Acids and Bases (Advanced) | IB

1. Advanced pH Calculations

Strong Acids and Bases

For strong monoprotic acids, $[\mathrm{H}^+]$ equals the acid concentration. For strong diprotic Acids:

[\mathrm{H}^+] = 2 \times [\mathrm{acid}]

Example $0.050\mathrm{ M}$ $\mathrm{H}_2\mathrm{SO}_4$ (assuming complete first dissociation and significant Second dissociation):

[\mathrm{H}^+] \approx 2 \times 0.050 = 0.100\mathrm{ M}

\mathrm{pH} = -\log(0.100) = 1.00

Weak Acids

For a weak acid $\mathrm{HA}$ with acid dissociation constant $K_a$ :

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

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

Assuming $[\mathrm{H}^+] = [\mathrm{A}^-] = x$ and $[\mathrm{HA}] \approx c_0$ (the 5% rule):

X = \sqrt{K_a \cdot c_0}

\mathrm{pH} = -\log x

Very Dilute Weak Acids

When the acid is very dilute ( $c_0$ is small) or $K_a$ is large, the approximation $[\mathrm{HA}] \approx c_0$ breaks down. Solve the full quadratic:

K_a = \frac{x^2}{c_0 - x}

X^2 + K_a x - K_a c_0 = 0

Example Calculate the $\mathrm{pH}$ of $0.010\mathrm{ M}$ ethanoic acid ( $K_a = 1.8 \times 10^{-5}$ ).

Using the approximation:

X = \sqrt{1.8 \times 10^{-5} \times 0.010} = \sqrt{1.8 \times 10^{-7}} = 4.24 \times 10^{-4}

Check: $\dfrac{4.24 \times 10^{-4}}{0.010} \times 100\% = 4.24\% \lt 5\%$ . The approximation is Valid.

\mathrm{pH} = -\log(4.24 \times 10^{-4}) = 3.37

Polyprotic Acids

For diprotic acids ( $\mathrm{H}_2\mathrm{A}$ ), the first dissociation dominates:

K_{a1} \gg K_{a2}

The $\mathrm{pH}$ is determined primarily by $K_{a1}$ . The second dissociation contributes Additional $\mathrm{H}^+$ but is negligible.

For $\mathrm{H}_2\mathrm{SO}_4$ : the first dissociation is complete ( $K_{a1}$ very large), but $K_{a2} = 1.2 \times 10^{-2}$ .

Common Pitfalls

The 5% rule: if $x/c_0 \gt 5\%$ Use the quadratic formula.
For very dilute strong acids ( $c_0 \lt 10^{-6}\mathrm{ M}$ ), the contribution of water”s autoionization ( $[\mathrm{H}^+] = 10^{-7}\mathrm{ M}$ ) becomes significant.
$\mathrm{pOH} = 14 - \mathrm{pH}$ only holds at $25\degree\mathrm{C}$ .

2. Buffer Solutions

Definition

A buffer solution resists changes in $\mathrm{pH}$ 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).

The Henderson-Hasselbalch Equation

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

Where $\mathrm{p}K_a = -\log K_a$ .

Buffer Capacity

Buffer capacity is maximised when $[\mathrm{HA}] = [\mathrm{A}^-]$ I.e., when $\mathrm{pH} = \mathrm{p}K_a$ . A buffer is effective within $\pm 1$ unit of its $\mathrm{p}K_a$ .

Preparing a Buffer

Example Prepare an ethanoic acid/sodium ethanoate buffer with $\mathrm{pH} = 5.00$ . Given $\mathrm{p}K_a = 4.76$ .

5.00 = 4.76 + \log\frac{[\mathrm{CH}_3\mathrm{COO}^-]}{[\mathrm{CH}_3\mathrm{COOH}]}

\log\frac{[\mathrm{CH}_3\mathrm{COO}^-]}{[\mathrm{CH}_3\mathrm{COOH}]} = 0.24

\frac{[\mathrm{CH}_3\mathrm{COO}^-]}{[\mathrm{CH}_3\mathrm{COOH}]} = 10^{0.24} = 1.74

Mix ethanoic acid and sodium ethanoate in a molar ratio of $1 : 1.74$ (or approximately $1 : 1.7$ ).

Calculating pH Changes in a Buffer

When acid is added:

[\mathrm{HA}]_{\mathrm{new}} = [\mathrm{HA}]_{\mathrm{old}} + [\mathrm{H}^+]_{\mathrm{added}}

[\mathrm{A}^-]_{\mathrm{new}} = [\mathrm{A}^-]_{\mathrm{old}} - [\mathrm{H}^+]_{\mathrm{added}}

When base is added:

[\mathrm{HA}]_{\mathrm{new}} = [\mathrm{HA}]_{\mathrm{old}} - [\mathrm{OH}^-]_{\mathrm{added}}

[\mathrm{A}^-]_{\mathrm{new}} = [\mathrm{A}^-]_{\mathrm{old}} + [\mathrm{OH}^-]_{\mathrm{added}}

Example A buffer contains $0.10\mathrm{ M}$ $\mathrm{CH}_3\mathrm{COOH}$ and $0.10\mathrm{ M}$ $\mathrm{CH}_3\mathrm{COO}^-$ ( $\mathrm{pH} = 4.76$ ). Add $0.01\mathrm{ mol}$ of $\mathrm{HCl}$ to $1.0\mathrm{ L}$ of buffer.

[\mathrm{CH}_3\mathrm{COOH}] = 0.10 + 0.01 = 0.11\mathrm{ M}

[\mathrm{CH}_3\mathrm{COO}^-] = 0.10 - 0.01 = 0.09\mathrm{ M}

\mathrm{pH} = 4.76 + \log\frac{0.09}{0.11} = 4.76 + \log(0.818) = 4.76 - 0.09 = 4.67

The $\mathrm{pH}$ changed by only $0.09$ units. Without the buffer, $0.01\mathrm{ M}$ $\mathrm{HCl}$ Would give $\mathrm{pH} = 2.00$ .

Common Pitfalls

A buffer cannot withstand addition of large amounts of strong acid or base — it has finite capacity.
The Henderson-Hasselbalch equation assumes concentrations equal activities (valid for dilute solutions).
$\mathrm{pH} = \mathrm{p}K_a$ does not mean the solution is neutral; it means the acid and conjugate base are present in equal amounts.

3. Indicator Theory

How Indicators Work

Acid-base indicators are weak acids ( $\mathrm{HIn}$ ) where the acid and conjugate base forms have Different colours:

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

The observed colour depends on the ratio $[\mathrm{In}^-]/[\mathrm{HIn}]$ :

Condition	Dominant species	Colour observed
$\mathrm{pH} \ll \mathrm{p}K_{\mathrm{In}}$	$\mathrm{HIn}$	Acid colour
$\mathrm{pH} \approx \mathrm{p}K_{\mathrm{In}}$	Both present	Transition
$\mathrm{pH} \gg \mathrm{p}K_{\mathrm{In}}$	$\mathrm{In}^-$	Base colour

Effective Range

An indicator changes colour over approximately $\pm 1$ unit around its $\mathrm{p}K_{\mathrm{In}}$ :

\mathrm{pH\ range} = \mathrm{p}K_{\mathrm{In}} \pm 1

Choosing the Right Indicator

The indicator’s transition range should overlap with the equivalence point of the titration.

Titration type	Equivalence point pH	Suitable indicator
Strong acid—strong base	$\mathrm{pH} = 7$	Bromothymol blue
Strong acid—weak base	$\mathrm{pH} \lt 7$	Methyl orange
Weak acid—strong base	$\mathrm{pH} \gt 7$	Phenolphthalein
Weak acid—weak base	No sharp change	No suitable indicator

Common Indicators

Indicator	$\mathrm{pH}$ range	Acid colour	Base colour
Methyl orange	$3.1$ — $4.4$	Red	Yellow
Bromothymol blue	$6.0$ — $7.6$	Yellow	Blue
Phenolphthalein	$8.3$ — $10.0$	Colourless	Pink

4. Acid Deposition

Formation of Acid Rain

Sulfur dioxide from combustion of fossil fuels is oxidised:

\mathrm{SO}_2 + \mathrm{H}_2\mathrm{O} \to \mathrm{H}_2\mathrm{SO}_3

2\mathrm{SO}_2 + \mathrm{O}_2 \to 2\mathrm{SO}_3 \quad \mathrm{(catalysed by } \mathrm{V}_2\mathrm{O}_5)

\mathrm{SO}_3 + \mathrm{H}_2\mathrm{O} \to \mathrm{H}_2\mathrm{SO}_4

Nitrogen oxides from high-temperature combustion:

\mathrm{N}_2 + \mathrm{O}_2 \to 2\mathrm{NO}

2\mathrm{NO} + \mathrm{O}_2 \to 2\mathrm{NO}_2

3\mathrm{NO}_2 + \mathrm{H}_2\mathrm{O} \to 2\mathrm{HNO}_3 + \mathrm{NO}

Environmental Effects

Effect	Mechanism
Soil acidification	Leaches $\mathrm{Ca}^{2+}$ , $\mathrm{Mg}^{2+}$ Releases $\mathrm{Al}^{3+}$
Lake acidification	Reduces biodiversity, mobilises toxic $\mathrm{Al}^{3+}$
Plant damage	Damages leaves, inhibits root growth
Building corrosion	$\mathrm{CaCO}_3$ (limestone/marble) reacts with acid
Respiratory issues	Fine sulfate/nitrate particles irritate lungs

Mitigation

Flue gas desulfurisation (FGD): $\mathrm{SO}_2$ removed by reaction with $\mathrm{CaO}$ or $\mathrm{CaCO}_3$ .
Catalytic converters: reduce $\mathrm{NO}_x$ emissions.
Switching to low-sulfur fuels or renewable energy.

5. Solubility Product ( $K_{sp}$ )

Definition

For a sparingly soluble salt $\mathrm{M}_x\mathrm{A}_y$ :

\mathrm{M}_x\mathrm{A}_y(s) \rightleftharpoons x\mathrm{M}^{y+}(aq) + y\mathrm{A}^{x-}(aq)

K_`\{sp}` = [\mathrm{M}^{y+}]^x[\mathrm{A}^{x-}]^y

$K_{sp}$ is the equilibrium constant for the dissolution of a solid. It is temperature-dependent.

Common $K_{sp}$ Values

Compound	$K_{sp}$
$\mathrm{AgCl}$	$1.8 \times 10^{-10}$
$\mathrm{AgBr}$	$5.0 \times 10^{-13}$
$\mathrm{AgI}$	$8.3 \times 10^{-17}$
$\mathrm{BaSO}_4$	$1.1 \times 10^{-10}$
$\mathrm{CaCO}_3$	$3.4 \times 10^{-9}$
$\mathrm{PbI}_2$	$7.1 \times 10^{-9}$
$\mathrm{Mg(OH)}_2$	$1.8 \times 10^{-11}$

Solubility Calculations

Example Calculate the solubility of $\mathrm{AgCl}$ in $0.10\mathrm{ M}$ $\mathrm{NaCl}$ .

K_`\{sp}` = [\mathrm{Ag}^+][\mathrm{Cl}^-] = s \times 0.10

S = \frac{1.8 \times 10^{-10}}{0.10} = 1.8 \times 10^{-9}\mathrm{ mol/L}

Compared to pure water ( $1.3 \times 10^{-5}\mathrm{ mol/L}$ ), the solubility decreased by a factor Of about $7000$ .

Precipitation

A precipitate forms when the ion product exceeds $K_{sp}$ :

Q = [\mathrm{M}^{y+}]^x[\mathrm{A}^{x-}]^y

Condition	Result
$Q \lt K_{sp}$	No precipitate (unsaturated)
$Q = K_{sp}$	Saturated, at equilibrium
$Q \gt K_{sp}$	Precipitate forms

Common Pitfalls

$K_{sp}$ expressions do not include the concentration of the solid.
Solids and pure liquids are excluded from equilibrium expressions.
The common ion effect does not change $K_{sp}$ itself — it shifts the equilibrium position.

Practice Problems

Problem 1

Calculate the $\mathrm{pH}$ of a buffer prepared by mixing $100\mathrm{ mL}$ of $0.20\mathrm{ M}$ $\mathrm{CH}_3\mathrm{COOH}$ with $50\mathrm{ mL}$ of $0.20\mathrm{ M}$ $\mathrm{NaOH}$ . ( $\mathrm{p}K_a = 4.76$ )

Solution:

Moles of $\mathrm{CH}_3\mathrm{COOH} = 0.100 \times 0.20 = 0.020\mathrm{ mol}$

Moles of $\mathrm{NaOH} = 0.050 \times 0.20 = 0.010\mathrm{ mol}$

$\mathrm{NaOH}$ neutralises half the acid:

N(\mathrm{CH}_3\mathrm{COOH})_{\mathrm{remaining}} = 0.020 - 0.010 = 0.010\mathrm{ mol}

N(\mathrm{CH}_3\mathrm{COO}^-)_{\mathrm{formed}} = 0.010\mathrm{ mol}

Total volume $= 150\mathrm{ mL} = 0.150\mathrm{ L}$ :

[\mathrm{CH}_3\mathrm{COOH}] = \frac{0.010}{0.150} = 0.0667\mathrm{ M}

[\mathrm{CH}_3\mathrm{COO}^-] = \frac{0.010}{0.150} = 0.0667\mathrm{ M}

\mathrm{pH} = 4.76 + \log\frac{0.0667}{0.0667} = 4.76 + 0 = 4.76

Problem 2

Will a precipitate form when $50.0\mathrm{ mL}$ of $1.0 \times 10^{-4}\mathrm{ M}$ $\mathrm{AgNO}_3$ Is mixed with $50.0\mathrm{ mL}$ of $1.0 \times 10^{-4}\mathrm{ M}$ $\mathrm{NaCl}$ ? ( $K_{sp}(\mathrm{AgCl}) = 1.8 \times 10^{-10}$ )

Solution:

Total volume $= 100\mathrm{ mL}$ . Diluted concentrations:

[\mathrm{Ag}^+] = \frac{0.050 \times 1.0 \times 10^{-4}}{0.100} = 5.0 \times 10^{-5}\mathrm{ M}

[\mathrm{Cl}^-] = \frac{0.050 \times 1.0 \times 10^{-4}}{0.100} = 5.0 \times 10^{-5}\mathrm{ M}

Q = (5.0 \times 10^{-5})^2 = 2.5 \times 10^{-9}

Since $Q = 2.5 \times 10^{-9} \gt K_{sp} = 1.8 \times 10^{-10}$ A precipitate of $\mathrm{AgCl}$ Will form.

Problem 3

Calculate the $\mathrm{pH}$ of $0.050\mathrm{ M}$ $\mathrm{NH}_3$ ( $K_b = 1.8 \times 10^{-5}$ ).

Solution:

\mathrm{NH}_3 + \mathrm{H}_2\mathrm{O} \rightleftharpoons \mathrm{NH}_4^+ + \mathrm{OH}^-

K_b = \frac{[\mathrm{NH}_4^+][\mathrm{OH}^-]}{[\mathrm{NH}_3]} = \frac{x^2}{0.050 - x}

Approximation: $x = \sqrt{1.8 \times 10^{-5} \times 0.050} = \sqrt{9.0 \times 10^{-7}} = 9.5 \times 10^{-4}$

Check: $9.5 \times 10^{-4} / 0.050 = 1.9\% \lt 5\%$ . Valid.

\mathrm{pOH} = -\log(9.5 \times 10^{-4}) = 3.02

\mathrm{pH} = 14.00 - 3.02 = 10.98

Problem 4

Explain why adding $\mathrm{NH}_4\mathrm{Cl}$ to an $\mathrm{NH}_3$ solution decreases the $\mathrm{pH}$ .

Solution:

$\mathrm{NH}_4\mathrm{Cl}$ dissociates completely to give $\mathrm{NH}_4^+$ The conjugate acid of $\mathrm{NH}_3$ . This is the common ion effect:

\mathrm{NH}_3 + \mathrm{H}_2\mathrm{O} \rightleftharpoons \mathrm{NH}_4^+ + \mathrm{OH}^-

Adding $\mathrm{NH}_4^+$ shifts the equilibrium to the left (Le Chatelier’s principle), decreasing $[\mathrm{OH}^-]$ and therefore increasing $[\mathrm{H}^+]$ Which lowers the $\mathrm{pH}$ .

Worked Examples

Worked Example: pH of a very dilute strong acid

Calculate the $\mathrm{pH}$ of $1.0 \times 10^{-8}\mathrm{ M}$ $\mathrm{HCl}$ at $25\degree\mathrm{C}$ .

Solution

At this concentration, the contribution from water autoionization ( $[\mathrm{H}^+] = 10^{-7}\mathrm{ M}$ ) is Significant and cannot be ignored. Let $x = [\mathrm{OH}^-]$ from water autoionization:

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

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

$x^2 + 1.0 \times 10^{-8}x - 1.0 \times 10^{-14} = 0$

Using the quadratic formula:

$x = \frac{-1.0 \times 10^{-8} + \sqrt{(1.0 \times 10^{-8})^2 + 4(1.0 \times 10^{-14})}}{2} = \frac{-1.0 \times 10^{-8} + 2.00 \times 10^{-7}}{2} = 9.5 \times 10^{-8}$

$[\mathrm{H}^+]_{\mathrm{total}} = 1.0 \times 10^{-8} + 9.5 \times 10^{-8} = 1.05 \times 10^{-7}\mathrm{ M}$

$\mathrm{pH} = -\log(1.05 \times 10^{-7}) = 6.98$

The $\mathrm{pH}$ is close to 7 but slightly acidic, as expected for a very dilute strong acid. Ignoring Water autoionization would give the incorrect result $\mathrm{pH} = 8.00$ (a basic $\mathrm{pH}$ from Adding acid), which violates chemical intuition.

Worked Example: pH at the equivalence point of a weak acid—strong base titration

Calculate the $\mathrm{pH}$ at the equivalence point when $25.0\mathrm{ mL}$ of $0.100\mathrm{ M}$ $\mathrm{CH}_3\mathrm{COOH}$ ( $K_a = 1.8 \times 10^{-5}$ ) is titrated with $0.100\mathrm{ M}$ $\mathrm{NaOH}$ .

Solution

At the equivalence point, all of the weak acid has been converted to its conjugate base:

$n(\mathrm{CH}_3\mathrm{COOH}) = 0.0250 \times 0.100 = 0.00250\mathrm{ mol}$

$V(\mathrm{NaOH}) = \frac{0.00250}{0.100} = 25.0\mathrm{ mL}$

$V_{\mathrm{total}} = 50.0\mathrm{ mL} = 0.0500\mathrm{ L}$

$[\mathrm{CH}_3\mathrm{COO}^-] = \frac{0.00250}{0.0500} = 0.0500\mathrm{ M}$

The conjugate base hydrolyses water:

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

$K_b = \frac{K_w}{K_a} = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} = 5.56 \times 10^{-10}$

$[\mathrm{OH}^-] = \sqrt{K_b \times [\mathrm{CH}_3\mathrm{COO}^-]} = \sqrt{5.56 \times 10^{-10} \times 0.0500} = 5.27 \times 10^{-6}\mathrm{ M}$

$\mathrm{pOH} = -\log(5.27 \times 10^{-6}) = 5.28$

$\mathrm{pH} = 14.00 - 5.28 = 8.72$

The equivalence point is basic because the conjugate base of a weak acid is itself a weak base. Phenolphthalein (transition range 8.3—10.0) is a suitable indicator. Bromothymol blue (6.0—7.6) would change colour Before reaching the equivalence point and would give a systematically low titre reading.

Worked Example: Selective precipitation using $K_{sp}$

A solution contains $0.020\mathrm{ M}$ $\mathrm{Cl}^-$ and $0.020\mathrm{ M}$ $\mathrm{CrO}_4^{2-}$ . Solid $\mathrm{AgNO}_3$ is added gradually. $K_{sp}(\mathrm{AgCl}) = 1.8 \times 10^{-10}$ $K_{sp}(\mathrm{Ag}_2\mathrm{CrO}_4) = 1.1 \times 10^{-12}$ . Determine which salt precipitates first, the $[\mathrm{Ag}^+]$ at which precipitation begins, and the $[\mathrm{Ag}^+]$ at which the second salt begins To precipitate.

Solution

Step 1: Calculate the threshold $[\mathrm{Ag}^+]$ for each salt.

For $\mathrm{AgCl}$ :

$[\mathrm{Ag}^+] = \frac{K_{sp}}{[\mathrm{Cl}^-]} = \frac{1.8 \times 10^{-10}}{0.020} = 9.0 \times 10^{-9}\mathrm{ M}$

For $\mathrm{Ag}_2\mathrm{CrO}_4$ :

$[\mathrm{Ag}^+] = \sqrt{\frac{K_{sp}}{[\mathrm{CrO}_4^{2-}]}} = \sqrt{\frac{1.1 \times 10^{-12}}{0.020}} = 7.4 \times 10^{-6}\mathrm{ M}$

Step 2: Identify which precipitates first.

$\mathrm{AgCl}$ precipitates first since it requires a lower $[\mathrm{Ag}^+]$ ( $9.0 \times 10^{-9}\mathrm{ M}$ Vs $7.4 \times 10^{-6}\mathrm{ M}$ ).

Step 3: Calculate $[\mathrm{Cl}^-]$ remaining when $\mathrm{Ag}_2\mathrm{CrO}_4$ begins to precipitate.

When $[\mathrm{Ag}^+] = 7.4 \times 10^{-6}\mathrm{ M}$ :

$[\mathrm{Cl}^-]_{\mathrm{remaining}} = \frac{K_{sp}}{[\mathrm{Ag}^+]} = \frac{1.8 \times 10^{-10}}{7.4 \times 10^{-6}} = 2.4 \times 10^{-5}\mathrm{ M}$

Fraction of $\mathrm{Cl}^-$ precipitated: $\dfrac{0.020 - 2.4 \times 10^{-5}}{0.020} \times 100\% = 99.9\%$

This analysis underpins Mohr’s method for argentometric determination of chloride: the red colour of $\mathrm{Ag}_2\mathrm{CrO}_4$ appears only after virtually all $\mathrm{Cl}^-$ has been removed.

Worked Example: Buffer preparation from a weak base

Prepare an $\mathrm{NH}_3$ / $\mathrm{NH}_4\mathrm{Cl}$ buffer with $\mathrm{pH} = 9.50$ . Given $K_b(\mathrm{NH}_3) = 1.8 \times 10^{-5}$ . Calculate the mass of $\mathrm{NH}_4\mathrm{Cl}$ ( $M = 53.49\mathrm{ g/mol}$ ) that must be dissolved in $500\mathrm{ mL}$ of $0.200\mathrm{ M}$ $\mathrm{NH}_3$ .

Solution

Step 1: Find the $\mathrm{p}K_a$ of the conjugate acid $\mathrm{NH}_4^+$ .

$\mathrm{p}K_b = -\log(1.8 \times 10^{-5}) = 4.74$

$\mathrm{p}K_a = 14.00 - 4.74 = 9.26$

Step 2: Apply the Henderson-Hasselbalch equation.

$9.50 = 9.26 + \log\frac{[\mathrm{NH}_3]}{[\mathrm{NH}_4^+]}$

$\log\frac{[\mathrm{NH}_3]}{[\mathrm{NH}_4^+]} = 0.24$

$\frac{[\mathrm{NH}_3]}{[\mathrm{NH}_4^+]} = 10^{0.24} = 1.74$

Step 3: Solve for the required $[\mathrm{NH}_4^+]$ .

$[\mathrm{NH}_4^+] = \frac{0.200}{1.74} = 0.115\mathrm{ M}$

Step 4: Calculate the mass of $\mathrm{NH}_4\mathrm{Cl}$ .

$n(\mathrm{NH}_4\mathrm{Cl}) = 0.115 \times 0.500 = 0.0575\mathrm{ mol}$

$m(\mathrm{NH}_4\mathrm{Cl}) = 0.0575 \times 53.49 = 3.08\mathrm{ g}$

Worked Example: Titration curve analysis for a polyprotic acid

$25.0\mathrm{ mL}$ of $0.100\mathrm{ M}$ $\mathrm{H}_3\mathrm{PO}_4$ ( $K_{a1} = 7.5 \times 10^{-3}$ $K_{a2} = 6.2 \times 10^{-8}$ , $K_{a3} = 4.8 \times 10^{-13}$ ) is titrated with $0.100\mathrm{ M}$ $\mathrm{NaOH}$ . Calculate the $\mathrm{pH}$ at the first and second equivalence points and identify Suitable indicators.

Solution

First equivalence point ( $25.0\mathrm{ mL}$ $\mathrm{NaOH}$ added):

All $\mathrm{H}_3\mathrm{PO}_4$ is converted to $\mathrm{H}_2\mathrm{PO}_4^-$ (an amphoteric species).

$[\mathrm{H}_2\mathrm{PO}_4^-] = \frac{0.00250}{0.0500} = 0.0500\mathrm{ M}$

For an amphoteric species, the $\mathrm{pH}$ is approximately the average of the two relevant $\mathrm{p}K_a$ values:

$\mathrm{pH} \approx \frac{\mathrm{p}K_{a1} + \mathrm{p}K_{a2}}{2} = \frac{2.12 + 7.21}{2} = 4.67$

A suitable indicator: bromocresol green (3.8—5.4) or methyl red (4.4—6.2).

Second equivalence point ( $50.0\mathrm{ mL}$ $\mathrm{NaOH}$ added):

All $\mathrm{H}_2\mathrm{PO}_4^-$ is converted to $\mathrm{HPO}_4^{2-}$ (also amphoteric).

$[\mathrm{HPO}_4^{2-}] = \frac{0.00250}{0.0750} = 0.0333\mathrm{ M}$

$\mathrm{pH} \approx \frac{\mathrm{p}K_{a2} + \mathrm{p}K_{a3}}{2} = \frac{7.21 + 12.32}{2} = 9.76$

A suitable indicator: phenolphthalein (8.3—10.0).

Third equivalence point: Not achievable in aqueous solution because $K_{a3}$ is too small ( $4.8 \times 10^{-13}$ ) for complete neutralisation of the third proton.

Common Pitfalls

Assuming complete dissociation for all diprotic acids: $\mathrm{H}_2\mathrm{SO}_4$ has a complete first dissociation but a partial second ( $K_{a2} = 1.2 \times 10^{-2}$ ), so $[\mathrm{H}^+] \neq 2[\mathrm{acid}]$ . For $\mathrm{H}_2\mathrm{CO}_3$ Both dissociation steps are weak. Always check the magnitude of each $K_a$ before making simplifying assumptions.
Using $\mathrm{pOH} = 14 - \mathrm{pH}$ without specifying temperature: $K_w = 1.0 \times 10^{-14}$ only at $25\degree\mathrm{C}$ . At $37\degree\mathrm{C}$ , $K_w \approx 2.4 \times 10^{-14}$ So $\mathrm{pH} + \mathrm{pOH} = 13.62$ . Always state the temperature assumption explicitly.
Applying Henderson-Hasselbalch to strong acid—base mixtures: The equation requires a weak acid and its conjugate base. For a strong acid titrated with a strong base, calculate the excess $[\mathrm{H}^+]$ or $[\mathrm{OH}^-]$ directly from stoichiometry.
Forgetting dilution when mixing buffer components: When two solutions are combined to make a buffer, the total volume changes. Convert all quantities to moles first, then recalculate concentrations in the combined volume before applying the Henderson-Hasselbalch equation.
Comparing $K_{sp}$ values across different stoichiometries: $K_{sp}(\mathrm{AgCl}) = 1.8 \times 10^{-10}$ and $K_{sp}(\mathrm{Ag}_2\mathrm{CrO}_4) = 1.1 \times 10^{-12}$ But $\mathrm{Ag}_2\mathrm{CrO}_4$ is more soluble in water because its $K_{sp}$ expression contains $[\mathrm{Ag}^+]^2$ . Always calculate molar solubility from $K_{sp}$ before comparing.
Ignoring water autoionization for dilute solutions: When the calculated $[\mathrm{H}^+]$ from the acid or base alone is below $10^{-6}\mathrm{ M}$ The contribution from water ( $10^{-7}\mathrm{ M}$ ) is comparable and must be included via the full quadratic.
Confusing buffer capacity with buffer range: Buffer capacity (total moles of acid or base that can be absorbed) depends on the absolute concentrations of the buffer components, not their ratio. A $0.01\mathrm{ M}$ buffer at $\mathrm{pH} = \mathrm{p}K_a$ has far less capacity than a $1.0\mathrm{ M}$ buffer at the same $\mathrm{pH}$ .
Assuming the 5% rule is always valid: The approximation $[\mathrm{HA}] \approx c_0$ fails when $K_a$ is large relative to $c_0$ . Always verify by computing $x/c_0 \times 100\%$ . If it exceeds 5%, solve the full quadratic.
Using the wrong $\mathrm{p}K_a$ in Henderson-Hasselbalch: When working with a weak base (e.g., $\mathrm{NH}_3$ ), use the $\mathrm{p}K_a$ of its conjugate acid ( $\mathrm{NH}_4^+$ ), not the $\mathrm{p}K_b$ of the base itself. The relationship is $\mathrm{p}K_a + \mathrm{p}K_b = 14.00$ .
Assuming the common ion effect changes $K_{sp}$ : Adding a common ion shifts the equilibrium position (decreasing solubility), but $K_{sp}$ itself is a thermodynamic constant that depends only on temperature.

Exam-Style Problems

Calculate the $\mathrm{pH}$ of the solution formed when $15.0\mathrm{ mL}$ of $0.100\mathrm{ M}$ $\mathrm{H}_2\mathrm{SO}_4$ is added to $35.0\mathrm{ mL}$ of $0.100\mathrm{ M}$ $\mathrm{NaOH}$ at $25\degree\mathrm{C}$ . ( $K_{a2}$ of $\mathrm{HSO}_4^-$ = $1.2 \times 10^{-2}$ .) State all assumptions and justify their validity. [Medium]
A buffer is prepared by dissolving $4.10\mathrm{ g}$ of sodium ethanoate ( $\mathrm{CH}_3\mathrm{COONa}$ , $M = 82.03\mathrm{ g/mol}$ ) in $250\mathrm{ mL}$ of $0.200\mathrm{ M}$ ethanoic acid ( $\mathrm{p}K_a = 4.76$ ). (a) Calculate the buffer $\mathrm{pH}$ . (b) Calculate the new $\mathrm{pH}$ after adding $5.0\mathrm{ mL}$ of $0.100\mathrm{ M}$ $\mathrm{HCl}$ . (c) Calculate the percentage change in $\mathrm{pH}$ and comment on the effectiveness of the buffer. [Medium]
Will a precipitate form when $25.0\mathrm{ mL}$ of $2.0 \times 10^{-4}\mathrm{ M}$ $\mathrm{Pb}(\mathrm{NO}_3)_2$ is mixed with $25.0\mathrm{ mL}$ of $1.0 \times 10^{-3}\mathrm{ M}$ $\mathrm{NaI}$ ? $K_{sp}(\mathrm{PbI}_2) = 7.1 \times 10^{-9}$ . If a precipitate forms, calculate $[\mathrm{Pb}^{2+}]$ and $[\mathrm{I}^-]$ remaining at equilibrium. [Hard]
$20.0\mathrm{ mL}$ of $0.150\mathrm{ M}$ $\mathrm{NH}_3$ ( $K_b = 1.8 \times 10^{-5}$ ) is titrated with $0.150\mathrm{ M}$ $\mathrm{HCl}$ . Calculate the $\mathrm{pH}$ at each of the following volumes of $\mathrm{HCl}$ added: (a) $0.0\mathrm{ mL}$ (b) $10.0\mathrm{ mL}$ (c) $20.0\mathrm{ mL}$ (equivalence point), (d) $25.0\mathrm{ mL}$ . Sketch the approximate titration curve and label the buffer region. [Hard]
The solubility of $\mathrm{CaF}_2$ in pure water is $2.15 \times 10^{-4}\mathrm{ M}$ at $25\degree\mathrm{C}$ . (a) Calculate $K_{sp}$ . (b) Calculate the solubility of $\mathrm{CaF}_2$ in $0.050\mathrm{ M}$ $\mathrm{CaCl}_2$ . (c) Determine the maximum concentration of $\mathrm{NaF}$ that can coexist with $0.010\mathrm{ M}$ $\mathrm{CaCl}_2$ without precipitation. [Medium]
Explain why phenolphthalein is a suitable indicator for the titration of ethanoic acid with sodium hydroxide, but methyl orange is not. Support your answer with a quantitative calculation of the equivalence point $\mathrm{pH}$ and reference to the transition ranges of both indicators. [Medium]
A student prepares a buffer by mixing $100\mathrm{ mL}$ of $0.200\mathrm{ M}$ $\mathrm{CH}_3\mathrm{COOH}$ with $100\mathrm{ mL}$ of $0.100\mathrm{ M}$ $\mathrm{NaOH}$ . (a) Calculate the buffer $\mathrm{pH}$ . (b) Determine the maximum volume of $0.100\mathrm{ M}$ $\mathrm{HCl}$ that can be added before the $\mathrm{pH}$ drops below $4.00$ . (c) Comment on whether this buffer would be effective at $\mathrm{pH} = 4.00$ given the $\mathrm{p}K_a$ of ethanoic acid. [Hard]
An environmental scientist measures the $\mathrm{pH}$ of a lake at $4.50$ . (a) Calculate $[\mathrm{H}^+]$ and $[\mathrm{SO}_4^{2-}]$ Assuming the acidity is entirely from dissolved $\mathrm{H}_2\mathrm{SO}_4$ with complete dissociation of both protons. (b) Determine whether $\mathrm{CaSO}_4$ would precipitate if $[\mathrm{Ca}^{2+}] = 1.5 \times 10^{-3}\mathrm{ M}$ . $K_{sp}(\mathrm{CaSO}_4) = 2.4 \times 10^{-5}$ . (c) Calculate the minimum $[\mathrm{Ca}^{2+}]$ required to initiate precipitation of $\mathrm{CaSO}_4$ . [Medium]

If You Get These Wrong, Revise

Equilibrium principles (Le Chatelier, $K$ expressions) → Review …/7-equilibrium/1_equilibrium
Stoichiometry and mole calculations → Review …/1-stoichiometry/1_stoichiometric-relationships
Uncertainty propagation in titrations → Review …/11-measurement-and-data-processing/1_measurement-and-data-processing
Electron configurations and ion formation → Review …/2-atomic-structure/1_atomic-theory

flowchart TD
    A[2_Acids And Bases Advanced] --> 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

This topic covers the essential chemistry of acids and bases (advanced), including key reactions, underlying theories, and practical applications.

Key concepts include:

Brønsted-Lowry theory
strong and weak acids/bases
pH calculations
titration curves and indicators
hydrolysis of salts

Mastery of these concepts requires both theoretical understanding and the ability to apply knowledge to unfamiliar contexts, particularly in calculation and practical questions.

Cross-References

Equilibrium: Acid-base equilibria involve dynamic balance
Redox: Acid-base reactions can involve redox
Energetics: Neutralisation releases enthalpy

Intuition

Acids and bases are about proton transfer. An acid is a proton donor; a base is a proton acceptor. The strength of an acid depends on how readily it gives up its proton — strong acids dissociate completely, weak acids only partially. pH is the logarithmic measure of hydrogen ion concentration, so each unit change in pH represents a tenfold change in acidity.

Buffer solutions resist pH changes by having both an acid and its conjugate base present. When you add acid, the conjugate base neutralizes it. When you add base, the acid neutralizes it. This is why blood maintains pH 7.4 despite metabolic acids being constantly produced. Titration curves show the pH change as base is added to acid — the steep part near the equivalence point is where the indicator changes color. The Henderson-Hasselbalch equation ( $\text{pH} = \text{p}K_a + \log\frac{[\text{A}^-]}{[\text{HA}]}$ ) connects pH to the ratio of conjugate base to acid.

Common Mistakes

Confusing strong/weak with concentrated/diluted. A strong acid dissociates completely in water regardless of concentration; a weak acid only partially dissociates. A dilute strong acid has a higher pH than a concentrated strong acid, but a dilute weak acid may have a similar pH to a concentrated weak acid. These are independent dimensions — strength is about extent of dissociation, concentration is about amount per unit volume.
Misidentifying the equivalence point pH. For a strong acid–strong base titration, the equivalence point is at pH 7. For a weak acid–strong base titration, the equivalence point is above pH 7 (basic) because the conjugate base of the weak acid hydrolyses. For a weak base–strong acid titration, it is below pH 7. Students often assume pH 7 always.
Using the Henderson-Hasselbalch equation outside its valid range. The equation assumes that $[\text{HA}]$ and $[\text{A}^-]$ are approximately equal to their initial concentrations. This breaks down near the equivalence point or when the acid/base is very dilute. At those points, use an ICE table instead.
Choosing the wrong indicator for a titration. The indicator’s pKa should be close to the pH at the equivalence point. Phenolphthalein (pKa ~9) is suitable for weak acid–strong base titrations but not for strong acid–weak base titrations, where methyl red (pKa ~5) is appropriate.
Confusing $K_a$ with $\text{p}K_a$ . A larger $K_a$ means a stronger acid, but a smaller $\text{p}K_a$ means a stronger acid (since $\text{p}K_a = -\log K_a$ ). Confusing the two leads to incorrect comparisons of acid strength.

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