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 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.

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^+].

\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 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{ MSo $K_w = 10^{-7} \times 10^{-7} = 10^{-14}$.

$K_w$ 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

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\%

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

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

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.

\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{ MFind 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] = xThen [\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.

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.

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

Equivalence point at pH < 7 (acidic).

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 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

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.

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

:::caution The original question 8 in the practice section stated that “phenolphthalein is suitable 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.

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.

1. Strong vs. Weak acids: A strong acid is completely dissociated; a weak acid is partially dissociated. Concentration and strength are independent.

2. [\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 $c/K_a > 100$).

3. pH of water: Pure water has pH 7 at $25°C$But this changes with temperature because $K_w$ changes.

4. Buffer calculations: The Henderson-Hasselbalch equation uses concentrations, not moles (unless the acid and base are in the same volume).

5. Choosing an indicator: The indicator range must include the pH at the equivalence point.

6. 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.

7. Diprotic acids: The second dissociation is always weaker than the first. For \mathrm{H_2\mathrm{SO_4The first proton dissociates completely but the second does not ($K_{a2} = 1.2 \times 10^{-2}$).

1. 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).

2. 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.

3. 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{ MFind the concentrations of each component.

4. 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.

5. 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.

6. 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.

7. Calculate $K_a$ for a 0.050 \mathrm{ M weak acid solution with pH 2.80.

8. 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.

9. 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.

10. 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.

11. 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.

12. 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.

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

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

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

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

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).

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

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

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

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

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:

1. Initial pH: ~2.87 (weak acid)
2. 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^-).
3. Half-equivalence point: At 12.5 \mathrm{ mL (half the acid neutralised), \mathrm{pH = pK_a = 4.76.
4. Equivalence point: At 25 \mathrm{ mLPH = 8.73.
5. Beyond equivalence: pH approaches that of the excess \mathrm{NaOH.

The solubility of some salts is affected by pH.

\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).

\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).

13. Calculate the pH of a 0.050 \mathrm{ M solution of \mathrm{AlCl_3Given that K_a(\mathrm{Al^{3+}) = 1.0 \times 10^{-5}. (Hint: \mathrm{Al^{3+} acts as a weak acid.)

14. 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.

15. 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.

16. 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.

17. Explain why \mathrm{Na_2\mathrm{CO_3 solution is alkaline, writing the relevant hydrolysis equations.

18. 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.

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}) = 8This 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.

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 $K_{a1}$.

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.

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 $[\text{H}^+] = 0.050\,\text{mol\,dm}^{-3}$.

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

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