pH — Added 0.00 mL Indicator —

Titration curve

Results notebook

RunInitial / mLFinal / mLTitre / mL
No runs yet. Titrate to the colour change, then record the end point.

Use the mean of your concordant titres (within 0.10 mL of each other).

Neutralisation and pH

In an acid–base titration a solution of known concentration (the titrant, in the burette) is added to a measured volume of the solution being analysed (the analyte, pipetted into the conical flask) until the reaction is exactly complete. For a strong acid and a strong base the net reaction is simply

$$\mathrm{H_3O^+(aq) + OH^-(aq) \longrightarrow 2\,H_2O(l)}\qquad \Delta H \approx -57\ \mathrm{kJ\,mol^{-1}}$$

Acidity is reported on the logarithmic pH scale. Water always self-ionises, which links the hydronium and hydroxide concentrations:

$$\begin{gathered}\mathrm{pH} = -\log_{10}[\mathrm{H_3O^+}]\\ K_w=[\mathrm{H_3O^+}][\mathrm{OH^-}] = 1.0\times10^{-14}\ \text{at }25\,{}^\circ\mathrm{C}\\ \mathrm{pH+pOH}=14.00\end{gathered}$$

Weak acids and bases: $K_a$ and $K_b$

A weak acid is only partly ionised. Its strength is measured by the acid dissociation constant $K_a$ (bigger $K_a$, smaller p$K_a$ = stronger acid); for a weak base we use $K_b$. When HA loses H⁺ it becomes A⁻, its conjugate base; HA and A⁻ are a conjugate pair, and their constants are linked through $K_w$:

$$\begin{gathered}\mathrm{HA + H_2O \rightleftharpoons H_3O^+ + A^-}\\ K_a=\frac{[\mathrm{H_3O^+}][\mathrm{A^-}]}{[\mathrm{HA}]}\\ \mathrm{p}K_a=-\log K_a,\qquad K_a\,K_b = K_w\end{gathered}$$

Ethanoic (acetic) acid has $K_a = 1.8\times10^{-5}$ ($\mathrm{p}K_a = 4.74$); ammonia has $K_b = 1.8\times10^{-5}$, so its conjugate acid NH₄⁺ has $\mathrm{p}K_a = 9.26$.

Equivalence point vs end point

  • The equivalence point is where the amount of titrant added is exactly the stoichiometric amount (the amount the balanced equation requires): $n_\text{titrant} = \nu\,n_\text{analyte}$, where $\nu$ is the mole ratio from the equation (1 for HCl + NaOH, 2 for H₂C₂O₄ + 2 NaOH). It is a property of the chemistry.
  • The end point is where you observe the indicator change colour (or the pH meter shows the steepest rise). A good indicator makes the end point fall within one drop of the equivalence point.

At equivalence the pH is not always 7. The salt formed can hydrolyse (react with water, releasing OH⁻ or H₃O⁺): CH₃COO⁻ is a weak base (pH > 7 at equivalence); NH₄⁺ is a weak acid (pH < 7).

Good practice. Rinse the burette with the titrant and the pipette with the analyte; leftover water would dilute them. Rinse the flask with distilled water only, because that does not change the moles in it. Read the bottom of the meniscus at eye level, to ±0.05 mL. A titre is the difference of two readings, so its uncertainty is ±0.10 mL, which is ±0.4 % of a 25 mL titre.

Buffers and the half-equivalence point

Between the start and the equivalence point of a weak acid titration, the flask contains both HA and A⁻: a buffer. The Henderson–Hasselbalch equation is an excellent approximation here:

$$\begin{gathered}\mathrm{pH} = \mathrm{p}K_a + \log_{10}\frac{[\mathrm{A^-}]}{[\mathrm{HA}]}\\ \text{at half-equivalence }[\mathrm{A^-}]=[\mathrm{HA}]\ \text{so}\ \mathrm{pH}=\mathrm{p}K_a\end{gathered}$$

That is why the curve is flattest at half-equivalence: buffer capacity (the ability to absorb added OH⁻ with little pH change) is maximum there. It also gives a practical way to measure $\mathrm{p}K_a$ from a titration curve.

Why the sudden jump? pH is a logarithm, so each pH unit is a tenfold change in [H₃O⁺]. Just before equivalence almost all the acid is used up; one more drop of NaOH removes most of what is left, and the next drop leaves excess OH⁻. [H₃O⁺] therefore falls by many powers of ten within a few drops, which is the steep vertical section an indicator detects.

How this simulator calculates pH optional, advanced

Rather than stitching together approximations for each region, the lab solves the full equilibrium problem at every drop. For a mixture containing Na⁺, Cl⁻, a weak acid HnA (total $C_A$) and a weak base B (total $C_B$), electroneutrality requires

$$\begin{gathered}[\mathrm{H_3O^+}] + [\mathrm{Na^+}] + C_B\,\frac{[\mathrm{H_3O^+}]}{[\mathrm{H_3O^+}]+K_{a,\mathrm{BH^+}}}\\ = [\mathrm{OH^-}] + [\mathrm{Cl^-}] + C_A\sum_{j=1}^{n} j\,\alpha_j\end{gathered}$$
$$\alpha_j=\frac{[\mathrm{H^+}]^{n-j}\prod_{i\le j}K_{a,i}}{\sum_{k=0}^{n}[\mathrm{H^+}]^{n-k}\prod_{i\le k}K_{a,i}},\qquad [\mathrm{OH^-}]=\frac{K_w}{[\mathrm{H_3O^+}]}$$

Every term is monotonic in pH, so the equation has exactly one root, found numerically by bisection to better than 10⁻⁶ pH units. Dilution by the added titrant is included. The model uses concentrations rather than activities and assumes 25 °C, which is the usual first-year treatment; real pH readings in concentrated solutions can differ by a few hundredths.

Polyprotic acids. H₃PO₄ has three p$K_a$ values (2.12, 7.21, 12.32). The first two equivalence points are clear, but the third is not, because HPO₄²⁻ is too weak an acid relative to water at 0.1 M. Oxalic acid (p$K_a$ 1.23 and 4.19) shows only a weak first inflection because its two p$K_a$ values are close.

Choosing an indicator

An indicator is itself a weak acid, HIn, whose two forms have different colours. The colour change is seen over roughly $\mathrm{p}K_\text{In}\pm1$:

$$\mathrm{HIn \rightleftharpoons H^+ + In^-},\qquad \mathrm{pH}=\mathrm{p}K_\text{In}+\log\frac{[\mathrm{In^-}]}{[\mathrm{HIn}]}$$

Pick an indicator whose colour-change range lies entirely within the vertical part of the titration curve. Turn on Indicator range under the curve to see the band.

TitrationpH at equivalence (0.1 M)Vertical sectionSuitable indicators
Strong acid – strong base7.00≈ 3 to 11Any of them (phenolphthalein, methyl orange, methyl red, BTB)
Weak acid – strong base8.72≈ 7 to 11Phenolphthalein
Strong acid – weak base5.28≈ 3 to 7Methyl orange, methyl red
Weak acid – weak base7.0No sharp jumpNone reliable: use a pH meter

Universal indicator is a blend that changes colour continuously across the whole scale. It is great for estimating pH but useless for a sharp end point.

Example 1: pH along a weak acid titration

25.00 mL of 0.100 M CH₃COOH titrated with 0.100 M NaOH (the default set-up; check each value in the lab).

  1. Start (0 mL). $x^2/(0.100-x)=1.8\times10^{-5}$ gives $x=[\mathrm{H_3O^+}]=1.33\times10^{-3}$ M, so pH = 2.88.
  2. Equivalence volume. $n(\mathrm{CH_3COOH}) = 0.100\times0.02500 = 2.50\times10^{-3}$ mol, so $V_\text{eq}=2.50\times10^{-3}/0.100 = $ 25.00 mL.
  3. Half-equivalence (12.50 mL). $[\mathrm{A^-}]\approx[\mathrm{HA}]$ so pH ≈ p$K_a$ = 4.74. The exact charge-balance value shown on the curve is 4.75: a little extra acid ionises, so $[\mathrm{A^-}]$ is very slightly larger than $[\mathrm{HA}]$.
  4. Equivalence (25.00 mL). All acid is now CH₃COO⁻ at $2.50\times10^{-3}/0.0500 = 0.0500$ M. $K_b = K_w/K_a = 5.56\times10^{-10}$; $[\mathrm{OH^-}]=\sqrt{K_b\,c}=5.27\times10^{-6}$ M; pOH = 5.28, so pH = 8.72.
  5. Past equivalence (30.00 mL). Excess OH⁻ $=(3.00-2.50)\times10^{-3}/0.0550 = 9.09\times10^{-3}$ M; pOH = 2.04, pH = 11.96.

Example 2: Finding an unknown concentration

A student pipettes 25.00 mL of hydrochloric acid and titrates with 0.100 M NaOH. Titres: rough 23.6, then 23.15, 23.20, 23.40 mL.

  1. Concordant titres agree within 0.10 mL: 23.15 and 23.20 mL. Discard the rough and the 23.40 mL.
  2. Mean titre = (23.15 + 23.20)/2 = 23.18 mL (keep 23.175 in the calculation).
  3. $n(\mathrm{NaOH}) = 0.100 \times 0.023175 = 2.3175\times10^{-3}$ mol. The ratio HCl : NaOH is 1 : 1.
  4. $c(\mathrm{HCl}) = 2.3175\times10^{-3}/0.02500 = $ 0.0927 mol L⁻¹.

Example 3: A diprotic acid

20.00 mL of oxalic acid needs 24.80 mL of 0.100 M NaOH to reach the phenolphthalein end point.

  1. Phenolphthalein changes at the second equivalence point: $\mathrm{H_2C_2O_4 + 2\,NaOH \to Na_2C_2O_4 + 2\,H_2O}$.
  2. $n(\mathrm{NaOH}) = 0.100\times0.02480 = 2.48\times10^{-3}$ mol, so $n(\mathrm{H_2C_2O_4}) = 1.24\times10^{-3}$ mol.
  3. $c = 1.24\times10^{-3}/0.02000 = $ 0.0620 mol L⁻¹.

Example 4: Strong acid – weak base at equivalence

25.00 mL of 0.100 M NH₃ with 0.100 M HCl: at 25.00 mL, $[\mathrm{NH_4^+}] = 0.0500$ M, $K_a = 10^{-14}/1.8\times10^{-5} = 5.56\times10^{-10}$, $[\mathrm{H_3O^+}] = \sqrt{K_a c} = 5.27\times10^{-6}$ M, pH = 5.28. So methyl red (4.4–6.2) is the right indicator and phenolphthalein would change far too early.

Challenges (checked automatically)

    Guided experiments

    1. Rough, then accurate. Use Fast mode for a rough titre, record it, take a new flask, then run fast to about 1 mL before the rough value and finish dropwise. Repeat until two titres agree within 0.10 mL.
    2. Compare the four curve shapes. Run each monoprotic pair with 0.100 M solutions and note the starting pH, the pH at equivalence and the height of the vertical jump. Why is the weak–weak jump so small?
    3. Dilution. Set both concentrations to 0.010 M for HCl/NaOH. What happens to the vertical section? Which indicators would now fail?
    4. Measure p$K_a$. Titrate CH₃COOH and read the pH at exactly half the equivalence volume. Compare with p$K_a$ = 4.74.
    5. Two steps. Titrate H₃PO₄ with the derivative plot on. Measure both equivalence volumes and confirm $V_2 = 2V_1$. Which indicator detects each one?
    6. Swap roles. Put NH₃ in the flask and HCl in the burette. The curve now falls. Where is the half-equivalence pH, and what does it tell you?