Learn › Chemistry

Acids, bases, pH and titration

pH decides whether a metal corrodes, whether an enzyme works or what color an indicator takes. This topic explains the pH scale, strong and weak acids and bases, salts that change the pH, buffer solutions and titration curves, with the same exact calculation the tool performs.

Go to the simulator

The pH scale

Water self-ionizes: $2 H_{2} O ⇌ H_{3} O^{+} + OH^{-}$ , with $K_{w} = [H_{3} O^{+}] [OH^{-}] = 1.0 \times 1 0^{- 14}$ at 25 °C. Since the concentrations range from 1 to $1 0^{- 14}$ , logarithms are used:

pH = - lo g [H_{3} O^{+}] pOH = - lo g [OH^{-}] pH + pOH = 14

pH scale: at 25 °C, below 7 is acidic and above 7 is basic.

Strong and weak acids and bases

A strong acid (HCl, HNO₃) dissociates completely: $[H_{3} O^{+}] = C$ . A weak one only partly, according to its acidity constant:

HA + H_{2} O ⇌ H_{3} O^{+} + A^{-} K_{a} = \frac{[ H _{3} O ^{+} ] [ A ^{-} ]}{[ HA ]} p K_{a} = - lo g K_{a}

The smaller the p $K_{a}$ , the stronger the acid. For a weak acid that is not too dilute, $[H_{3} O^{+}] \approx K_{a} C$ . Weak bases are treated the same way with $K_{b}$ , and an acid and its conjugate base satisfy $K_{a} K_{b} = K_{w}$ . That is why the salt of a weak acid (NaCN, CH₃COONa) is basic: its anion takes protons from water.

The tool does not use these approximations: it writes the exact charge balance (the sum of positive charges equal to that of negative ones), with the fraction of each form of each acid, and solves it by bisection. That way it treats a mixture of several acids, a polyprotic acid or a point of a titration the same way. It assumes an ideal solution (no activity coefficients) and additive volumes.

[H_{3} O^{+}] + [Na^{+}] = [OH^{-}] + [A^{-}] + \dots [A^{-}] = C \frac{K _{a}}{K _{a} + [ H _{3} O ^{+} ]}

Buffer solutions

A mixture of a weak acid and its conjugate base (CH₃COOH + CH₃COO⁻) resists changes in pH: added acid is consumed by the anion, and added base by the acid. Its pH is given by the Henderson–Hasselbalch equation:

pH = p K_{a} + lo g \frac{[ A ^{-} ]}{[ HA ]}

Fraction of acetic acid and acetate against pH: they cross at the pKa, and around it (pKa ± 1) is the buffering zone.

It buffers best with equal amounts of both forms (pH = p $K_{a}$ ) and with high concentrations, and stops working far from the p $K_{a}$ (more than one unit), when one of the two forms is almost used up.

Titration

In a titration a solution of known concentration (the titrant, in the burette) is added until it reacts exactly with the analyte: the equivalence point, where $C_{a} V_{a} = C_{b} V_{b}$ (for 1:1 stoichiometry). There the pH changes abruptly, and the curve has its maximum slope.

Strong acid with strong base: the equivalence pH is 7 and the jump is large.
Weak acid with strong base: halfway pH = p $K_{a}$ (there is as much HA as A⁻), and at equivalence the pH is basic, because the conjugate base remains.
Polyprotic (H₃PO₄): one equivalence point for each proton that can be told apart (their p $K_{a}$ separated by about 3 units or more).

The indicator is chosen so that it changes color within the jump: phenolphthalein (8.2–10.0) for a weak acid with strong base, methyl orange (3.1–4.4) for a weak base with strong acid, and methyl red (4.4–6.2) or bromothymol blue (6.0–7.6) for strong with strong.

Common mistakes

Taking $[H_{3} O^{+}] = C$ for a weak acid (that only holds for strong ones).
Forgetting dilution when mixing: the moles spread over the total volume.
Thinking the pH at equivalence is always 7: only for strong with strong.
Using Henderson–Hasselbalch with a strong acid, or far from the pKa.
Applying pH + pOH = 14 at another temperature: $K_{w}$ changes with T.

Worked examples

Example 1pH of a weak acid

Find the pH of a 0.1 M acetic acid solution (p

K_{a}

= 4.756) and what fraction is dissociated.

Solution

$K_{a} = 1 0^{- 4.756} = 1.754 \times 1 0^{- 5}$ . With $x = [H_{3} O^{+}]$ : $\frac{x ^{2}}{0.1 - x} = K_{a}$ .
Approximation: $x \approx K_{a} C = 1.324 \times 1 0^{- 3}$ → pH 2.878. Exact (quadratic): $x = 1.316 \times 1 0^{- 3}$ → pH 2.881.
Dissociated fraction: $x / C = 1.316%$ ; 98.68% remains as CH₃COOH. A 0.1 M HCl would have pH 1.

pH = 2.881; 1.316% dissociated.

Solve it in Simuveliq

In Chemistry → Acid–base equilibrium (pH), a single row with 0.1 M acetic acid (the button sets it). The card gives pH 2.881 and the species table, HA 98.68% and A⁻ 1.316%.

Open in the simulator →

Example 2Buffer solution

50 mL of 0.1 M CH₃COOH are mixed with 50 mL of 0.1 M CH₃COONa. Find the pH.

Solution

In the 100 mL there are 5 mmol of each form: $[HA] = [A^{-}] = 0.05$ M.
Henderson–Hasselbalch: pH $= 4.756 + lo g (0.05/0.05) = 4.756$ .

pH = 4.756 = p

K_{a}

Solve it in Simuveliq

In Examples, «Buffer: 0.1 M CH3COOH + 0.1 M CH3COONa»: pH 4.756, with HA and A⁻ almost 50% each (49.98% and 50.02%, from the acid's small dissociation).

Open in the simulator →

Example 3The buffer against an acid

1 mL of 0.1 M HCl is added to the previous buffer. How much does the pH change? And if that same amount of HCl were added to 100 mL of pure water?

Solution

The HCl (0.1 mmol) reacts with acetate: A⁻ goes from 5 to 4.9 mmol and HA from 5 to 5.1 mmol.
pH $= 4.756 + lo g (4.9/5.1) = 4.739$ : it drops only 0.017.
In 100 mL of water: $[H_{3} O^{+}] = 0.1/101 = 9.9 \times 1 0^{- 4}$ M → pH 3.00, four units below neutral water.

With the buffer the pH goes from 4.756 to 4.739; in pure water it would drop from 7 to 3.00.

Solve it in Simuveliq

The button adds a third row, 0.1 M HCl and 1 mL: the card gives pH 4.739 and the table, HA 50.98% and A⁻ 49.02%.

Open in the simulator →

Example 4A basic salt

Find the pH of a 0.1 M NaCN solution (p

K_{a}

of HCN = 9.21).

Solution

CN⁻ is the conjugate base of HCN: $K_{b} = K_{w} / K_{a} = 1 0^{- 14} /1 0^{- 9.21} = 1.62 \times 1 0^{- 5}$ .
$[OH^{-}] \approx K_{b} C = 1.27 \times 1 0^{- 3}$ M → pOH 2.89 → pH 11.10.

pH = 11.10 (1.27% of the cyanide takes a proton).

Solve it in Simuveliq

«Basic salt: 0.1 M NaCN»: pH 11.10, pOH 2.898 and in the table HA (HCN) 1.265% and A⁻ 98.73%.

Open in the simulator →

Example 5Titration curve of a weak acid

25 mL of 0.1 M acetic acid are titrated with 0.1 M NaOH. Find the pH at the start, halfway, at equivalence and a little after, and choose the indicator.

Solution

Start: it is the acid alone, pH 2.881 (example 1).
Halfway (12.5 mL): half the acid became acetate, HA = A⁻: pH = p $K_{a}$ = 4.756.
Equivalence (25 mL): 2.5 mmol of acetate in 50 mL, 0.05 M. $K_{b} = 1 0^{- 9.244}$ , $[OH^{-}] = K_{b} \cdot 0.05 = 5.34 \times 1 0^{- 6}$ → pH 8.728.
After (26 mL): 0.1 mmol of NaOH is left over in 51 mL, $[OH^{-}] = 1.96 \times 1 0^{- 3}$ → pH 11.29. Between 24 and 26 mL the pH jumps from 6.14 to 11.29: phenolphthalein (8.2–10.0) changes right there; methyl orange would not work.

pH 2.88 → 4.76 (halfway) → 8.73 (equivalence, 25 mL) → 11.29 (26 mL); indicator: phenolphthalein.

Solve it in Simuveliq

«Titration: 0.1 M acetic acid with 0.1 M NaOH»: with 25 mL of NaOH the card gives pH 8.728, and the Titration curve marks «Eq. 1» at V = 25.00 mL, pH = 8.73. Change the NaOH volume to 12.5 or 26 mL to read the other points.
In Chemistry → Interactive titration the same is done drop by drop, with a burette, a flask and the indicator's color.

Open in the simulator →

Example 6Strong acid with strong base

25 mL of 0.1 M HCl are titrated with 0.1 M NaOH. Find the pH with 20 mL added and at equivalence.

Solution

With 20 mL: $2.5 - 2.0 = 0.5$ mmol of H₃O⁺ are left in 45 mL → $0.01111$ M → pH 1.954.
At 25 mL everything is NaCl, which does not hydrolyze: pH 7. Just 0.1 mL before or after takes it from pH 3.70 to 10.30.

pH 1.954 with 20 mL; equivalence at 25 mL with pH 7.

Solve it in Simuveliq

«Titration: 0.1 M HCl with 0.1 M NaOH» (with 20 mL): pH 1.954, and the curve marks the equivalence at 25.00 mL with pH 7.00.

Open in the simulator →

Example 7A polyprotic acid

20 mL of 0.1 M H₃PO₄ (p

K_{a}

= 2.15; 7.20; 12.35) are titrated with 0.1 M NaOH. Where are the equivalence points and what pH do they have?

Solution

Each proton needs 2 mmol of NaOH = 20 mL. First equivalence at 20 mL (all H₂PO₄⁻) and second at 40 mL (all HPO₄²⁻). The third is not seen: p $K_{a}$ 12.35 is too close to the pH of the base itself.
An amphoteric species has pH ≈ (p $K_{a 1}$ + p $K_{a 2}$ )/2: H₂PO₄⁻ ≈ 4.68 (exact 4.70) and HPO₄²⁻ ≈ (7.20 + 12.35)/2 = 9.78 (exact 9.66, because at that dilution water matters).
Halfway through the second stage (30 mL) the pH equals p $K_{a 2}$ = 7.20. Halfway through the first (10 mL) it gives 2.29, not 2.15: H₃PO₄ is already highly dissociated by itself and the «pH = p $K_{a}$ » rule no longer holds.

Equivalences at 20 mL (pH ≈ 4.70) and 40 mL (pH ≈ 9.66).

Solve it in Simuveliq

«Polyprotic (custom): H3PO4 with NaOH»: the acid is a «Custom» row with the three pKa separated by «;». The curve marks Eq. 1 at 20.00 mL and Eq. 2 at 40.00 mL.

Open in the simulator →

Acids, bases, pH and titration

The pH scale

Strong and weak acids and bases

Buffer solutions

Titration

Common mistakes

Worked examples

Example 1pH of a weak acid

Solution

Solve it in Simuveliq

Example 2Buffer solution

Solution

Solve it in Simuveliq

Example 3The buffer against an acid

Solution

Solve it in Simuveliq

Example 4A basic salt

Solution

Solve it in Simuveliq

Example 5Titration curve of a weak acid

Solution

Solve it in Simuveliq

Example 6Strong acid with strong base

Solution

Solve it in Simuveliq

Example 7A polyprotic acid

Solution

Solve it in Simuveliq

Related topics