. Na⁺ and Cl⁻ are spectator ions that do not react with water, so [H₃O⁺] = [OH⁻] = 1.0 × 10⁻⁷ mol/L.
425.00 mL of KOH(aq) needs 31.25 mL of 0.0800 mol/L HNO₃ to reach equivalence. Find [KOH].
n(HNO₃) = 0.0800 × 0.03125 = 2.500 × 10⁻³ mol = n(KOH). c = 2.500 × 10⁻³ ÷ 0.02500 = 0.100 mol/L.
5In a titration of 20.00 mL of 0.150 mol/L NaOH with 0.100 mol/L HCl, what volume of HCl is needed to reach equivalence?
n(NaOH) = 3.000 × 10⁻³ mol; V(HCl) = 3.000 × 10⁻³ ÷ 0.100 = 30.0 mL.
6Sketch (in words) the curve for a strong base titrated with a strong acid.
Starts at high pH (about 13), decreases slowly, then drops steeply through pH 7 at the equivalence point, then levels off at low pH. It is the mirror image of the acid-with-base curve.
More practice: warm-up
1What is plotted on each axis of a titration curve?
pH on the vertical axis and volume of titrant added on the horizontal axis.
2What is the pH at the equivalence point of HCl titrated with NaOH at 25 °C?
3In region B (before equivalence) of an acid-with-base titration, which ion is in excess?
4What is the starting pH of 0.0100 mol/L HCl?
5What mole ratio of NaOH to H₂SO₄ is needed for complete neutralization?
More practice: on your own
150.00 mL of 0.0500 mol/L HNO₃ is titrated with 0.100 mol/L NaOH. Find the equivalence volume and the pH after 12.50 mL of NaOH.
n(acid) = 2.500 × 10⁻³ mol, so V(eq) = 25.00 mL. At 12.50 mL: H₃O⁺ left = 2.500 × 10⁻³ − 1.250 × 10⁻³ = 1.250 × 10⁻³ mol in 0.06250 L = 0.0200 mol/L, pH = 1.70.
2For the titration in the previous question, find the pH after 30.00 mL of NaOH.
OH⁻ left = 3.000 × 10⁻³ − 2.500 × 10⁻³ = 5.00 × 10⁻⁴ mol in 0.08000 L = 6.25 × 10⁻³ mol/L. pOH = 2.20, pH = 11.80.
3A 0.500 g sample of solid NaOH contaminated with an unreactive solid is dissolved in water and needs 24.50 mL of 0.500 mol/L HCl to reach equivalence. Find the percentage of NaOH in the sample.
n(HCl) = 0.500 × 0.02450 = 0.01225 mol = n(NaOH). m = 0.01225 × 40.00 = 0.490 g. Percentage = 0.490 ÷ 0.500 × 100% = 98.0%.
4Explain why the pH changes so little between 0 and 20 mL in the titration in this lesson, but so much between 24.9 and 25.1 mL.
pH is a logarithmic scale. From 0 to 20 mL, [H₃O⁺] only falls from 0.100 to 0.0111 mol/L, less than a factor of 10, so pH changes by less than 1. Near equivalence almost all the H₃O⁺ is gone, so each drop changes [H₃O⁺] (or [OH⁻]) by large factors, and the pH jumps.
5A student's curve for 25.00 mL of HCl shows the equivalence point at 26.80 mL of 0.0950 mol/L NaOH. Find [HCl] and its percentage difference from the labelled value of 0.100 mol/L.
n(NaOH) = 0.0950 × 0.02680 = 2.546 × 10⁻³ mol; c(HCl) = 2.546 × 10⁻³ ÷ 0.02500 = 0.102 mol/L. Difference = (0.102 − 0.100) ÷ 0.100 × 100% ≈ 2% (1.8% unrounded).
Challenge
120.00 mL of 0.100 mol/L Ba(OH)₂ is titrated with 0.100 mol/L HCl. Find the equivalence volume and the pH after 30.00 mL of HCl.
n(OH⁻) = 2 × 0.100 × 0.02000 = 4.000 × 10⁻³ mol, so V(eq) = 4.000 × 10⁻³ ÷ 0.100 = 40.00 mL. At 30.00 mL: OH⁻ left = 4.000 × 10⁻³ − 3.000 × 10⁻³ = 1.000 × 10⁻³ mol in 0.05000 L = 0.0200 mol/L; pOH = 1.70, pH = 12.30.
2Show that at 24.95 mL in this lesson's titration the pH is 4.00, and at 25.05 mL it is 10.00.
At 24.95 mL: H₃O⁺ left = 2.500 × 10⁻³ − 2.495 × 10⁻³ = 5.0 × 10⁻⁶ mol in 0.04995 L = 1.0 × 10⁻⁴ mol/L, pH 4.00. At 25.05 mL: OH⁻ left = 5.0 × 10⁻⁶ mol in 0.05005 L = 1.0 × 10⁻⁴ mol/L, pOH 4.00, pH 10.00.
3A student mixes 15.0 mL of 0.200 mol/L HCl and 10.0 mL of 0.100 mol/L H₂SO₄ (assume both protons react). What volume of 0.250 mol/L NaOH will neutralize the mixture?
n(H₃O⁺) from HCl = 3.00 × 10⁻³ mol; from H₂SO₄ = 2 × 0.100 × 0.0100 = 2.00 × 10⁻³ mol. Total 5.00 × 10⁻³ mol. V(NaOH) = 5.00 × 10⁻³ ÷ 0.250 = 20.0 mL.
Quiz
1. In the titration of HCl with NaOH, what is in the flask at the equivalence point?
2. 25.00 mL of 0.100 mol/L HCl is mixed with 15.00 mL of 0.100 mol/L NaOH. Which calculation gives [H₃O⁺]?
3. Where is the equivalence point on a titration curve?
4. 10.00 mL of H₂SO₄ needs 25.00 mL of 0.200 mol/L NaOH. What is [H₂SO₄]?
5. How does using more dilute acid and base change a strong acid-strong base titration curve?
6. What is the pH of the flask after 30.00 mL of 0.100 mol/L NaOH has been added to 25.00 mL of 0.100 mol/L HCl?
Big idea (E2.5, E3.5): a titration curve plots pH against the volume of titrant. For a strong acid with a strong base, the pH rises slowly, jumps steeply at the equivalence point (pH 7.00), then levels off. The pH at any point comes from stoichiometry first (what is left over?) and then the total volume.
How to help: Ask your teen to point to each region of the curve and say what is in the flask there. When checking a calculation, look for three things: subtraction of moles, division by the total volume (in litres), and converting pOH to pH after equivalence. Graph paper and a ruler are all the activity needs.
Common difficulties:
- Dividing by only the starting volume instead of the total volume.
- Forgetting the mole ratio for diprotic acids (H₂SO₄) or Group 2 hydroxides (Ba(OH)₂).
- Reporting pOH as pH after the equivalence point.
- Believing every equivalence point is at pH 7. That is only true for strong acid-strong base titrations; the next lesson shows why.