Acids, Bases & Buffers

Acid-Base Definitions

Bronsted-Lowry Definition

An acid is a proton ($\mathrm{H}^+$) donor. A base is a proton acceptor.

When an acid donates a proton, the remaining species is its conjugate base. When a base accepts a proton, the resulting species is its conjugate acid.

$$\mathrm{HA} + \mathrm{H}_2\mathrm{O} \rightleftharpoons \mathrm{H}_3\mathrm{O}^+ + \mathrm{A}^-$$

• $\mathrm{HA}$: acid; $\mathrm{A}^-$: conjugate base
• $\mathrm{H}_2\mathrm{O}$: base; $\mathrm{H}_3\mathrm{O}^+$: conjugate acid

A conjugate pair differs by a single proton. For example, $\mathrm{NH}_4^+/\mathrm{NH}_3$ and $\mathrm{H}_2\mathrm{O}/\mathrm{OH}^-$ are conjugate pairs.

Lewis Definition

A Lewis acid is an electron pair acceptor. A Lewis base is an electron pair donor. This definition is broader than Bronsted-Lowry and includes reactions that do not involve proton transfer.

Example: $\mathrm{BF}_3$ is a Lewis acid (accepts an electron pair into its empty $p$ orbital). $\mathrm{NH}_3$ is a Lewis base (donates its lone pair).

$$\mathrm{BF}_3 + \mathrm{NH}_3 \to \mathrm{F}_3\mathrm{B}\mathrm{--}\mathrm{NH}_3$$

Strong and Weak Acids and Bases

Strong Acids

A strong acid is completely dissociated in aqueous solution:

$$\mathrm{HCl}(aq) \to \mathrm{H}^+(aq) + \mathrm{Cl}^-(aq)$$

Common strong acids: $\mathrm{HCl}$, $\mathrm{HBr}$, $\mathrm{HI}$, $\mathrm{HNO}_3$, $\mathrm{H}_2\mathrm{SO}_4$ (first dissociation), $\mathrm{HClO}_4$.

For a strong monoprotic acid of concentration $c$: $[\mathrm{H}^+] = c$.

Weak Acids

A weak acid is partially dissociated in aqueous solution:

$$\mathrm{HA}(aq) \rightleftharpoons \mathrm{H}^+(aq) + \mathrm{A}^-(aq)$$

The position of equilibrium lies to the left. The degree of dissociation $\alpha$ is the fraction of acid molecules that dissociate. For weak acids, $\alpha \ll 1$.

Common weak acids: $\mathrm{CH}_3\mathrm{COOH}$, $\mathrm{HCOOH}$, $\mathrm{HCN}$, $\mathrm{H}_2\mathrm{CO}_3$, $\mathrm{NH}_4^+$.

Strong Bases

Common strong bases: Group 1 hydroxides ($\mathrm{NaOH}$, $\mathrm{KOH}$), $\mathrm{Ba(OH)}_2$, and $\mathrm{Ca(OH)}_2$ (sparingly soluble but fully dissociated).

Weak Bases

Common weak bases: $\mathrm{NH}_3$, amines, $\mathrm{CO}_3^{2-}$, $\mathrm{HCO}_3^-$.

The pH Scale

The pH is defined as:

$$\mathrm{pH} = -\log_{10}[\mathrm{H}^+]$$

where $[\mathrm{H}^+]$ is in $\mathrm{mol/dm}^3$.

• At $25^\circ\mathrm{C}$, pure water has $[\mathrm{H}^+] = 1.0 \times 10^{-7}\,\mathrm{mol/dm}^3$, giving $\mathrm{pH} = 7.0$.
• Acidic solutions: $\mathrm{pH} \lt 7$ ($[\mathrm{H}^+] \gt [\mathrm{OH}^-]$).
• Alkaline solutions: $\mathrm{pH} \gt 7$ ($[\mathrm{H}^+] \lt [\mathrm{OH}^-]$).
• pH is typically measured on a scale of 0--14, though values outside this range are possible.

The Ionic Product of Water ($K_w$)

Water undergoes autoionisation:

$$2\mathrm{H}_2\mathrm{O}(l) \rightleftharpoons \mathrm{H}_3\mathrm{O}^+(aq) + \mathrm{OH}^-(aq)$$

Simplified:

$$\mathrm{H}_2\mathrm{O}(l) \rightleftharpoons \mathrm{H}^+(aq) + \mathrm{OH}^-(aq)$$ $$K_w = [\mathrm{H}^+][\mathrm{OH}^-]$$

At $25^\circ\mathrm{C}$: $K_w = 1.0 \times 10^{-14}\,\mathrm{mol^2\,dm^{-6}}$.

Since $[\mathrm{H}^+] = [\mathrm{OH}^-]$ in pure water: $[\mathrm{H}^+] = \sqrt{K_w} = 1.0 \times 10^{-7}\,\mathrm{mol/dm}^3$.

$K_w$ is temperature-dependent. At $50^\circ\mathrm{C}$, $K_w = 5.5 \times 10^{-14}$, so neutral $\mathrm{pH} = 6.63$. The solution is still neutral ($[\mathrm{H}^+] = [\mathrm{OH}^-]$) but the pH is lower because $K_w$ has increased.

Acid Dissociation Constant ($K_a$)

For a weak acid $\mathrm{HA}$:

$$\mathrm{HA}(aq) \rightleftharpoons \mathrm{H}^+(aq) + \mathrm{A}^-(aq)$$ $$K_a = \frac◆LB◆[\mathrm{H}^+][\mathrm{A}^-]◆RB◆◆LB◆[\mathrm{HA}]◆RB◆$$

Units of $K_a$: $\mathrm{mol\,dm^{-3}}$.

p$K_a$

$$\mathrm{p}K_a = -\log_{10} K_a$$

Lower $\mathrm{p}K_a$ = stronger acid. Typical values:

Acid	$K_a$ ($\mathrm{mol/dm}^3$)	$\mathrm{p}K_a$
$\mathrm{CH}_3\mathrm{COOH}$	$1.74 \times 10^{-5}$	4.76
$\mathrm{HCOOH}$	$1.78 \times 10^{-4}$	3.75
$\mathrm{HCN}$	$6.2 \times 10^{-10}$	9.21
$\mathrm{HF}$	$6.3 \times 10^{-4}$	3.20

Base Dissociation Constant ($K_b$)

For a weak base $\mathrm{B}$:

$$\mathrm{B}(aq) + \mathrm{H}_2\mathrm{O}(l) \rightleftharpoons \mathrm{BH}^+(aq) + \mathrm{OH}^-(aq)$$ $$K_b = \frac◆LB◆[\mathrm{BH}^+][\mathrm{OH}^-]◆RB◆◆LB◆[\mathrm{B}]◆RB◆$$

The relationship between $K_a$ and $K_b$ for a conjugate pair:

$$K_a \times K_b = K_w$$ $$\mathrm{p}K_a + \mathrm{p}K_b = \mathrm{p}K_w = 14.00\ (\mathrm{at } 25^\circ\mathrm{C})$$

pH Calculations

Strong Acid

For a strong monoprotic acid of concentration $c$:

$$[\mathrm{H}^+] = c$$ $$\mathrm{pH} = -\log_{10} c$$

Example. $0.050\,\mathrm{mol/dm}^3$ $\mathrm{HCl}$: $\mathrm{pH} = -\log_{10}(0.050) = 1.30$.

Strong Base

For a strong base $\mathrm{MOH}$ of concentration $c$:

$$[\mathrm{OH}^-] = c$$ $$[\mathrm{H}^+] = \frac◆LB◆K_w◆RB◆◆LB◆[\mathrm{OH}^-]◆RB◆ = \frac◆LB◆1.0 \times 10^{-14}◆RB◆◆LB◆c◆RB◆$$ $$\mathrm{pH} = 14 + \log_{10} c$$

Example. $0.020\,\mathrm{mol/dm}^3$ $\mathrm{NaOH}$: $\mathrm{pH} = 14 + \log_{10}(0.020) = 14 - 1.70 = 12.30$.

Weak Acid

For a weak acid $\mathrm{HA}$ of concentration $c$ with $K_a$:

$$K_a = \frac◆LB◆[\mathrm{H}^+][\mathrm{A}^-]◆RB◆◆LB◆[\mathrm{HA}]◆RB◆$$

Assuming $[\mathrm{H}^+] \approx [\mathrm{A}^-]$ and $[\mathrm{HA}] \approx c$ (valid when $\alpha \lt 5\%$):

$$K_a = \frac◆LB◆[\mathrm{H}^+]^2◆RB◆◆LB◆c◆RB◆$$ $$[\mathrm{H}^+] = \sqrt◆LB◆K_a \times c◆RB◆$$ $$\mathrm{pH} = -\log_{10}\sqrt{K_a c} = \frac{1}{2}(\mathrm{p}K_a - \log_{10} c)$$

Worked Example. Calculate the pH of $0.100\,\mathrm{mol/dm}^3$ ethanoic acid ($K_a = 1.74 \times 10^{-5}$).

$$[\mathrm{H}^+] = \sqrt◆LB◆1.74 \times 10^{-5} \times 0.100◆RB◆ = \sqrt◆LB◆1.74 \times 10^{-6}◆RB◆ = 1.32 \times 10^{-3}\,\mathrm{mol/dm}^3$$ $$\mathrm{pH} = -\log_{10}(1.32 \times 10^{-3}) = 2.88$$

Verification: $\alpha = 1.32 \times 10^{-3} / 0.100 = 1.32\% \lt 5\%$. The approximation is valid.

Weak Base

For a weak base $\mathrm{B}$ of concentration $c$ with $K_b$:

$$[\mathrm{OH}^-] = \sqrt◆LB◆K_b \times c◆RB◆$$ $$\mathrm{pOH} = -\log_{10}[\mathrm{OH}^-]$$ $$\mathrm{pH} = 14 - \mathrm{pOH}$$

Example. Calculate the pH of $0.200\,\mathrm{mol/dm}^3$ $\mathrm{NH}_3$ ($K_b = 1.78 \times 10^{-5}$).

$$[\mathrm{OH}^-] = \sqrt◆LB◆1.78 \times 10^{-5} \times 0.200◆RB◆ = \sqrt◆LB◆3.56 \times 10^{-6}◆RB◆ = 1.89 \times 10^{-3}\,\mathrm{mol/dm}^3$$ $$\mathrm{pOH} = -\log_{10}(1.89 \times 10^{-3}) = 2.72$$ $$\mathrm{pH} = 14 - 2.72 = 11.28$$

Buffer Solutions

Composition

A buffer solution resists changes in pH when small amounts of acid or base are added. An acidic buffer consists of:

1. A weak acid ($\mathrm{HA}$).
2. Its conjugate base ($\mathrm{A}^-$), typically supplied as a salt (e.g. $\mathrm{CH}_3\mathrm{COOH} + \mathrm{CH}_3\mathrm{COONa}$).

A basic buffer consists of a weak base and its conjugate acid (e.g. $\mathrm{NH}_3 + \mathrm{NH}_4\mathrm{Cl}$).

Mechanism

Adding acid ($\mathrm{H}^+$): The conjugate base $\mathrm{A}^-$ reacts with the added $\mathrm{H}^+$:

$$\mathrm{A}^- + \mathrm{H}^+ \to \mathrm{HA}$$

This consumes the added $\mathrm{H}^+$, minimising the pH change.

Adding base ($\mathrm{OH}^-$): The weak acid $\mathrm{HA}$ reacts with the added $\mathrm{OH}^-$:

$$\mathrm{HA} + \mathrm{OH}^- \to \mathrm{A}^- + \mathrm{H}_2\mathrm{O}$$

This consumes the added $\mathrm{OH}^-$, minimising the pH change.

Henderson-Hasselbalch Equation

Starting from the $K_a$ expression:

$$K_a = \frac◆LB◆[\mathrm{H}^+][\mathrm{A}^-]◆RB◆◆LB◆[\mathrm{HA}]◆RB◆$$

Rearranging:

$$[\mathrm{H}^+] = K_a \times \frac◆LB◆[\mathrm{HA}]◆RB◆◆LB◆[\mathrm{A}^-]◆RB◆$$

Taking $-\log_{10}$:

$$\mathrm{pH} = \mathrm{p}K_a + \log_{10}\frac◆LB◆[\mathrm{A}^-]◆RB◆◆LB◆[\mathrm{HA}]◆RB◆$$

This is the Henderson-Hasselbalch equation.

Key points:

• When $[\mathrm{A}^-] = [\mathrm{HA}]$: $\mathrm{pH} = \mathrm{p}K_a$.
• The buffer is most effective when $\mathrm{pH} \approx \mathrm{p}K_a$ (within $\pm 1$ unit).
• Buffer capacity depends on the absolute concentrations of $\mathrm{HA}$ and $\mathrm{A}^-$.

Worked Example. Calculate the pH of a buffer containing $0.100\,\mathrm{mol/dm}^3$ $\mathrm{CH}_3\mathrm{COOH}$ and $0.150\,\mathrm{mol/dm}^3$ $\mathrm{CH}_3\mathrm{COONa}$ ($\mathrm{p}K_a = 4.76$).

$$\mathrm{pH} = 4.76 + \log_{10}\left(\frac{0.150}{0.100}\right) = 4.76 + 0.176 = 4.94$$

Titration Curves

Strong Acid vs Strong Base

Example: $\mathrm{HCl}$ vs $\mathrm{NaOH}$.

• Initial pH: low ($\approx 1$ for $0.1\,\mathrm{mol/dm}^3$ $\mathrm{HCl}$).
• pH rises slowly, then very steeply near the equivalence point.
• Equivalence point at $\mathrm{pH} = 7.0$ (neutral).
• The pH jump is large (typically 3--11), allowing a wide choice of indicators.

Strong Acid vs Weak Base

Example: $\mathrm{HCl}$ vs $\mathrm{NH}_3$.

• Equivalence point at $\mathrm{pH} \lt 7$ (acidic, because the conjugate acid $\mathrm{NH}_4^+$ hydrolyses).
• Suitable indicator: methyl orange (range 3.1--4.4).

Weak Acid vs Strong Base

Example: $\mathrm{CH}_3\mathrm{COOH}$ vs $\mathrm{NaOH}$.

• Equivalence point at $\mathrm{pH} \gt 7$ (alkaline, because the conjugate base $\mathrm{CH}_3\mathrm{COO}^-$ hydrolyses).
• Suitable indicator: phenolphthalein (range 8.3--10.0).

Half-equivalence point: At half-neutralisation, $[\mathrm{HA}] = [\mathrm{A}^-]$, so $\mathrm{pH} = \mathrm{p}K_a$. This allows experimental determination of $\mathrm{p}K_a$ from a titration curve.

Weak Acid vs Weak Base

Example: $\mathrm{CH}_3\mathrm{COOH}$ vs $\mathrm{NH}_3$.

• No sharp equivalence point.
• The pH change is gradual, making it difficult to select an appropriate indicator.
• A pH meter is required for accurate endpoint determination.

Summary of Indicator Choices

Titration type	Equivalence point pH	Suitable indicator
Strong acid / Strong base	7.0	Any (e.g. bromothymol blue)
Strong acid / Weak base	< 7.0	Methyl orange (3.1--4.4)
Weak acid / Strong base	> 7.0	Phenolphthalein (8.3--10.0)
Weak acid / Weak base	$\approx 7.0$ (gradual)	None suitable; use pH meter

Neutralisation Enthalpy

The standard enthalpy of neutralisation is the enthalpy change when one mole of water is formed from the reaction of an acid and a base under standard conditions.

$$\mathrm{H}^+(aq) + \mathrm{OH}^-(aq) \to \mathrm{H}_2\mathrm{O}(l) \quad \Delta H = -57.1\,\mathrm{kJ/mol}$$

This value is approximately constant for strong acid-strong base reactions because the net ionic equation is always the same.

For reactions involving weak acids or weak bases, the enthalpy of neutralisation is less exothermic (e.g. $-51$ to $-55\,\mathrm{kJ/mol}$) because energy is absorbed to dissociate the weak acid or weak base.

Common Pitfalls

1. Confusing pH with [H+]. pH = 3 does not mean $[\mathrm{H}^+] = 3\,\mathrm{mol/dm}^3$. It means $[\mathrm{H}^+] = 10^{-3}\,\mathrm{mol/dm}^3$. A lower pH means a higher $[\mathrm{H}^+]$.

2. Applying the weak acid approximation when it is not valid. If $c/K_a \lt 100$, the assumption $[\mathrm{HA}] \approx c$ fails, and the quadratic formula must be used.

3. Using the wrong indicator. The indicator range must overlap with the steep portion of the titration curve at the equivalence point.

4. Forgetting that $K_w$ changes with temperature. At temperatures other than $25^\circ\mathrm{C}$, $\mathrm{p}K_w \ne 14$, so $\mathrm{pH} + \mathrm{pOH} \ne 14$.

5. Adding strong acid/base to a buffer in quantities exceeding its capacity. The buffer can only resist small additions; large additions will overwhelm it.

Practice Problems

Problem 1

Calculate the pH of a solution formed by mixing $25.0\,\mathrm{cm}^3$ of $0.100\,\mathrm{mol/dm}^3$ $\mathrm{CH}_3\mathrm{COOH}$ ($\mathrm{p}K_a = 4.76$) with $10.0\,\mathrm{cm}^3$ of $0.100\,\mathrm{mol/dm}^3$ $\mathrm{NaOH}$.

Solution:

$n(\mathrm{CH}_3\mathrm{COOH}) = 0.100 \times 0.0250 = 2.50 \times 10^{-3}\,\mathrm{mol}$

$n(\mathrm{NaOH}) = 0.100 \times 0.0100 = 1.00 \times 10^{-3}\,\mathrm{mol}$

$\mathrm{NaOH}$ is the limiting reagent. After reaction:

$n(\mathrm{CH}_3\mathrm{COOH}) = 2.50 \times 10^{-3} - 1.00 \times 10^{-3} = 1.50 \times 10^{-3}\,\mathrm{mol}$

$n(\mathrm{CH}_3\mathrm{COO}^-) = 1.00 \times 10^{-3}\,\mathrm{mol}$

Total volume = $35.0\,\mathrm{cm}^3$.

$[\mathrm{CH}_3\mathrm{COOH}] = 1.50 \times 10^{-3} / 0.0350 = 0.0429\,\mathrm{mol/dm}^3$

$[\mathrm{CH}_3\mathrm{COO}^-] = 1.00 \times 10^{-3} / 0.0350 = 0.0286\,\mathrm{mol/dm}^3$

$$\mathrm{pH} = 4.76 + \log_{10}\left(\frac{0.0286}{0.0429}\right) = 4.76 - 0.176 = 4.58$$

Problem 2

The pH of a $0.050\,\mathrm{mol/dm}^3$ solution of a weak acid $\mathrm{HX}$ is 3.00. Calculate $K_a$ and $\mathrm{p}K_a$.

Solution:

$$[\mathrm{H}^+] = 10^{-3.00} = 1.00 \times 10^{-3}\,\mathrm{mol/dm}^3$$$$K_a = \frac◆LB◆[\mathrm{H}^+]^2◆RB◆◆LB◆c◆RB◆ = \frac◆LB◆(1.00 \times 10^{-3})^2◆RB◆◆LB◆0.050◆RB◆ = \frac◆LB◆1.00 \times 10^{-6}◆RB◆◆LB◆0.050◆RB◆ = 2.0 \times 10^{-5}\,\mathrm{mol/dm}^3$$$$\mathrm{p}K_a = -\log_{10}(2.0 \times 10^{-5}) = 4.70$$

Problem 3

A buffer solution is prepared by adding $50.0\,\mathrm{cm}^3$ of $0.200\,\mathrm{mol/dm}^3$ $\mathrm{NaOH}$ to $100\,\mathrm{cm}^3$ of $0.200\,\mathrm{mol/dm}^3$ $\mathrm{CH}_3\mathrm{COOH}$ ($\mathrm{p}K_a = 4.76$). Calculate the pH of the buffer. What is the pH change when $5.0\,\mathrm{cm}^3$ of $0.100\,\mathrm{mol/dm}^3$ $\mathrm{HCl}$ is added to $25.0\,\mathrm{cm}^3$ of this buffer?

Solution:

Part 1: Buffer pH

$n(\mathrm{CH}_3\mathrm{COOH})_\mathrm{initial} = 0.200 \times 0.100 = 0.0200\,\mathrm{mol}$

$n(\mathrm{NaOH}) = 0.200 \times 0.0500 = 0.0100\,\mathrm{mol}$

After reaction: $n(\mathrm{CH}_3\mathrm{COOH}) = 0.0200 - 0.0100 = 0.0100\,\mathrm{mol}$

$n(\mathrm{CH}_3\mathrm{COO}^-) = 0.0100\,\mathrm{mol}$

Total volume = $150\,\mathrm{cm}^3 = 0.150\,\mathrm{dm}^3$

$[\mathrm{CH}_3\mathrm{COOH}] = 0.0100/0.150 = 0.0667\,\mathrm{mol/dm}^3$

$[\mathrm{CH}_3\mathrm{COO}^-] = 0.0100/0.150 = 0.0667\,\mathrm{mol/dm}^3$

$$\mathrm{pH} = 4.76 + \log_{10}\left(\frac{0.0667}{0.0667}\right) = 4.76 + \log_{10}(1) = 4.76 + 0 = 4.76$$

The buffer pH equals the $\mathrm{p}K_a$ because $[\mathrm{acid}] = [\mathrm{salt}]$. This is the most effective buffer composition.

Part 2: pH change on adding $\mathrm{HCl}$

$n(\mathrm{HCl}) = 0.100 \times 0.0050 = 5.0 \times 10^{-4}\,\mathrm{mol}$

The added $\mathrm{H}^+$ reacts with $\mathrm{CH}_3\mathrm{COO}^-$:

$n(\mathrm{CH}_3\mathrm{COO}^-)$ after = $0.0667 \times 0.0250 - 5.0 \times 10^{-4} = 1.668 \times 10^{-3} - 5.0 \times 10^{-4} = 1.168 \times 10^{-3}\,\mathrm{mol}$

$n(\mathrm{CH}_3\mathrm{COOH})$ after = $0.0667 \times 0.0250 + 5.0 \times 10^{-4} = 1.668 \times 10^{-3} + 5.0 \times 10^{-4} = 2.168 \times 10^{-3}\,\mathrm{mol}$

Volume = $25.0 + 5.0 = 30.0\,\mathrm{cm}^3 = 0.0300\,\mathrm{dm}^3$

$[\mathrm{CH}_3\mathrm{COOH}] = 2.168 \times 10^{-3}/0.0300 = 0.0723\,\mathrm{mol/dm}^3$

$[\mathrm{CH}_3\mathrm{COO}^-] = 1.168 \times 10^{-3}/0.0300 = 0.0389\,\mathrm{mol/dm}^3$

$$\mathrm{pH} = 4.76 + \log_{10}\left(\frac{0.0389}{0.0723}\right) = 4.76 - 0.269 = 4.49$$

pH change = $4.76 - 4.49 = 0.27$ units. The buffer resists the pH change effectively.

For comparison, adding the same amount of $\mathrm{HCl}$ to $25.0\,\mathrm{cm}^3$ of pure water would give:

$[\mathrm{H}^+] = 5.0 \times 10^{-4} / 0.0300 = 0.0167\,\mathrm{mol/dm}^3$, $\mathrm{pH} = 1.78$

A change from pH 7.00 to pH 1.78 -- 5.22 units. The buffer reduces the pH change by a factor of approximately 19.

Problem 4

Explain why the pH at the half-equivalence point of a weak acid-strong base titration equals the $\mathrm{p}K_a$ of the weak acid.

Solution:

At the half-equivalence point, exactly half the weak acid has been neutralised by the strong base:

$$\mathrm{HA} + \mathrm{OH}^- \to \mathrm{A}^- + \mathrm{H}_2\mathrm{O}$$

If the initial amount of $\mathrm{HA}$ is $n_0$, then at the half-equivalence point:

$n(\mathrm{HA})_\mathrm{remaining} = n_0/2$

$n(\mathrm{A}^-)\_\mathrm{formed} = n_0/2$

Therefore $[\mathrm{HA}] = [\mathrm{A}^-]$ (both are in the same total volume, so the ratio is $1:1$).

Applying the Henderson-Hasselbalch equation:

$$\mathrm{pH} = \mathrm{p}K_a + \log_{10}\left(\frac◆LB◆[\mathrm{A}^-]◆RB◆◆LB◆[\mathrm{HA}]◆RB◆\right) = \mathrm{p}K_a + \log_{10}(1) = \mathrm{p}K_a + 0 = \mathrm{p}K_a$$

This is a useful experimental method for determining the $\mathrm{p}K_a$ of a weak acid: read the pH from the titration curve at the point where half the equivalence volume has been added.

Indicators

An acid-base indicator is a weak acid that has a different colour from its conjugate base. The colour change occurs over a specific pH range (typically about 2 pH units wide).

Indicator	Colour in acid	Colour in alkali	pH range
Methyl orange	Red	Yellow	3.1--4.4
Bromothymol blue	Yellow	Blue	6.0--7.6
Phenolphthalein	Colourless	Pink	8.2--10.0
Universal indicator	Red/orange	Green/blue/purple	1--14 (multiple colours)

Choosing the Right Indicator

The indicator must change colour at the pH of the equivalence point:

Titration type	Equivalence point pH	Suitable indicator
Strong acid vs strong base	$\mathrm{pH} = 7$	Bromothymol blue, phenol red
Strong acid vs weak base	$\mathrm{pH} \lt 7$	Molecular orange, bromophenol blue
Weak acid vs strong base	$\mathrm{pH} \gt 7$	Phenolphthalein

pH Curves in Detail

Strong acid-strong base (e.g. $\mathrm{HCl}$ vs $\mathrm{NaOH}$):

• Initial pH is low (e.g. pH 1 for $0.1\,\mathrm{mol/dm}^3$ $\mathrm{HCl}$).
• pH rises slowly at first, then very steeply near the equivalence point.
• The equivalence point is at pH 7 (neutral, because neither cation nor anion hydrolyses appreciably in water).
• The vertical portion of the curve is very steep (pH jumps from approximately 3 to 11 over a very small volume), allowing a wide choice of indicators.

Weak acid-strong base (e.g. $\mathrm{CH}_3\mathrm{COOH}$ vs $\mathrm{NaOH}$):

• Initial pH is higher than for a strong acid of the same concentration (e.g. pH 2.9 for $0.1\,\mathrm{mol/dm}^3$ $\mathrm{CH}_3\mathrm{COOH}$).
• The initial rise is more gradual (buffer region).
• The equivalence point is at $\mathrm{pH} \gt 7$ (basic, because the conjugate base $\mathrm{CH}_3\mathrm{COO}^-$ is a weak base and hydrolyses water).
• Phenolphthalein is the appropriate indicator.

Strong acid-weak base (e.g. $\mathrm{HCl}$ vs $\mathrm{NH}_3$):

• The equivalence point is at $\mathrm{pH} \lt 7$ (acidic, because $\mathrm{NH}_4^+$ is acidic and hydrolyses water).
• Methyl orange is the appropriate indicator.

Weak acid-weak base:

• No sharp equivalence point; the pH change is gradual throughout.
• No single indicator is suitable; a pH meter must be used.

The Buffer Region on a Titration Curve

In a weak acid-strong base titration, the buffer region is the flat portion of the curve before the equivalence point. In this region, added base converts $\mathrm{HA}$ to $\mathrm{A}^-$, and the pH changes only slowly because the buffer resists pH change. At the half-equivalence point, $\mathrm{pH} = \mathrm{p}K_a$.

The pH Scale and pH Calculations

Calculating pH of Strong Acids and Bases

Strong monoprotic acid:

$$[\mathrm{H}^+] = c_0 \implies \mathrm{pH} = -\log_{10}(c_0)$$

Strong diprotic acid (e.g. $\mathrm{H}_2\mathrm{SO}_4$):

$$[\mathrm{H}^+] = 2c_0 \implies \mathrm{pH} = -\log_{10}(2c_0)$$

Strong base:

$$[\mathrm{OH}^-] = c_0 \implies \mathrm{pOH} = -\log_{10}(c_0) \implies \mathrm{pH} = 14 - \mathrm{pOH}$$

Worked Example. Calculate the pH of $0.050\,\mathrm{mol/dm}^3$ $\mathrm{H}_2\mathrm{SO}_4$.

$$[\mathrm{H}^+] = 2 \times 0.050 = 0.100\,\mathrm{mol/dm}^3$$ $$\mathrm{pH} = -\log_{10}(0.100) = 1.00$$

Calculating pH of Weak Acids

For a weak acid $\mathrm{HA}$:

$$\mathrm{HA} \rightleftharpoons \mathrm{H}^+ + \mathrm{A}^-$$ $$K_a = \frac◆LB◆[\mathrm{H}^+][\mathrm{A}^-]◆RB◆◆LB◆[\mathrm{HA}]◆RB◆$$

If the acid is weak enough that $[\mathrm{H}^+] = [\mathrm{A}^-] \ll c_0$ (the initial concentration):

$$K_a \approx \frac◆LB◆[\mathrm{H}^+]^2◆RB◆◆LB◆c_0◆RB◆$$ $$[\mathrm{H}^+] = \sqrt{K_a c_0}$$ $$\mathrm{pH} = \frac{1}{2}(\mathrm{p}K_a - \log_{10} c_0)$$

Worked Example. Calculate the pH of $0.100\,\mathrm{mol/dm}^3$ ethanoic acid ($K_a = 1.74 \times 10^{-5}\,\mathrm{mol/dm}^3$).

[\mathrm{H}^+] = \sqrt◆LB◆1.74 \times 10^{-5} \times 0.100◆RB◆ = \sqrt◆LB◆1.74 \times 10^{-6}◆RB◆ = 1.32 \times 10^{-3}\,\mathrm{mol/dm}^3} $$\mathrm{pH} = -\log_{10}(1.32 \times 10^{-3}) = 2.88$$

Note that $1.32 \times 10^{-3}$ is approximately $1.3\%$ of $0.100$, so the approximation is valid. For weaker acids or higher concentrations, the full quadratic expression must be solved:

$$K_a = \frac{x^2}{c_0 - x} \implies x^2 + K_a x - K_a c_0 = 0$$

where $x = [\mathrm{H}^+]$.

Calculating pH of Very Dilute Solutions

At very low concentrations ($c_0 \lt 10^{-6}\,\mathrm{mol/dm}^3$), the contribution of $\mathrm{H}^+$ from water autodissociation becomes significant:

$$[\mathrm{H}^+]_\mathrm{total} = [\mathrm{H}^+]_\mathrm{acid} + 10^{-7}$$

For a $10^{-8}\,\mathrm{mol/dm}^3$ solution of $\mathrm{HCl}$:

$$[\mathrm{H}^+] = 10^{-8} + 10^{-7} = 1.1 \times 10^{-7}\,\mathrm{mol/dm}^3$$ $$\mathrm{pH} = -\log_{10}(1.1 \times 10^{-7}) = 6.96$$

The pH is close to 7 despite the solution being acidic, because the acid is so dilute that water's contribution dominates.

Calculating pH of Polyprotic Acids

Sulphuric acid is a diprotic acid:

\mathrm{H}_2\mathrm{SO}_4 \rightleftharpoons \mathrm{H}^+ + \mathrm{HSO}_4^-} \quad K_{a1} \approx \text{very large (strong)} $$\mathrm{HSO}_4^- \rightleftharpoons \mathrm{H}^+ + \mathrm{SO}_4^{2-} \quad K_{a2} = 1.2 \times 10^{-2}$$

The first dissociation is essentially complete. The second dissociation contributes additional $\mathrm{H}^+$:

$$[\mathrm{H}^+]_\mathrm{total} = c_0 + [\mathrm{H}^+]_{K_{a2}}$$

For $0.100\,\mathrm{mol/dm}^3$ $\mathrm{H}_2\mathrm{SO}_4$:

$$[\mathrm{H}^+]_{K_{a2}} \approx \sqrt◆LB◆1.2 \times 10^{-2} \times 0.100◆RB◆ = 0.0346\,\mathrm{mol/dm}^3$$ $$[\mathrm{H}^+]_\mathrm{total} = 0.100 + 0.0346 = 0.135\,\mathrm{mol/dm}^3$$ $$\mathrm{pH} = -\log_{10}(0.135) = 0.87$$

(Using the exact solution of the quadratic gives $0.110\,\mathrm{mol/dm}^3$ and pH = 0.96.)

Strong Acid-Strong Base Titration Calculations

Worked Example: Full Titration Curve pH Calculation

$25.0\,\mathrm{cm}^3$ of $0.100\,\mathrm{mol/dm}^3$ $\mathrm{HCl}$ is titrated with $0.100\,\mathrm{mol/dm}^3$ $\mathrm{NaOH}$. Calculate the pH at each key stage.

Initial (before any $\mathrm{NaOH}$ added):

$$[\mathrm{H}^+] = 0.100 \implies \mathrm{pH} = 1.00$$

After $10.0\,\mathrm{cm}^3$ $\mathrm{NaOH}$ added:

Moles of $\mathrm{HCl} = 0.0250 \times 0.100 = 2.50 \times 10^{-3}\,\mathrm{mol}$

Moles of $\mathrm{NaOH} = 0.0100 \times 0.100 = 1.00 \times 10^{-3}\,\mathrm{mol}$

Excess $\mathrm{HCl} = 2.50 - 1.00 = 1.50 \times 10^{-3}\,\mathrm{mol}$

Total volume $= 25.0 + 10.0 = 35.0\,\mathrm{cm}^3 = 0.0350\,\mathrm{dm}^3$

$$[\mathrm{H}^+] = \frac◆LB◆1.50 \times 10^{-3}◆RB◆◆LB◆0.0350◆RB◆ = 0.0429\,\mathrm{mol/dm}^3 \implies \mathrm{pH} = 1.37$$

After $24.9\,\mathrm{cm}^3$ $\mathrm{NaOH}$ added (near equivalence):

Moles of $\mathrm{NaOH} = 0.0249 \times 0.100 = 2.49 \times 10^{-3}\,\mathrm{mol}$

Excess $\mathrm{HCl} = 2.50 - 2.49 = 0.01 \times 10^{-3}\,\mathrm{mol}$

$$[\mathrm{H}^+] = \frac◆LB◆0.01 \times 10^{-3}◆RB◆◆LB◆0.0499◆RB◆ = 2.00 \times 10^{-4}\,\mathrm{mol/dm}^3 \implies \mathrm{pH} = 3.70$$

At equivalence point ($25.0\,\mathrm{cm}^3$):

$\mathrm{HCl}$ and $\mathrm{NaOH}$ are in stoichiometric ratio. The solution contains $\mathrm{NaCl}$ and water. pH = 7.00 (neutral).

After $25.1\,\mathrm{cm}^3$ $\mathrm{NaOH}$ added (just past equivalence):

Moles of $\mathrm{NaOH}$ added = $0.0251 \times 0.100 = 2.51 \times 10^{-3}\,\mathrm{mol}$

Excess $\mathrm{NaOH} = 2.51 - 2.50 = 0.01 \times 10^{-3}\,\mathrm{mol}$

$$[\mathrm{OH}^-] = \frac◆LB◆0.01 \times 10^{-3}◆RB◆◆LB◆0.0501◆RB◆ = 2.00 \times 10^{-4}\,\mathrm{mol/dm}^3$$ $$\mathrm{pOH} = -\log_{10}(2.00 \times 10^{-4}) = 3.70 \implies \mathrm{pH} = 14.0 - 3.70 = 10.30$$

Note the symmetry: the pH jumps from 3.70 to 10.30 over just $0.2\,\mathrm{cm}^3$ of added $\mathrm{NaOH}$.

Salt Hydrolysis

When a salt is dissolved in water, the pH of the solution depends on whether the cation and anion come from a strong or weak acid/base:

Salt	Cation from	Anion from	Solution pH	Explanation
$\mathrm{NaCl}$	Strong base	Strong acid	7 (neutral)	Neither ion hydrolyses
$\mathrm{NH}_4\mathrm{Cl}$	Weak base	Strong acid	$< 7$ (acidic)	$\mathrm{NH}_4^+$ hydrolyses: $\mathrm{NH}_4^+ + \mathrm{H}_2\mathrm{O} \rightleftharpoons \mathrm{NH}_3 + \mathrm{H}_3\mathrm{O}^+$
$\mathrm{CH}_3\mathrm{COONa}$	Strong base	Weak acid	$> 7$ (basic)	$\mathrm{CH}_3\mathrm{COO}^-$ hydrolyses: $\mathrm{CH}_3\mathrm{COO}^- + \mathrm{H}_2\mathrm{O} \rightleftharpoons \mathrm{CH}_3\mathrm{COOH} + \mathrm{OH}^-$
$\mathrm{NH}_4\mathrm{CH}_3\mathrm{COO}$	Weak base	Weak acid	$\approx 7$	Both hydrolyse; pH depends on relative $K_a$ and $K_b$

Worked Example. Calculate the pH of a $0.050\,\mathrm{mol/dm}^3$ solution of $\mathrm{NH}_4\mathrm{Cl}$. ($K_b(\mathrm{NH}_3) = 1.8 \times 10^{-5}$)

$$K_a(\mathrm{NH}_4^+) = \frac{K_w}{K_b} = \frac◆LB◆1.0 \times 10^{-14}◆RB◆◆LB◆1.8 \times 10^{-5}◆RB◆ = 5.6 \times 10^{-10}$$ $$\mathrm{NH}_4^+ + \mathrm{H}_2\mathrm{O} \rightleftharpoons \mathrm{NH}_3 + \mathrm{H}_3\mathrm{O}^+$$ $$K_a = \frac◆LB◆[\mathrm{NH}_3][\mathrm{H}_3\mathrm{O}^+]◆RB◆◆LB◆[\mathrm{NH}_4^+]◆RB◆ \approx \frac{x^2}{0.050}$$ $$x = \sqrt◆LB◆5.6 \times 10^{-10} \times 0.050◆RB◆ = \sqrt◆LB◆2.8 \times 10^{-11}◆RB◆ = 5.3 \times 10^{-6}\,\mathrm{mol/dm}^3$$ $$\mathrm{pH} = -\log_{10}(5.3 \times 10^{-6}) = 5.28$$

$K_w$ and the Ionic Product of Water

$$\mathrm{H}_2\mathrm{O}(l) \rightleftharpoons \mathrm{H}^+(aq) + \mathrm{OH}^-(aq)$$ $$K_w = [\mathrm{H}^+][\mathrm{OH}^-] = 1.0 \times 10^{-14}\,\mathrm{mol}^2\,\mathrm{dm}^{-6} \quad \text{at } 298\,\mathrm{K}$$

$K_w$ is temperature-dependent. At higher temperatures, the endothermic autodissociation of water is favoured, and $K_w$ increases:

Temperature (K)	$K_w$	pH of pure water
273	$0.11 \times 10^{-14}$	7.47
298	$1.00 \times 10^{-14}$	7.00
323	$5.5 \times 10^{-14}$	6.63
373	$51.3 \times 10^{-14}$	6.14

Note: pure water is always neutral ($[\mathrm{H}^+] = [\mathrm{OH}^-]$), but its pH changes with temperature because $K_w$ changes.

Additional Practice Problems

Problem 5

Calculate the pH of a buffer solution prepared by mixing $100\,\mathrm{cm}^3$ of $0.200\,\mathrm{mol/dm}^3$ ethanoic acid ($K_a = 1.74 \times 10^{-5}$) with $50.0\,\mathrm{cm}^3$ of $0.100\,\mathrm{mol/dm}^3$ $\mathrm{NaOH}$.

Solution:

Step 1: Calculate moles of acid and base.

$n(\mathrm{CH}_3\mathrm{COOH}) = 0.200 \times 0.100 = 0.0200\,\mathrm{mol}$

$n(\mathrm{NaOH}) = 0.100 \times 0.0500 = 0.00500\,\mathrm{mol}$

Step 2: The NaOH partially neutralises the ethanoic acid:

$\mathrm{CH}_3\mathrm{COOH} + \mathrm{OH}^- \to \mathrm{CH}_3\mathrm{COO}^- + \mathrm{H}_2\mathrm{O}$

Remaining $\mathrm{CH}_3\mathrm{COOH} = 0.0200 - 0.00500 = 0.0150\,\mathrm{mol}$

$\mathrm{CH}_3\mathrm{COO}^-$ formed = $0.00500\,\mathrm{mol}$

Step 3: Calculate concentrations in the total volume ($150\,\mathrm{cm}^3 = 0.150\,\mathrm{dm}^3$).

$[\mathrm{HA}] = 0.0150/0.150 = 0.100\,\mathrm{mol/dm}^3$

$[\mathrm{A}^-] = 0.00500/0.150 = 0.0333\,\mathrm{mol/dm}^3$

Step 4: Apply the Henderson-Hasselbalch equation.

$$\mathrm{pH} = \mathrm{p}K_a + \log\frac◆LB◆[\mathrm{A}^-]◆RB◆◆LB◆[\mathrm{HA}]◆RB◆ = 4.76 + \log\frac{0.0333}{0.100} = 4.76 + \log(0.333) = 4.76 - 0.478 = 4.28$$

Problem 6

Explain why a mixture of $\mathrm{HCl}$ and $\mathrm{NaCl}$ in water does not function as an effective buffer, whereas a mixture of $\mathrm{CH}_3\mathrm{COOH}$ and $\mathrm{CH}_3\mathrm{COONa}$ does.

Solution:

A buffer requires a weak acid and its conjugate base (or a weak base and its conjugate acid). The buffer action relies on the equilibrium:

$$\mathrm{HA} \rightleftharpoons \mathrm{H}^+ + \mathrm{A}^-$$

When acid is added, $\mathrm{A}^-$ consumes the added $\mathrm{H}^+$, shifting the equilibrium left and minimising pH change. When base is added, HA donates $\mathrm{H}^+$ to neutralise the added $\mathrm{OH}^-$, shifting the equilibrium right.

$\mathrm{HCl}$ and $\mathrm{NaCl}$: $\mathrm{HCl}$ is a strong acid that dissociates completely. $\mathrm{Cl}^-$ is the conjugate base of a strong acid and has negligible basicity ($\mathrm{Cl}^-$ does not accept protons appreciably). When $\mathrm{H}^+$ is added, there is no mechanism to consume it (no weak base present). The solution's pH changes dramatically with small additions of acid or base.

$\mathrm{CH}_3\mathrm{COOH}$ and $\mathrm{CH}_3\mathrm{COONa}$: Ethanoic acid is a weak acid ($K_a = 1.74 \times 10^{-5}$) that exists in equilibrium with its conjugate base ($\mathrm{CH}_3\mathrm{COO}^-$). Added $\mathrm{H}^+$ is consumed by $\mathrm{CH}_3\mathrm{COO}^-$; added $\mathrm{OH}^-$ is consumed by $\mathrm{CH}_3\mathrm{COOH}$. The pH changes only slightly because the ratio $[\mathrm{A}^-]/[\mathrm{HA}]$ changes only slightly.

Advanced Acid-Base Calculations

pH of Weak Acids: Beyond the Approximation

The standard approximation for weak acid pH assumes $[\mathrm{H}^+] \ll [\mathrm{HA}]_0$. When this assumption fails (very dilute or very strong weak acids), the quadratic formula must be used.

Worked Example: Calculate the pH of a $1.00 \times 10^{-4}\,\mathrm{mol\,dm^{-3}}$ solution of ethanoic acid ($K_a = 1.74 \times 10^{-5}$).

Let $x = [\mathrm{H}^+] = [\mathrm{CH}_3\mathrm{COO}^-]$.

$K_a = \frac◆LB◆x^2◆RB◆◆LB◆1.00 \times 10^{-4} - x◆RB◆ = 1.74 \times 10^{-5}$

$x^2 = 1.74 \times 10^{-5}(1.00 \times 10^{-4} - x) = 1.74 \times 10^{-9} - 1.74 \times 10^{-5}x$

$x^2 + 1.74 \times 10^{-5}x - 1.74 \times 10^{-9} = 0$

Using the quadratic formula:

$x = \frac◆LB◆-1.74 \times 10^{-5} + \sqrt◆LB◆(1.74 \times 10^{-5})^2 + 4(1.74 \times 10^{-9})◆RB◆◆RB◆◆LB◆2◆RB◆$

$x = \frac◆LB◆-1.74 \times 10^{-5} + \sqrt◆LB◆3.03 \times 10^{-10} + 6.96 \times 10^{-9}◆RB◆◆RB◆◆LB◆2◆RB◆ = \frac◆LB◆-1.74 \times 10^{-5} + \sqrt◆LB◆7.26 \times 10^{-9}◆RB◆◆RB◆◆LB◆2◆RB◆$

$x = \frac◆LB◆-1.74 \times 10^{-5} + 8.52 \times 10^{-5}◆RB◆◆LB◆2◆RB◆ = \frac◆LB◆6.78 \times 10^{-5}◆RB◆◆LB◆2◆RB◆ = 3.39 \times 10^{-5}\,\mathrm{mol\,dm^{-3}}$

$\mathrm{pH} = -\log(3.39 \times 10^{-5}) = 4.47$

Check: $\frac◆LB◆3.39 \times 10^{-5}◆RB◆◆LB◆1.00 \times 10^{-4}◆RB◆ \times 100 = 33.9\%$ dissociation. Since this exceeds 5%, the approximation was not valid and the quadratic solution was necessary.

Buffer Capacity and pH Range

A buffer is most effective when $\mathrm{pH} = \mathrm{p}K_a$ (where $[\mathrm{HA}] = [\mathrm{A}^-]$). The useful range is $\mathrm{p}K_a \pm 1$.

Worked Example: How many moles of $\mathrm{NaOH}$ must be added to $500\,\mathrm{cm}^3$ of $0.200\,\mathrm{mol\,dm^{-3}}$ ethanoic acid ($\mathrm{p}K_a = 4.76$) to produce a buffer with pH $= 5.00$?

$\mathrm{pH} = \mathrm{p}K_a + \log\frac◆LB◆[\mathrm{A}^-]◆RB◆◆LB◆[\mathrm{HA}]◆RB◆$

$5.00 = 4.76 + \log\frac◆LB◆[\mathrm{A}^-]◆RB◆◆LB◆[\mathrm{HA}]◆RB◆$

$\log\frac◆LB◆[\mathrm{A}^-]◆RB◆◆LB◆[\mathrm{HA}]◆RB◆ = 0.24$

$\frac◆LB◆[\mathrm{A}^-]◆RB◆◆LB◆[\mathrm{HA}]◆RB◆ = 10^{0.24} = 1.74$

Initial moles of ethanoic acid: $n = 0.200 \times 0.500 = 0.100\,\mathrm{mol}$

Let $x$ = moles of $\mathrm{NaOH}$ added. Then $[\mathrm{A}^-] = x$ and $[\mathrm{HA}] = 0.100 - x$.

$\frac{x}{0.100 - x} = 1.74$

$x = 1.74(0.100 - x) = 0.174 - 1.74x$

$2.74x = 0.174$

$x = 0.0635\,\mathrm{mol}$

So $0.0635\,\mathrm{mol}$ of $\mathrm{NaOH}$ must be added. This converts $0.0635\,\mathrm{mol}$ of ethanoic acid to sodium ethanoate, leaving $0.0365\,\mathrm{mol}$ of ethanoic acid unreacted.

pH Curves: Titration of a Weak Acid with a Strong Base

The titration of ethanoic acid with $\mathrm{NaOH}$ produces a characteristic S-shaped pH curve:

Key regions:

1. Initial pH: pH of the weak acid (higher than for a strong acid of the same concentration because the weak acid is only partially dissociated).

2. Buffer region: After some $\mathrm{NaOH}$ has been added, the solution contains both $\mathrm{CH}_3\mathrm{COOH}$ and $\mathrm{CH}_3\mathrm{COO}^-$, forming a buffer. The pH changes slowly in this region.

3. Half-equivalence point: When half the acid has been neutralised, $[\mathrm{HA}] = [\mathrm{A}^-]$ and $\mathrm{pH} = \mathrm{p}K_a$. This is a useful experimental method for determining $\mathrm{p}K_a$.

4. Equivalence point: pH $> 7$ (alkaline) because the salt of a weak acid and strong base hydrolyses: $\mathrm{CH}_3\mathrm{COO}^- + \mathrm{H}_2\mathrm{O} \rightleftharpoons \mathrm{CH}_3\mathrm{COOH} + \mathrm{OH}^-$.

5. Beyond equivalence: pH is determined by the excess $\mathrm{OH}^-$.

Worked Example: Calculate the pH at the equivalence point when $25.0\,\mathrm{cm}^3$ of $0.100\,\mathrm{mol\,dm^{-3}}$ $\mathrm{NaOH}$ is added to $25.0\,\mathrm{cm}^3$ of $0.100\,\mathrm{mol\,dm^{-3}}$ ethanoic acid ($K_a = 1.74 \times 10^{-5}$).

At equivalence: $n(\mathrm{CH}_3\mathrm{COO}^-) = 0.100 \times 0.0250 = 2.50 \times 10^{-3}\,\mathrm{mol}$

Total volume $= 50.0\,\mathrm{cm}^3 = 0.0500\,\mathrm{dm}^3$

$[\mathrm{CH}_3\mathrm{COO}^-] = \frac◆LB◆2.50 \times 10^{-3}◆RB◆◆LB◆0.0500◆RB◆ = 0.0500\,\mathrm{mol\,dm^{-3}}$

The ethanoate ion hydrolyses:

$\mathrm{CH}_3\mathrm{COO}^- + \mathrm{H}_2\mathrm{O} \rightleftharpoons \mathrm{CH}_3\mathrm{COOH} + \mathrm{OH}^-$

$K_b = \frac{K_w}{K_a} = \frac◆LB◆1.00 \times 10^{-14}◆RB◆◆LB◆1.74 \times 10^{-5}◆RB◆ = 5.75 \times 10^{-10}$

$K_b = \frac◆LB◆[\mathrm{OH}^-]^2◆RB◆◆LB◆[\mathrm{CH}_3\mathrm{COO}^-]◆RB◆ = \frac{x^2}{0.0500} = 5.75 \times 10^{-10}$

$x = \sqrt◆LB◆5.75 \times 10^{-10} \times 0.0500◆RB◆ = 5.37 \times 10^{-6}\,\mathrm{mol\,dm^{-3}}$

$\mathrm{pOH} = -\log(5.37 \times 10^{-6}) = 5.27$

$\mathrm{pH} = 14 - 5.27 = 8.73$

Choosing an Indicator

An indicator is chosen so that its colour change range overlaps with the steep part of the titration curve at the equivalence point.

Titration type	Equivalence point pH	Suitable indicator
Strong acid / strong base	7.0	Bromothymol blue (6.0--7.6)
Weak acid / strong base	8--9	Phenolphthalein (8.2--10.0)
Strong acid / weak base	4--6	Methyl orange (3.1--4.4)
Weak acid / weak base	$\approx 7$	None ideal; use pH meter

Polyprotic Acids

Sulfuric acid ($\mathrm{H}_2\mathrm{SO}_4$) is a diprotic acid:

• First dissociation: complete ($\mathrm{H}_2\mathrm{SO}_4 \to \mathrm{H}^+ + \mathrm{HSO}_4^-$, strong acid).
• Second dissociation: partial ($\mathrm{HSO}_4^- \rightleftharpoons \mathrm{H}^+ + \mathrm{SO}_4^{2-}$, $K_{a2} = 1.02 \times 10^{-2}$).

Worked Example: Calculate the pH of $0.050\,\mathrm{mol\,dm^{-3}}$ $\mathrm{H}_2\mathrm{SO}_4$.

The first dissociation gives [\mathrm{H}^+] = 0.050\,\mathrm{mol\,dm^{-3} and [\mathrm{HSO}_4^-] = 0.050\,\mathrm{mol\,dm^{-3}.

For the second dissociation: let $x$ be the additional $[\mathrm{H}^+]$ from $\mathrm{HSO}_4^-$.

$K_{a2} = \frac{(0.050 + x)(x)}{0.050 - x} = 1.02 \times 10^{-2}$

Approximation: $0.050 + x \approx 0.050$, $0.050 - x \approx 0.050$:

$x = 1.02 \times 10^{-2} = 0.0102\,\mathrm{mol\,dm^{-3}}$

Check: $\frac{0.0102}{0.050} \times 100 = 20.4\%$ -- the approximation is marginal. For greater accuracy, use the quadratic formula. However, for A-Level purposes:

$[\mathrm{H}^+]_\text{total} \approx 0.050 + 0.010 = 0.060\,\mathrm{mol\,dm^{-3}}$

$\mathrm{pH} = -\log(0.060) = 1.22$

Exam-Style Questions with Full Mark Schemes

Q1 (5 marks)

Calculate the pH of $0.050\,\mathrm{mol\,dm^{-3}}$ $\mathrm{NaOH}$. Then calculate the pH after $10.0\,\mathrm{cm}^3$ of this $\mathrm{NaOH}$ is added to $25.0\,\mathrm{cm}^3$ of $0.100\,\mathrm{mol\,dm^{-3}}$ $\mathrm{HCl}$.

Mark Scheme:

Part 1: $\mathrm{NaOH}$ is a strong base: $[\mathrm{OH}^-] = 0.050\,\mathrm{mol\,dm^{-3}}$.

$\mathrm{pOH} = -\log(0.050) = 1.30$

$\mathrm{pH} = 14 - 1.30 = 12.70$ (1 mark).

Part 2: $n(\mathrm{HCl}) = 0.100 \times 0.0250 = 2.50 \times 10^{-3}\,\mathrm{mol}$

$n(\mathrm{NaOH}) = 0.050 \times 0.0100 = 5.00 \times 10^{-4}\,\mathrm{mol}$

$\mathrm{NaOH}$ is the limiting reagent. After reaction: $n(\mathrm{HCl})_\text{excess} = 2.50 \times 10^{-3} - 5.00 \times 10^{-4} = 2.00 \times 10^{-3}\,\mathrm{mol}$

Total volume $= 35.0\,\mathrm{cm}^3 = 0.0350\,\mathrm{dm}^3$

$[\mathrm{H}^+] = \frac◆LB◆2.00 \times 10^{-3}◆RB◆◆LB◆0.0350◆RB◆ = 0.0571\,\mathrm{mol\,dm^{-3}}$

$\mathrm{pH} = -\log(0.0571) = 1.24$ (1 mark for moles, 1 mark for excess, 1 mark for concentration, 1 mark for pH.)

Q2 (6 marks)

A buffer solution is prepared by adding $0.100\,\mathrm{mol}$ of sodium ethanoate to $250\,\mathrm{cm}^3$ of $0.400\,\mathrm{mol\,dm^{-3}}$ ethanoic acid ($\mathrm{p}K_a = 4.76$).

(a) Calculate the pH of the buffer. (3 marks)

(b) Calculate the pH after $1.00\,\mathrm{cm}^3$ of $1.00\,\mathrm{mol\,dm^{-3}}$ $\mathrm{HCl}$ is added to $25.0\,\mathrm{cm}^3$ of the buffer. (3 marks)

Mark Scheme:

(a) $n(\mathrm{CH}_3\mathrm{COOH}) = 0.400 \times 0.250 = 0.100\,\mathrm{mol}$ (1 mark).

$n(\mathrm{CH}_3\mathrm{COO}^-) = 0.100\,\mathrm{mol}$

$\mathrm{pH} = \mathrm{p}K_a + \log\frac◆LB◆[\mathrm{CH}_3\mathrm{COO}^-]◆RB◆◆LB◆[\mathrm{CH}_3\mathrm{COOH}]◆RB◆ = 4.76 + \log\frac{0.100}{0.100} = 4.76 + 0 = 4.76$ (1 mark for expression, 1 mark for answer.)

(b) $n(\mathrm{HCl}) = 1.00 \times 0.001 = 1.00 \times 10^{-3}\,\mathrm{mol}$

In $25.0\,\mathrm{cm}^3$ of buffer: $n(\mathrm{CH}_3\mathrm{COOH}) = \frac{0.100}{250} \times 25 = 0.0100\,\mathrm{mol}$

$n(\mathrm{CH}_3\mathrm{COO}^-) = 0.0100\,\mathrm{mol}$

After adding $\mathrm{HCl}$: $n(\mathrm{CH}_3\mathrm{COOH}) = 0.0100 + 0.001 = 0.0110\,\mathrm{mol}$

$n(\mathrm{CH}_3\mathrm{COO}^-) = 0.0100 - 0.001 = 0.0090\,\mathrm{mol}$

$\mathrm{pH} = 4.76 + \log\frac{0.0090}{0.0110} = 4.76 + \log(0.818) = 4.76 - 0.087 = 4.67$

(1 mark for calculating moles after reaction, 1 mark for expression, 1 mark for answer.)

Q3 (5 marks)

Define the term $\mathrm{p}K_a$. Explain why the $\mathrm{p}K_a$ of chloroethanoic acid ($\mathrm{CH}_2\mathrm{ClCOOH}$, $\mathrm{p}K_a = 2.86$) is lower than that of ethanoic acid ($\mathrm{CH}_3\mathrm{COOH}$, $\mathrm{p}K_a = 4.76$).

Mark Scheme:

$\mathrm{p}K_a = -\log K_a$, where $K_a$ is the acid dissociation constant (1 mark). A lower $\mathrm{p}K_a$ means a stronger acid (greater dissociation).

Chlorine is more electronegative than hydrogen, so it withdraws electron density from the carboxyl group through the inductive effect (1 mark). This destabilises the undissociated acid (makes the O--H bond more polar and easier to break) (1 mark) and stabilises the carboxylate anion by delocalising the negative charge more effectively (1 mark). The net effect is to increase $K_a$ (and decrease $\mathrm{p}K_a$), making chloroethanoic acid a stronger acid than ethanoic acid (1 mark).

Q4 (4 marks)

Explain why the equivalence point in the titration of ethanoic acid with sodium hydroxide has a pH greater than 7.

Mark Scheme:

At the equivalence point, all the ethanoic acid has been converted to sodium ethanoate (2 marks). The ethanoate ion is the conjugate base of a weak acid and therefore hydrolyses in water: $\mathrm{CH}_3\mathrm{COO}^- + \mathrm{H}_2\mathrm{O} \rightleftharpoons \mathrm{CH}_3\mathrm{COOH} + \mathrm{OH}^-$ (1 mark). The production of $\mathrm{OH}^-$ makes the solution alkaline, so the pH is greater than 7 (1 mark).

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tip

Diagnostic Test Ready to test your understanding of Acids, Bases and Buffers? The diagnostic test contains the hardest questions within the A-Level specification for this topic, each with a full worked solution.

Unit tests probe edge cases and common misconceptions. Integration tests combine Acids, Bases and Buffers with other chemistry topics to test synthesis under exam conditions.

See Diagnostic Guide for instructions on self-marking and building a personal test matrix.