Acids and Bases | Highers - Wyatt's Notes

Acids and Bases

Higher Acids and Bases

Definitions

Arrhenius: An acid produces $\mathrm{H^+$ ions in solution; a base produces $\mathrm{OH^-$ ions.

Bronsted-Lowry: An acid is a proton ( $\mathrm{H^+$ ) donor; a base is a proton acceptor.

Conjugate pairs: When an acid donates a proton, the remaining species is its conjugate base.

$\mathrm{HA + \mathrm{B \rightleftharpoons \mathrm{A^- + \mathrm{BH^+$

$\mathrm{HA/A^-$ and $\mathrm{B/BH^+$ are conjugate acid-base pairs.

Example: Identify the conjugate acid-base pairs in:

$\mathrm{NH_3 + \mathrm{H_2\mathrm{O \rightleftharpoons \mathrm{NH_4^+ + \mathrm{OH^-$

$\mathrm{NH_3/\mathrm{NH_4^+$ (base/conjugate acid) and $\mathrm{H_2\mathrm{O/\mathrm{OH^-$ (acid/conjugate base).

Worked Example 1: Identify the conjugate acid-base pairs in the reaction of $\mathrm{HSO_4^-$ With $\mathrm{H_2\mathrm{O$ :

$\mathrm{HSO_4^- + \mathrm{H_2\mathrm{O \rightleftharpoons \mathrm{SO_4^{2-} + \mathrm{H_3\mathrm{O^+$

$\mathrm{HSO_4^-/\mathrm{SO_4^{2-}$ (acid/conjugate base) and $\mathrm{H_2\mathrm{O/\mathrm{H_3\mathrm{O^+$ (base/conjugate acid). Note that $\mathrm{HSO_4^-$ is Acting as an acid (donating a proton) and $\mathrm{H_2\mathrm{O$ is acting as a base (accepting a Proton).

Strong and Weak Acids

Strong acids are completely dissociated in aqueous solution.

$\mathrm{HCl \to \mathrm{H^+ + \mathrm{Cl^-$

Common strong acids: $\mathrm{HCl$$\mathrm{HNO_3$$\mathrm{H_2\mathrm{SO_4$ (first dissociation), $\mathrm{HClO_4$ .

Weak acids are partially dissociated in aqueous solution.

$\mathrm{CH_3\mathrm{COOH \rightleftharpoons \mathrm{CH_3\mathrm{COO^- + \mathrm{H^+$

Common weak acids: $\mathrm{CH_3\mathrm{COOH$$\mathrm{H_2\mathrm{CO_3$$\mathrm{HF$ $\mathrm{H_3\mathrm{PO_4$ .

Comparison of Strong and Weak Acids

Property	Strong acid	Weak acid
Dissociation	Complete	Partial
Equilibrium	Not established	Dynamic equilibrium
pH at 0.1 M	1.0	Approximately 2.9
Conductivity	High	Lower
Reaction rate with Mg	Faster	Slower (same at same $[\mathrm{H^+]$ )

Key misconception: Concentration and strength are independent. A 0.001 M strong acid and a 0.001 M weak acid have the same concentration but different $[\mathrm{H^+]$ .

The pH Scale

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

Where $[\mathrm{H^+]$ is the concentration of hydrogen ions in mol/L.

At $25°C$ : $\mathrm{pH = 7$ is neutral, $\mathrm{pH < 7$ is acidic, $\mathrm{pH > 7$ is alkaline.

Worked Example 2: Find the pH of $0.05 \mathrm{ M$ $\mathrm{HNO_3$ .

$\mathrm{pH = -\log_{10}(0.05) = 1.30$

Worked Example 3: Find $[\mathrm{H^+]$ for a solution of pH 3.40.

$[\mathrm{H^+] = 10^{-3.40} = 3.98 \times 10^{-4} \mathrm{ mol/L$

Worked Example 4: Find the pH of $0.005 \mathrm{ M$ $\mathrm{H_2\mathrm{SO_4$ (assume complete Dissociation of the first proton and ignore the second).

$[\mathrm{H^+] = 0.005 \mathrm{ M$

$\mathrm{pH = -\log_{10}(0.005) = 2.30$

Water and the Ionic Product

Water undergoes autoionisation:

$\mathrm{H_2\mathrm{O \rightleftharpoons \mathrm{H^+ + \mathrm{OH^-$

$K_w = [\mathrm{H^+][\mathrm{OH^-] = 1.0 \times 10^{-14} \mathrm{ mol^2\mathrm{L^{-2} \quad \mathrm{at 25°C$

Derivation of $K_w$ :

From the autoionisation equilibrium:

$K_w = [\mathrm{H^+][\mathrm{OH^-]$

In pure water at $25°C$ : $[\mathrm{H^+] = [\mathrm{OH^-] = 10^{-7} \mathrm{ M$ So $K_w = 10^{-7} \times 10^{-7} = 10^{-14}$ .

$K_w$ is temperature-dependent. At higher temperatures, more water molecules dissociate, so $K_w$ Increases. This means the pH of pure water decreases with temperature, but the water remains neutral (since $[\mathrm{H^+] = [\mathrm{OH^-]$ ).

Worked Example 5: Find the pH of $0.02 \mathrm{ M$ $\mathrm{NaOH$ .

$[\mathrm{OH^-] = 0.02 \mathrm{ M$

$[\mathrm{H^+] = \frac{K_w}{[\mathrm{OH^-]} = \frac{1.0 \times 10^{-14}}{0.02} = 5.0 \times 10^{-13} \mathrm{ M$

$\mathrm{pH = -\log_{10}(5.0 \times 10^{-13}) = 12.30$

Acid Dissociation Constant ( $K_a$ )

For a weak acid $\mathrm{HA \rightleftharpoons \mathrm{H^+ + \mathrm{A^-$ :

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

$pK_a = -\log_{10} K_a$

The lower the $pK_a$ The stronger the acid.

Worked Example 6: Ethanoic acid has $K_a = 1.74 \times 10^{-5} \mathrm{ mol/L$ . Find the pH of a $0.10 \mathrm{ M$ solution.

$K_a = \frac{[\mathrm{H^+][\mathrm{A^-]}{[\mathrm{HA]} = \frac{[\mathrm{H^+]^2}{0.10 - [\mathrm{H^+]} \approx \frac{[\mathrm{H^+]^2}{0.10}$

$[\mathrm{H^+] = \sqrt{1.74 \times 10^{-5} \times 0.10} = \sqrt{1.74 \times 10^{-6}} = 1.32 \times 10^{-3} \mathrm{ M$

$\mathrm{pH = -\log_{10}(1.32 \times 10^{-3}) = 2.88$

Worked Example 7: A weak acid $\mathrm{HX$ has $K_a = 4.2 \times 10^{-4}$ . Find the pH of a $0.25 \mathrm{ M$ solution and the percentage dissociation.

$[\mathrm{H^+] = \sqrt{4.2 \times 10^{-4} \times 0.25} = \sqrt{1.05 \times 10^{-4}} = 1.025 \times 10^{-2} \mathrm{ M$

$\mathrm{pH = -\log_{10}(1.025 \times 10^{-2}) = 1.99$

$\%\mathrm{ dissociation = \frac{1.025 \times 10^{-2}}{0.25} \times 100 = 4.1\%$

Base Dissociation Constant ( $K_b$ )

For a weak base $\mathrm{B + \mathrm{H_2\mathrm{O \rightleftharpoons \mathrm{BH^+ + \mathrm{OH^-$ :

$K_b = \frac{[\mathrm{BH^+][\mathrm{OH^-]}{[\mathrm{B]}$

Relationship:

$K_a \times K_b = K_w$

Proof: For a conjugate pair $\mathrm{HA/A^-$ :

$K_a = \frac{[\mathrm{H^+][\mathrm{A^-]}{[\mathrm{HA]} \quad \mathrm{and \quad K_b = \frac{[\mathrm{HA][\mathrm{OH^-]}{[\mathrm{A^-]}$

$K_a \times K_b = \frac{[\mathrm{H^+][\mathrm{A^-]}{[\mathrm{HA]} \times \frac{[\mathrm{HA][\mathrm{OH^-]}{[\mathrm{A^-]} = [\mathrm{H^+][\mathrm{OH^-] = K_w$

pH Calculations for Weak Bases

Worked Example 8: Ammonia has $K_b = 1.78 \times 10^{-5} \mathrm{ mol/L$ . Find the pH of a $0.15 \mathrm{ M$ solution.

$[\mathrm{OH^-] = \sqrt{K_b \times [\mathrm{B]} = \sqrt{1.78 \times 10^{-5} \times 0.15} = \sqrt{2.67 \times 10^{-6}} = 1.63 \times 10^{-3} \mathrm{ M$

$\mathrm{pOH = -\log_{10}(1.63 \times 10^{-3}) = 2.79$

$\mathrm{pH = 14 - 2.79 = 11.21$

Buffers

What is a Buffer?

A buffer solution resists changes in pH when small amounts of acid or base are added. It Consists of a weak acid and its conjugate base (or a weak base and its conjugate acid).

Acidic buffer: Weak acid ( $\mathrm{HA$ ) + salt of weak acid ( $\mathrm{A^-$ ).

Example: Ethanoic acid + sodium ethanoate.

Basic buffer: Weak base ( $\mathrm{B$ ) + salt of weak base ( $\mathrm{BH^+$ ).

Example: Ammonia + ammonium chloride.

Henderson-Hasselbalch Equation

$\mathrm{pH = pK_a + \log_{10}\left(\frac{[\mathrm{A^-]}{[\mathrm{HA]}\right)$

Derivation:

Starting from the acid dissociation expression:

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

Rearranging: $[\mathrm{H^+] = K_a \times \frac{[\mathrm{HA]}{[\mathrm{A^-]}$

Taking $-\log_{10}$ of both sides:

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

$\mathrm{pH = pK_a + \log\frac{[\mathrm{A^-]}{[\mathrm{HA]}$

Worked Example 9: Calculate the pH of a buffer containing $0.20 \mathrm{ M$ ethanoic acid ( $pK_a = 4.76$ ) and $0.15 \mathrm{ M$ sodium ethanoate.

$\mathrm{pH = 4.76 + \log_{10}\left(\frac{0.15}{0.20}\right) = 4.76 + \log_{10}(0.75) = 4.76 - 0.125 = 4.64$

Worked Example 10: Prepare a buffer at pH 5.00 using ethanoic acid ( $pK_a = 4.76$ ) and sodium Ethanoate. If the total concentration is $0.30 \mathrm{ M$ Find the concentrations of each Component.

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

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

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

Let $[\mathrm{HA] = x$ Then $[\mathrm{A^-] = 1.74x$ .

$x + 1.74x = 0.30 \implies 2.74x = 0.30 \implies x = 0.109 \mathrm{ M$

$[\mathrm{HA] = 0.109 \mathrm{ M$ , $[\mathrm{A^-] = 0.191 \mathrm{ M$ .

Buffer Capacity

The buffer capacity depends on:

The absolute concentrations of the weak acid and conjugate base (higher concentrations = greater capacity)
The ratio $[\mathrm{A^-]/[\mathrm{HA]$ (most effective when this ratio is close to 1, i.e., pH near $pK_a$ )

Worked Example 11: A buffer contains $0.10 \mathrm{ M$ $\mathrm{NH_3$ and $0.15 \mathrm{ M$ $\mathrm{NH_4\mathrm{Cl$ . Calculate its pH and the pH after adding $0.01 \mathrm{ mol$ of $\mathrm{HCl$ To $1 \mathrm{ L$ of the buffer.

First, find $K_a$ for $\mathrm{NH_4^+$ :

$K_a = \frac{K_w}{K_b} = \frac{1.0 \times 10^{-14}}{1.78 \times 10^{-5}} = 5.62 \times 10^{-10}$

$pK_a = 9.25$

$\mathrm{pH = 9.25 + \log\frac{0.10}{0.15} = 9.25 - 0.176 = 9.07$

After adding $0.01 \mathrm{ mol$ $\mathrm{HCl$ :

$\mathrm{NH_3$ reacts with $\mathrm{H^+$ : $[\mathrm{NH_3]$ decreases by $0.01$ and $[\mathrm{NH_4^+]$ Increases by $0.01$ .

$[\mathrm{NH_3] = 0.10 - 0.01 = 0.09 \mathrm{ M$ $[\mathrm{NH_4^+] = 0.15 + 0.01 = 0.16 \mathrm{ M$

$\mathrm{pH = 9.25 + \log\frac{0.09}{0.16} = 9.25 + \log(0.5625) = 9.25 - 0.250 = 9.00$

The pH changes by only 0.07 units, demonstrating the buffer”s effectiveness.

Titrations

Strong Acid-Strong Base Titration

Equivalence point at pH 7.

Worked Example 12: $25.0 \mathrm{ cm^3$ of $0.10 \mathrm{ M$ $\mathrm{HCl$ is titrated with $0.10 \mathrm{ M$ $\mathrm{NaOH$ . Find the pH at the equivalence point.

At the equivalence point: moles of acid = moles of base.

$n = 0.10 \times 0.0250 = 0.00250 \mathrm{ mol$

Total volume = $50.0 \mathrm{ cm^3$ .

$[\mathrm{NaCl] = 0.00250/0.0500 = 0.0500 \mathrm{ M$ (neutral salt).

$\mathrm{pH = 7$

Strong Acid-Weak Base Titration

Equivalence point at pH < 7 (acidic).

Weak Acid-Strong Base Titration

Equivalence point at pH > 7 (alkaline).

Worked Example 13: $25.0 \mathrm{ cm^3$ of $0.10 \mathrm{ M$ $\mathrm{CH_3\mathrm{COOH$ ( $K_a = 1.74 \times 10^{-5}$ ) is titrated with $0.10 \mathrm{ M$ $\mathrm{NaOH$ . Find the pH at the Equivalence point.

Moles of $\mathrm{CH_3\mathrm{COO^-$ formed = $0.00250 \mathrm{ mol$ .

Total volume = $50.0 \mathrm{ cm^3$ .

$[\mathrm{CH_3\mathrm{COO^-] = 0.0500 \mathrm{ M$ .

The ethanoate ion hydrolyses:

$\mathrm{CH_3\mathrm{COO^- + \mathrm{H_2\mathrm{O \rightleftharpoons \mathrm{CH_3\mathrm{COOH + \mathrm{OH^-$

$K_b = \frac{K_w}{K_a} = \frac{1.0 \times 10^{-14}}{1.74 \times 10^{-5}} = 5.75 \times 10^{-10}$

$[\mathrm{OH^-] = \sqrt{K_b \times [\mathrm{CH_3\mathrm{COO^-]} = \sqrt{5.75 \times 10^{-10} \times 0.0500} = \sqrt{2.875 \times 10^{-11}} = 5.36 \times 10^{-6} \mathrm{ M$

$\mathrm{pOH = -\log_{10}(5.36 \times 10^{-6}) = 5.27$

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

Worked Example: pH During a Titration

Worked Example 14: $20.0 \mathrm{ cm^3$ of $0.15 \mathrm{ M$ $\mathrm{NH_3$ ( $K_b = 1.78 \times 10^{-5}$ ) is titrated with $0.10 \mathrm{ M$ $\mathrm{HCl$ . Find the pH after Adding $15.0 \mathrm{ cm^3$ of $\mathrm{HCl$ .

Moles of $\mathrm{NH_3 = 0.15 \times 0.0200 = 0.00300 \mathrm{ mol$ Moles of $\mathrm{HCl$ added $= 0.10 \times 0.0150 = 0.00150 \mathrm{ mol$

After reaction: $[\mathrm{NH_3]$ remaining $= 0.00300 - 0.00150 = 0.00150 \mathrm{ mol$ $[\mathrm{NH_4^+]$ formed $= 0.00150 \mathrm{ mol$

Total volume $= 35.0 \mathrm{ cm^3 = 0.0350 \mathrm{ L$

$[\mathrm{NH_3] = 0.00150/0.0350 = 0.0429 \mathrm{ M$ $[\mathrm{NH_4^+] = 0.00150/0.0350 = 0.0429 \mathrm{ M$

This is a buffer solution with equal concentrations, so:

$\mathrm{pH = pK_a + \log\frac{[\mathrm{NH_3]}{[\mathrm{NH_4^+]} = 9.25 + \log(1) = 9.25$

Indicators

An indicator is a weak acid where $\mathrm{HIn$ and $\mathrm{In^-$ have different colours.

$\mathrm{HIn \rightleftharpoons \mathrm{H^+ + \mathrm{In^-$

The indicator changes colour over approximately $\mathrm{pK_{\mathrm{In} \pm 1$ .

Indicator	pH Range	Colour Change
Methyl orange	3.1 - 4.4	Red to yellow
Bromothymol blue	6.0 - 7.6	Yellow to blue
Phenolphthalein	8.3 - 10.0	Colourless to pink

Choosing an Indicator

The indicator range must overlap with the steep part of the titration curve at the equivalence Point.

Titration type	Equivalence pH	Suitable indicator
Strong acid + strong base	pH = 7	Bromothymol blue
Strong acid + weak base	pH < 7	Methyl orange
Weak acid + strong base	pH > 7	Phenolphthalein

For a strong acid-weak base titration” — this is incorrect. Phenolphthalein (pH 8.3-10.0) is Suitable for weak acid-strong base titrations where the equivalence pH is above 7. For strong Acid-weak base titrations (equivalence pH below 7), methyl orange is the correct choice.

Polyprotic Acids

Diprotic Acids

Acids that can donate two protons, e.g., $\mathrm{H_2\mathrm{SO_4$ , $\mathrm{H_2\mathrm{CO_3$ $\mathrm{H_3\mathrm{PO_4$ (triprotic).

For carbonic acid:

$\mathrm{H_2\mathrm{CO_3 \rightleftharpoons \mathrm{H^+ + \mathrm{HCO_3^- \quad K_{a1} = 4.3 \times 10^{-7}$

$\mathrm{HCO_3^- \rightleftharpoons \mathrm{H^+ + \mathrm{CO_3^{2-} \quad K_{a2} = 4.8 \times 10^{-11}$

Note that $K_{a1} \gg K_{a2}$ : the first dissociation is much stronger than the second.

flowchart TD
    A[3_Acids Bases] --> B[Key Concepts]
    A --> C[Core Principles]
    A --> D[Practical Applications]
    B --> E[Fundamental definitions]
    C --> F[Design patterns]
    D --> G[Real-world usage]

Summary Table: Acid-Base Concepts

Concept	Formula/Definition	Key Point
pH	$\mathrm{pH = -\log[\mathrm{H^+]$	Lower pH = more acidic
$K_w$	$[\mathrm{H^+][\mathrm{OH^-] = 10^{-14}$	At 25°C
$K_a$	$\frac{[\mathrm{H^+][\mathrm{A^-]}{[\mathrm{HA]}$	Higher $K_a$ = stronger acid
$K_b$	$\frac{[\mathrm{BH^+][\mathrm{OH^-]}{[\mathrm{B]}$	Higher $K_b$ = stronger base
Henderson-Hasselbalch	$\mathrm{pH = pK_a + \log\frac{[\mathrm{A^-]}{[\mathrm{HA]}$	For buffer solutions
$K_a \times K_b$	$= K_w$	For conjugate pairs

Common Pitfalls

Strong vs. Weak acids: A strong acid is completely dissociated; a weak acid is partially dissociated. Concentration and strength are independent.
$[\mathrm{H^+]$ approximation: For weak acids, $[\mathrm{H^+] = \sqrt{K_a \times c}$ is valid only when $[\mathrm{H^+]$ is small compared to $c$ ( when $c/K_a > 100$ ).
pH of water: Pure water has pH 7 at $25°C$ But this changes with temperature because $K_w$ changes.
Buffer calculations: The Henderson-Hasselbalch equation uses concentrations, not moles (unless the acid and base are in the same volume).
Choosing an indicator: The indicator range must include the pH at the equivalence point.
Conjugate base strength: The weaker the acid, the stronger its conjugate base. $\mathrm{Cl^-$ is an extremely weak base because $\mathrm{HCl$ is a very strong acid.
Diprotic acids: The second dissociation is always weaker than the first. For $\mathrm{H_2\mathrm{SO_4$ The first proton dissociates completely but the second does not ( $K_{a2} = 1.2 \times 10^{-2}$ ).

Practice Questions

Calculate the pH of $0.005 \mathrm{ M$ $\mathrm{H_2\mathrm{SO_4$ (assume complete dissociation of the first proton and ignore the second).
A weak acid $\mathrm{HX$ has $K_a = 4.2 \times 10^{-4}$ . Find the pH of a $0.25 \mathrm{ M$ solution and the percentage dissociation.
Prepare a buffer at pH 5.00 using ethanoic acid ( $pK_a = 4.76$ ) and sodium ethanoate. If the total concentration is $0.30 \mathrm{ M$ Find the concentrations of each component.
$20.0 \mathrm{ cm^3$ of $0.15 \mathrm{ M$ $\mathrm{NH_3$ ( $K_b = 1.78 \times 10^{-5}$ ) is titrated with $0.10 \mathrm{ M$ $\mathrm{HCl$ . Find the pH after adding $15.0 \mathrm{ cm^3$ of $\mathrm{HCl$ .
Explain why the pH of a $0.01 \mathrm{ M$ solution of $\mathrm{HCl$ is 2.0, but the pH of a $0.01 \mathrm{ M$ solution of $\mathrm{CH_3\mathrm{COOH$ is approximately 3.4.
A buffer contains $0.10 \mathrm{ M$ $\mathrm{NH_3$ and $0.15 \mathrm{ M$ $\mathrm{NH_4\mathrm{Cl$ . Calculate its pH and the pH after adding $0.01 \mathrm{ mol$ of $\mathrm{HCl$ to $1 \mathrm{ L$ of the buffer.
Calculate $K_a$ for a $0.050 \mathrm{ M$ weak acid solution with pH 2.80.
Explain why phenolphthalein is a suitable indicator for a weak acid-strong base titration but methyl orange is suitable for a strong acid-weak base titration.
Calculate the pH of a $0.15 \mathrm{ M$ solution of $\mathrm{HF$ ( $K_a = 6.8 \times 10^{-4}$ ). Check whether the approximation $[\mathrm{H^+] \approx \sqrt{K_a \times c}$ is valid.
50 mL of $0.10 \mathrm{ M$ $\mathrm{NaOH$ is added to 50 mL of $0.10 \mathrm{ M$ $\mathrm{CH_3\mathrm{COOH$ . Calculate the pH of the resulting solution.
A student prepares a buffer by mixing $100 \mathrm{ mL$ of $0.20 \mathrm{ M$ ethanoic acid with $50 \mathrm{ mL$ of $0.20 \mathrm{ M$ $\mathrm{NaOH$ . Calculate the pH of the resulting buffer.
Explain why the pH of a solution of $\mathrm{NaCl$ is 7, but the pH of a solution of $\mathrm{NH_4\mathrm{Cl$ is less than 7.

Salt Hydrolysis

When a salt dissolves in water, its ions may react with water (hydrolyse), affecting the pH of the Solution.

Salts from Strong Acid + Strong Base

Example: $\mathrm{NaCl$ , $\mathrm{KNO_3$ .

Neither ion reacts with water. The solution is neutral (pH = 7).

Salts from Strong Acid + Weak Base

Example: $\mathrm{NH_4\mathrm{Cl$ .

The cation ( $\mathrm{NH_4^+$ ) is the conjugate acid of a weak base and hydrolyses:

$\mathrm{NH_4^+ + \mathrm{H_2\mathrm{O \rightleftharpoons \mathrm{NH_3 + \mathrm{H_3\mathrm{O^+$

This produces $\mathrm{H^+$ ions, making the solution acidic (pH < 7).

Worked Example 15: Calculate the pH of a $0.10 \mathrm{ M$ $\mathrm{NH_4\mathrm{Cl$ solution. ( $K_b(\mathrm{NH_3) = 1.78 \times 10^{-5}$ )

$K_a(\mathrm{NH_4^+) = \frac{K_w}{K_b} = \frac{1.0 \times 10^{-14}}{1.78 \times 10^{-5}} = 5.62 \times 10^{-10}$

$[\mathrm{H^+] = \sqrt{K_a \times c} = \sqrt{5.62 \times 10^{-10} \times 0.10} = \sqrt{5.62 \times 10^{-11}} = 7.50 \times 10^{-6} \mathrm{ M$

$\mathrm{pH = -\log(7.50 \times 10^{-6}) = 5.12$

Salts from Weak Acid + Strong Base

Example: $\mathrm{CH_3\mathrm{COONa$ .

The anion ( $\mathrm{CH_3\mathrm{COO^-$ ) is the conjugate base of a weak acid and hydrolyses:

$\mathrm{CH_3\mathrm{COO^- + \mathrm{H_2\mathrm{O \rightleftharpoons \mathrm{CH_3\mathrm{COOH + \mathrm{OH^-$

This produces $\mathrm{OH^-$ ions, making the solution alkaline (pH > 7).

Salts from Weak Acid + Weak Base

Example: $\mathrm{NH_4\mathrm{CH_3\mathrm{COO$ .

Both ions hydrolyse. The pH depends on the relative strengths of the acid and base:

If $K_a > K_b$ : solution is acidic
If $K_a < K_b$ : solution is alkaline
If $K_a \approx K_b$ : solution is approximately neutral

Summary Table: Salt Hydrolysis

Salt type	Hydrolysis	pH	Example
Strong acid + strong base	Neither ion hydrolyses	7	$\mathrm{NaCl$
Strong acid + weak base	Cation hydrolyses	< 7	$\mathrm{NH_4\mathrm{Cl$
Weak acid + strong base	Anion hydrolyses	> 7	$\mathrm{CH_3\mathrm{COONa$
Weak acid + weak base	Both hydrolyse	Depends	$\mathrm{NH_4\mathrm{CH_3\mathrm{COO$

pH of Mixtures

Mixing Two Strong Acids

$[\mathrm{H^+]_{\mathrm{total} = \frac{n_1 + n_2}{V_{\mathrm{total}}$

Mixing a Strong Acid and a Strong Base

Determine which is in excess, then calculate $[\mathrm{H^+]$ or $[\mathrm{OH^-]$ of the remaining Excess.

Worked Example 16: $50 \mathrm{ mL$ of $0.10 \mathrm{ M$ $\mathrm{NaOH$ is added to $50 \mathrm{ mL$ Of $0.10 \mathrm{ M$ $\mathrm{CH_3\mathrm{COOH$ . Calculate the pH.

Moles of $\mathrm{NaOH = 0.10 \times 0.050 = 0.0050 \mathrm{ mol$ Moles of $\mathrm{CH_3\mathrm{COOH = 0.10 \times 0.050 = 0.0050 \mathrm{ mol$

This is the equivalence point: all $\mathrm{CH_3\mathrm{COOH$ is converted to $\mathrm{CH_3\mathrm{COO^-$ .

$[\mathrm{CH_3\mathrm{COO^-] = 0.0050/0.100 = 0.050 \mathrm{ M$

$K_b = \frac{K_w}{K_a} = \frac{1.0 \times 10^{-14}}{1.74 \times 10^{-5}} = 5.75 \times 10^{-10}$

$[\mathrm{OH^-] = \sqrt{5.75 \times 10^{-10} \times 0.050} = 5.36 \times 10^{-6} \mathrm{ M$

$\mathrm{pOH = 5.27, \quad \mathrm{pH = 8.73$

Mixing Two Acids (One Strong, One Weak)

If both acids are present, the strong acid dominates the pH because it contributes far more $\mathrm{H^+$ than the weak acid.

Titration Curves

Shape of Titration Curves

Strong acid-strong base: The pH changes slowly at first, then rapidly near the equivalence point (almost vertical), then slowly again. The equivalence point is at pH 7.

Weak acid-strong base: The initial pH is higher than for a strong acid at the same Concentration. There is a buffering region in the middle of the curve (where the weak acid and its Conjugate base coexist). The equivalence point is at pH > 7.

Strong acid-weak base: The equivalence point is at pH < 7.

Worked Example 17: Sketch and describe the titration curve for $25 \mathrm{ mL$ of $0.10 \mathrm{ M$ $\mathrm{CH_3\mathrm{COOH$ titrated with $0.10 \mathrm{ M$ $\mathrm{NaOH$ .

Key features:

Initial pH: ~2.87 (weak acid)
Buffer region: From ~0 to ~20 mL added. The pH changes slowly because the solution acts as a buffer ( $\mathrm{CH_3\mathrm{COOH/\mathrm{CH_3\mathrm{COO^-$ ).
Half-equivalence point: At $12.5 \mathrm{ mL$ (half the acid neutralised), $\mathrm{pH = pK_a = 4.76$ .
Equivalence point: At $25 \mathrm{ mL$ PH = 8.73.
Beyond equivalence: pH approaches that of the excess $\mathrm{NaOH$ .

Solubility Product and pH

The solubility of some salts is affected by pH.

Example: $\mathrm{CaCO_3$ in Acid

$\mathrm{CaCO_3$ is more soluble in acidic solutions because the $\mathrm{CO_3^{2-}$ ion reacts with $\mathrm{H^+$ :

$\mathrm{CO_3^{2-} + \mathrm{H^+ \to \mathrm{HCO_3^-$

$\mathrm{HCO_3^- + \mathrm{H^+ \to \mathrm{H_2\mathrm{CO_3 \to \mathrm{CO_2 + \mathrm{H_2\mathrm{O$

This removes $\mathrm{CO_3^{2-}$ from the equilibrium, shifting the dissolution to the right (Le Chatelier’s principle).

Example: $\mathrm{Mg(OH)_2$ in Acid and Base

$\mathrm{Mg(OH)_2$ dissolves in acid because $\mathrm{OH^-$ is neutralised by $\mathrm{H^+$ :

$\mathrm{Mg(OH)_2\mathrm{(s) + 2\mathrm{H^+ \to \mathrm{Mg^{2+} + 2\mathrm{H_2\mathrm{O$

$\mathrm{Mg(OH)_2$ is insoluble in base (common ion effect: additional $\mathrm{OH^-$ suppresses Dissolution).

Practice Questions (Extended)

Calculate the pH of a $0.050 \mathrm{ M$ solution of $\mathrm{AlCl_3$ Given that $K_a(\mathrm{Al^{3+}) = 1.0 \times 10^{-5}$ . (Hint: $\mathrm{Al^{3+}$ acts as a weak acid.)
A buffer is prepared by mixing $0.20 \mathrm{ mol$ of $\mathrm{CH_3\mathrm{COOH$ and $0.10 \mathrm{ mol$ of $\mathrm{NaOH$ in $1.0 \mathrm{ L$ of solution. Calculate the pH of this buffer.
Explain qualitatively how the shape of the pH titration curve for a weak acid-strong base titration differs from that of a strong acid-strong base titration. Include a discussion of the buffer region and the half-equivalence point.
Calculate the pH at each of the following points in the titration of $25.0 \mathrm{ mL$ of $0.10 \mathrm{ M$ $\mathrm{HCl$ with $0.10 \mathrm{ M$ $\mathrm{NaOH$ : (a) 0 mL, (b) 12.5 mL, (c) 24.9 mL, (d) 25.0 mL, (e) 25.1 mL, (f) 30.0 mL.
Explain why $\mathrm{Na_2\mathrm{CO_3$ solution is alkaline, writing the relevant hydrolysis equations.
A student is asked to prepare a buffer at pH 7.40 (physiological pH) using $\mathrm{H_2\mathrm{PO_4^-$ / $\mathrm{HPO_4^{2-}$ ( $pK_{a2} = 7.21$ ). Calculate the required ratio $[\mathrm{HPO_4^{2-}]/[\mathrm{H_2\mathrm{PO_4^-]$ and suggest why this buffer system is used in biological systems.

Advanced pH Calculations

Very Dilute Strong Acids

When the concentration of a strong acid is very low (comparable to $10^{-7}$ M), the contribution of $\mathrm{H^+$ from water autoionisation becomes significant and cannot be ignored.

Worked Example 19: Find the pH of $1.0 \times 10^{-8}$ M HCl.

If we use $\mathrm{pH = -\log(10^{-8}) = 8$ This gives a basic pH for an acid solution, which Is wrong. We must account for the autoionisation of water.

$[\mathrm{H^+]_{\mathrm{total} = [\mathrm{H^+]_{\mathrm{HCl} + [\mathrm{H^+]_{\mathrm{water} = 1.0 \times 10^{-8} + [\mathrm{OH^-]$

From $K_w = [\mathrm{H^+][\mathrm{OH^-]$ :

$[\mathrm{H^+] \times ([\mathrm{H^+] - 1.0 \times 10^{-8}) = 1.0 \times 10^{-14}$

$[\mathrm{H^+]^2 - 1.0 \times 10^{-8}[\mathrm{H^+] - 1.0 \times 10^{-14} = 0$

Using the quadratic formula:

$[\mathrm{H^+] = \frac{1.0 \times 10^{-8} + \sqrt{(1.0 \times 10^{-8})^2 + 4 \times 10^{-14}}}{2}$

$= \frac{1.0 \times 10^{-8} + \sqrt{10^{-16} + 4 \times 10^{-14}}}{2} = \frac{1.0 \times 10^{-8} + \sqrt{4.01 \times 10^{-14}}}{2}$

$= \frac{1.0 \times 10^{-8} + 2.0025 \times 10^{-7}}{2} = \frac{2.1025 \times 10^{-7}}{2} = 1.051 \times 10^{-7} \mathrm{ M$

$\mathrm{pH = -\log(1.051 \times 10^{-7}) = 6.98$

This makes sense: the pH is very slightly below 7, consistent with a very dilute acid.

pH of a Polyprotic Acid

For $\mathrm{H_3\mathrm{PO_4$ :

$K_{a1} = 7.5 \times 10^{-3}, \quad K_{a2} = 6.2 \times 10^{-8}, \quad K_{a3} = 4.8 \times 10^{-13}$

Since $K_{a1} \gg K_{a2} \gg K_{a3}$ The first dissociation dominates. For a $0.10 \mathrm{ M$ Solution, the pH is calculated using only $K_{a1}$ .

Indicators in Detail: Colour Change Theory

An indicator is a weak acid $\mathrm{HIn$ where the protonated and deprotonated forms have different Colours:

$\mathrm{HIn \rightleftharpoons \mathrm{H^+ + \mathrm{In^-$

$K_{\mathrm{In} = \frac{[\mathrm{H^+][\mathrm{In^-]}{[\mathrm{HIn]}$

The human eye sees the acid colour when $[\mathrm{HIn]/[\mathrm{In^-] \gt 10$ and the base colour when $[\mathrm{In^-]/[\mathrm{HIn] \gt 10$ . The indicator changes colour over approximately $\mathrm{pK_{\mathrm{In} \pm 1$ .

Worked Examples

Example 1: pH calculation

Calculate the pH of a $0.050\,\text{mol\,dm}^{-3}$ solution of HCl.

Solution:

HCl is a strong acid, so $[\text{H}^+] = 0.050\,\text{mol\,dm}^{-3}$ .

$\text{pH} = -\log_{10}[\text{H}^+] = -\log_{10}(0.050) = 1.30$

Intuition

Chemistry explains how atoms combine to form the substances that make up everything around you. Bonding is about how electrons are shared or transferred between atoms, determining properties like melting point and conductivity. Chemical reactions are rearrangements of atoms, where old bonds break and new ones form. The mole concept bridges the atomic world and the laboratory, letting you predict exactly how much product a reaction will produce. Understanding these principles lets you predict the behavior of matter before you even enter the lab.

Cross-References

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