Acids and Bases

Bronsted-Lowry Theory

Definitions

• Acid: A proton ($\mathrm{H}^+$) donor.
• Base: A proton ($\mathrm{H}^+$) acceptor.

Conjugate Acid-Base Pairs

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.

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

Left side	Right side
HA (acid)	A$^-$ (conjugate base)
H$_2$O (base)	H$_3$O$^+$ (conjugate acid)

Key relationship: The stronger the acid, the weaker its conjugate base, and vice versa.

Strong vs Weak Acids and Bases

Property	Strong Acid/Base	Weak Acid/Base
Dissociation	Complete (100%)	Partial (equilibrium)
Equilibrium	Lies far to the right	Lies to the left
Conductivity	High	Low
pH (same conc.)	Lower (acid) / Higher (base)	Less extreme
Reaction rate with metals	Faster	Slower
Examples (acids)	HCl, HNO$_3$, H$_2$SO$_4$, HClO$_4$	CH$_3$COOH, HF, HCN, H$_2$CO$_3$
Examples (bases)	NaOH, KOH, Ba(OH)$_2$	NH$_3$, CH$_3$COO$^-$, CO$_3^{2-}$

:::warning[Exam Tip] H$_2$SO$_4$ is a diprotic acid. The first dissociation is complete (strong), but the second dissociation is partial (weak): $\mathrm{HSO}_4^- \rightleftharpoons \mathrm{H}^+ + \mathrm{SO}_4^{2-}$ with $K_a \approx 1.0 \times 10^{-2}$. :::

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pH and pOH Scales

Definitions

$$\mathrm{pH} = -\log[\mathrm{H}^+]$$ $$\mathrm{pOH} = -\log[\mathrm{OH}^-]$$

Relationship

At $25\degree\mathrm{C}$:

$$\mathrm{pH} + \mathrm{pOH} = 14$$ $$[\mathrm{H}^+][\mathrm{OH}^-] = K_w = 1.0 \times 10^{-14}$$

pH Scale

pH	Nature	[H$^+$] (mol/L)
0	Strongly acidic	$1.0 \times 10^0$
1		$1.0 \times 10^{-1}$
3		$1.0 \times 10^{-3}$
7	Neutral	$1.0 \times 10^{-7}$
11		$1.0 \times 10^{-11}$
14	Strongly basic	$1.0 \times 10^{-14}$

Worked Example 1: pH of a Strong Acid

Calculate the pH of $0.050\mathrm{ mol/L}$ HCl.

HCl is a strong acid and dissociates completely:

$$[\mathrm{H}^+] = 0.050\mathrm{ mol/L}$$$$\mathrm{pH} = -\log(0.050) = -\log(5.0 \times 10^{-2}) = 2 - \log 5.0 = 2 - 0.699 = 1.30$$

Worked Example 2: pH of a Strong Base

Calculate the pH of $0.020\mathrm{ mol/L}$ NaOH at $25\degree\mathrm{C}$.

NaOH is a strong base:

$$[\mathrm{OH}^-] = 0.020\mathrm{ mol/L}$$$$\mathrm{pOH} = -\log(0.020) = -\log(2.0 \times 10^{-2}) = 2 - \log 2.0 = 2 - 0.301 = 1.70$$$$\mathrm{pH} = 14 - 1.70 = 12.30$$

Worked Example 3: pH of a Diprotic Strong Acid

Calculate the pH of $0.010\mathrm{ mol/L}$ H$_2$SO$_4$.

Since the first dissociation of H$_2$SO$_4$ is complete:

$$\mathrm{H}_2\mathrm{SO}_4 \to \mathrm{H}^+ + \mathrm{HSO}_4^-$$

Each mole of H$_2$SO$_4$ gives 1 mole of H$^+$ from the first dissociation. At this concentration, the second dissociation contributes additional H$^+$, but for most IB exam questions, it is acceptable to consider only the first dissociation unless told otherwise:

$$[\mathrm{H}^+] \approx 0.010\mathrm{ mol/L}$$$$\mathrm{pH} = -\log(0.010) = 2.00$$

If the second dissociation is considered (with $K_{a2} = 1.0 \times 10^{-2}$):

Let $x$ be the additional [$\mathrm{H}^+$] from the second dissociation:

$$K_{a2} = \frac{[\mathrm{H}^+][\mathrm{SO}_4^{2-}]}{[\mathrm{HSO}_4^-]} = \frac{(0.010 + x)(x)}{0.010 - x} = 1.0 \times 10^{-2}$$

This gives $x \approx 0.0045$, so $[\mathrm{H}^+] \approx 0.0145$ and $\mathrm{pH} \approx 1.84$.

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Ka and Kb Expressions

Acid Dissociation Constant

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

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

Base Dissociation Constant

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

$$\mathrm{B} + \mathrm{H}_2\mathrm{O} \rightleftharpoons \mathrm{BH}^+ + \mathrm{OH}^-$$ $$K_b = \frac{[\mathrm{BH}^+][\mathrm{OH}^-]}{[\mathrm{B}]}$$

Relationship Between Ka and Kb

For a conjugate acid-base pair:

$$K_a \times K_b = K_w = 1.0 \times 10^{-14}$$ $$\mathrm{p}K_a + \mathrm{p}K_b = 14$$

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

Common Ka Values at $25\degree\mathrm{C}$

Acid	Ka	pKa
CH$_3$COOH	$1.8 \times 10^{-5}$	4.74
HCOOH (formic)	$1.8 \times 10^{-4}$	3.74
HCN	$6.2 \times 10^{-10}$	9.21
HF	$6.8 \times 10^{-4}$	3.17
HNO$_2$	$4.5 \times 10^{-4}$	3.35
NH$_4^+$	$5.6 \times 10^{-10}$	9.25

Worked Example 4: pH of a Weak Acid

Calculate the pH of $0.100\mathrm{ mol/L}$ CH$_3$COOH ($K_a = 1.8 \times 10^{-5}$).

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

ICE table (initial, change, equilibrium):

Species	[Initial]	Change	[Equilibrium]
CH$_3$COOH	0.100	$-x$	$0.100 - x$
H$^+$	0	$+x$	$x$
CH$_3$COO$^-$	0	$+x$	$x$

$$K_a = \frac{x^2}{0.100 - x} = 1.8 \times 10^{-5}$$

Check the $5\%$ rule: $\frac{K_a}{c} = \frac{1.8 \times 10^{-5}}{0.100} = 1.8 \times 10^{-4} \lt 0.05$. The approximation $0.100 - x \approx 0.100$ is valid.

$$x^2 = 1.8 \times 10^{-6}$$$$x = \sqrt{1.8 \times 10^{-6}} = 1.34 \times 10^{-3}\mathrm{ mol/L}$$$$\mathrm{pH} = -\log(1.34 \times 10^{-3}) = 2.87$$

Worked Example 5: pH of a Weak Base

Calculate the pH of $0.150\mathrm{ mol/L}$ NH$_3$ ($K_b = 1.8 \times 10^{-5}$) at $25\degree\mathrm{C}$.

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

Species	[Initial]	Change	[Equilibrium]
NH$_3$	0.150	$-x$	$0.150 - x$
NH$_4^+$	0	$+x$	$x$
OH$^-$	0	$+x$	$x$

$$K_b = \frac{x^2}{0.150 - x} = 1.8 \times 10^{-5}$$

Approximation valid: $\frac{K_b}{c} = 1.2 \times 10^{-4} \lt 0.05$.

$$x^2 = 0.150 \times 1.8 \times 10^{-5} = 2.7 \times 10^{-6}$$$$x = \sqrt{2.7 \times 10^{-6}} = 1.64 \times 10^{-3}\mathrm{ mol/L} = [\mathrm{OH}^-]$$$$\mathrm{pOH} = -\log(1.64 \times 10^{-3}) = 2.78$$$$\mathrm{pH} = 14 - 2.78 = 11.22$$

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Kw (Ionic Product of Water)

Water undergoes autoionisation:

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

At $25\degree\mathrm{C}$: $K_w = 1.0 \times 10^{-14}$ (mol$^2$/L$^2$).

$K_w$ is temperature dependent:

Temperature ($\degree$C)	$K_w$
0	$1.14 \times 10^{-15}$
25	$1.00 \times 10^{-14}$
50	$5.48 \times 10^{-14}$
100	$5.13 \times 10^{-13}$

:::warning[Exam Tip] At $50\degree\mathrm{C}$, pure water has $\mathrm{pH} = 6.63$ (not 7). This is because $K_w$ is larger, so $[\mathrm{H}^+] = [\mathrm{OH}^-] = \sqrt{K_w} \gt 10^{-7}$. The water is still neutral because $[\mathrm{H}^+] = [\mathrm{OH}^-]$. Neutral does not always mean pH = 7; it depends on temperature. :::

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Polyprotic Acids

A polyprotic acid can donate more than one proton. Each dissociation has its own $K_a$ value.

Carbonic acid (H$_2$CO$_3$):

$$\mathrm{H}_2\mathrm{CO}_3 \rightleftharpoons \mathrm{H}^+ + \mathrm{HCO}_3^- \quad K_{a1} = 4.3 \times 10^{-7}$$ $$\mathrm{HCO}_3^- \rightleftharpoons \mathrm{H}^+ + \mathrm{CO}_3^{2-} \quad K_{a2} = 4.8 \times 10^{-11}$$

Phosphoric acid (H$_3$PO$_4$):

$$\mathrm{H}_3\mathrm{PO}_4 \rightleftharpoons \mathrm{H}^+ + \mathrm{H}_2\mathrm{PO}_4^- \quad K_{a1} = 7.5 \times 10^{-3}$$ $$\mathrm{H}_2\mathrm{PO}_4^- \rightleftharpoons \mathrm{H}^+ + \mathrm{HPO}_4^{2-} \quad K_{a2} = 6.2 \times 10^{-8}$$ $$\mathrm{HPO}_4^{2-} \rightleftharpoons \mathrm{H}^+ + \mathrm{PO}_4^{3-} \quad K_{a3} = 4.2 \times 10^{-13}$$

General rule: $K_{a1} \gg K_{a2} \gg K_{a3}$. Each successive proton is harder to remove because removing a proton from an increasingly negative ion requires more energy.

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Indicators

Common Acid-Base Indicators

Indicator	Colour (Acid)	Transition pH Range	Colour (Base)
Methyl orange	Red	3.1 -- 4.4	Yellow
Bromothymol blue	Yellow	6.0 -- 7.6	Blue
Phenolphthalein	Colourless	8.3 -- 10.0	Pink/Magenta
Universal indicator	Red (pH 1)	Full range 1--14	Violet (pH 14)

How Indicators Work

Indicators are weak organic acids (HIn) where the acid form and conjugate base form have different colours:

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

The colour observed depends on the ratio $[\mathrm{HIn}]/[\mathrm{In}^-]$, which depends on [H$^+$] (pH).

Choosing an Indicator for a Titration

An indicator is suitable when its transition range overlaps with the equivalence point of the titration (where pH changes most steeply).

Titration Type	Equivalence Point pH	Suitable Indicator
Strong acid -- strong base	pH = 7	Bromothymol blue, phenolphthalein, methyl orange
Weak acid -- strong base	pH \gt 7	Phenolphthalein
Strong acid -- weak base	pH \lt 7	Methyl orange

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pH Curves for Titrations

Strong Acid -- Strong Base

Example: HCl + NaOH

• Initial pH: low (strong acid).
• pH rises slowly, then sharply near the equivalence point (pH = 7).
• Beyond equivalence point: pH approaches 14 (excess strong base).

Weak Acid -- Strong Base

Example: CH$_3$COOH + NaOH

• Initial pH: higher than strong acid at same concentration (partial dissociation).
• Buffer region: gradual pH rise before the equivalence point.
• Equivalence point: pH \gt 7 (conjugate base of weak acid hydrolyses).
• Suitable indicator: phenolphthalein.

Strong Acid -- Weak Base

Example: HCl + NH$_3$

• Initial pH: low (strong acid).
• Equivalence point: pH \lt 7 (conjugate acid of weak base hydrolyses).
• Suitable indicator: methyl orange.

Key Features of pH Curves

• Equivalence point: The volume where stoichiometrically equivalent amounts of acid and base have reacted. The pH at this point depends on the salt formed.
• Half-equivalence point: The volume where half the acid/base has been neutralised. At this point, $\mathrm{pH} = \mathrm{p}K_a$ (for weak acid titrations), because $[\mathrm{HA}] = [\mathrm{A}^-]$.

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Buffer Solutions

Composition

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

1. A weak acid and its conjugate base, or
2. A weak base and its conjugate acid.

Examples:

• CH$_3$COOH / CH$_3$COONa$^-$ (acetic acid / acetate)
• NH$_3$ / NH$_4$Cl (ammonia / ammonium)
• H$_2$CO$_3$ / HCO$_3^-$ (carbonic acid / bicarbonate -- the blood buffer system)

Mechanism

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

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

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

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

In both cases, the ratio $[\mathrm{HA}]/[\mathrm{A}^-]$ changes only slightly, so pH remains nearly constant.

Henderson-Hasselbalch Equation

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

or equivalently:

$$\mathrm{pH} = \mathrm{p}K_a + \log\frac{[\mathrm{base}]}{[\mathrm{acid}]}$$

Buffer Capacity

Buffer capacity is the amount of acid or base that can be added before the pH changes significantly. A buffer is most effective when:

• The concentrations of the weak acid and conjugate base are large (more moles available to react).
• The ratio $[\mathrm{A}^-]/[\mathrm{HA}]$ is close to 1 (i.e., pH is close to p$K_a$).

The effective buffer range is approximately $\mathrm{p}K_a \pm 1$.

Worked Example 6: Buffer pH Calculation

Calculate the pH of a buffer solution containing $0.200\mathrm{ mol/L}$ CH$_3$COOH and $0.150\mathrm{ mol/L}$ CH$_3$COONa. ($K_a = 1.8 \times 10^{-5}$, p$K_a = 4.74$)

Using the Henderson-Hasselbalch equation:

$$\mathrm{pH} = 4.74 + \log\frac{0.150}{0.200}$$$$\mathrm{pH} = 4.74 + \log(0.750)$$$$\mathrm{pH} = 4.74 + (-0.125) = 4.62$$

Worked Example 7: pH Change When Adding Acid to a Buffer

To $500\mathrm{ mL}$ of the buffer from Worked Example 6, $0.010\mathrm{ mol}$ of HCl is added. Calculate the new pH.

Initial moles:

• $n(\mathrm{CH}_3\mathrm{COOH}) = 0.200 \times 0.500 = 0.100\mathrm{ mol}$
• $n(\mathrm{CH}_3\mathrm{COO}^-) = 0.150 \times 0.500 = 0.075\mathrm{ mol}$

Adding $0.010\mathrm{ mol}$ HCl:

• HCl reacts with CH$_3$COO$^-$: $\mathrm{CH}_3\mathrm{COO}^- + \mathrm{H}^+ \to \mathrm{CH}_3\mathrm{COOH}$
• $n(\mathrm{CH}_3\mathrm{COO}^-) = 0.075 - 0.010 = 0.065\mathrm{ mol}$
• $n(\mathrm{CH}_3\mathrm{COOH}) = 0.100 + 0.010 = 0.110\mathrm{ mol}$

$$\mathrm{pH} = 4.74 + \log\frac{0.065}{0.110}$$$$\mathrm{pH} = 4.74 + \log(0.591) = 4.74 + (-0.228) = 4.51$$

The pH changed from 4.62 to 4.51, a change of only 0.11 units, demonstrating the buffer's effectiveness.

Worked Example 8: Preparing a Buffer

What mass of sodium acetate (CH$_3$COONa, $M_r = 82.0\mathrm{ g/mol}$) must be added to $1.00\mathrm{ L}$ of $0.100\mathrm{ mol/L}$ CH$_3$COOH to produce a buffer with pH = 5.00? ($K_a = 1.8 \times 10^{-5}$)

Using the Henderson-Hasselbalch equation:

$$5.00 = 4.74 + \log\frac{[\mathrm{CH}_3\mathrm{COO}^-]}{0.100}$$$$\log\frac{[\mathrm{CH}_3\mathrm{COO}^-]}{0.100} = 0.26$$$$\frac{[\mathrm{CH}_3\mathrm{COO}^-]}{0.100} = 10^{0.26} = 1.82$$$$[\mathrm{CH}_3\mathrm{COO}^-] = 0.182\mathrm{ mol/L}$$

In $1.00\mathrm{ L}$:

$$n(\mathrm{CH}_3\mathrm{COONa}) = 0.182\mathrm{ mol}$$$$m = n \times M_r = 0.182 \times 82.0 = 14.9\mathrm{ g}$$

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Acid-Base Titrations

Titration Calculations

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

Strong acid -- strong base:

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

pH at equivalence point = 7 (neutral salt).

Weak acid -- strong base:

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

The conjugate base A$^-$ hydrolyses:

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

pH at equivalence point \gt 7.

Strong acid -- weak base:

$$\mathrm{H}^+ + \mathrm{B} \to \mathrm{BH}^+$$

The conjugate acid BH$^+$ hydrolyses:

$$\mathrm{BH}^+ + \mathrm{H}_2\mathrm{O} \rightleftharpoons \mathrm{B} + \mathrm{H}_3\mathrm{O}^+$$

pH at equivalence point \lt 7.

Worked Example 9: Titration of Weak Acid with Strong Base

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

At the equivalence point, moles of NaOH = moles of CH$_3$COOH:

$$n = 0.100 \times 0.0250 = 0.00250\mathrm{ mol}$$

Volume of NaOH required:

$$V = \frac{0.00250}{0.100} = 0.0250\mathrm{ L} = 25.0\mathrm{ mL}$$

Total volume at equivalence point: $25.0 + 25.0 = 50.0\mathrm{ mL} = 0.0500\mathrm{ L}$.

All CH$_3$COOH has been converted to CH$_3$COO$^-$:

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

The acetate ion hydrolyses:

$$K_b = \frac{K_w}{K_a} = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} = 5.56 \times 10^{-10}$$$$[\mathrm{OH}^-] = \sqrt{K_b \times [\mathrm{CH}_3\mathrm{COO}^-]} = \sqrt{5.56 \times 10^{-10} \times 0.0500}$$$$[\mathrm{OH}^-] = \sqrt{2.78 \times 10^{-11}} = 5.27 \times 10^{-6}\mathrm{ mol/L}$$$$\mathrm{pOH} = -\log(5.27 \times 10^{-6}) = 5.28$$$$\mathrm{pH} = 14 - 5.28 = 8.72$$

Worked Example 10: pH at Half-Equivalence Point

Using the titration from Worked Example 9, calculate the pH when $12.5\mathrm{ mL}$ of NaOH has been added (half-equivalence point).

At half-equivalence point, half the acid has been neutralised:

$$n(\mathrm{NaOH}) = 0.100 \times 0.0125 = 0.00125\mathrm{ mol}$$

Moles of CH$_3$COOH remaining: $0.00250 - 0.00125 = 0.00125\mathrm{ mol}$

Moles of CH$_3$COO$^-$ formed: $0.00125\mathrm{ mol}$

Since $[\mathrm{HA}] = [\mathrm{A}^-]$:

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

This is a general result: at the half-equivalence point, $\mathrm{pH} = \mathrm{p}K_a$.

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Common Pitfalls

1. Strong vs weak acid pH: A $0.10\mathrm{ mol/L}$ strong acid has pH = 1.0, but a $0.10\mathrm{ mol/L}$ weak acid has pH \gt 1.0 (typically 2--3) because it only partially dissociates. Do not assume $[\mathrm{H}^+] = c$ for weak acids.

2. Diprotic acid contribution: H$_2$SO$_4$ gives 2 H$^+$ per molecule, but H$_2$CO$_3$ does not give 2 H$^+$ at normal concentrations because $K_{a2}$ is very small. Only the first dissociation contributes significantly.

3. pH + pOH = 14 only at $25\degree\mathrm{C}$: At other temperatures, use the actual $K_w$ value. $\mathrm{pH} + \mathrm{pOH} = \mathrm{p}K_w$.

4. Buffer range: A buffer is only effective within $\pm 1$ pH unit of its p$K_a$. Outside this range, the buffer capacity is essentially zero.

5. Equivalence point pH: For weak acid -- strong base titrations, the equivalence point pH is \gt 7, not 7. For strong acid -- weak base, it is \lt 7.

6. Henderson-Hasselbalch validity: The equation assumes that the concentrations of HA and A$^-$ are much larger than $[\mathrm{H}^+]$ and $[\mathrm{OH}^-]$. It is not valid for very dilute solutions.

7. Neutral pH: Neutral means $[\mathrm{H}^+] = [\mathrm{OH}^-]$, which equals pH = 7 only at $25\degree\mathrm{C}$. At $50\degree\mathrm{C}$, neutral pH is approximately 6.63.

8. $K_a$ and $K_b$ relationship: Remember $K_a \times K_b = K_w$. This connects a conjugate acid-base pair. The conjugate base of a weak acid has a calculable $K_b$.

9. Titration indicator choice: Methyl orange (3.1--4.4) is suitable for strong acid -- weak base titrations (equivalence pH \lt 7). Phenolphthalein (8.3--10.0) is suitable for weak acid -- strong base (equivalence pH \gt 7). Do not use methyl orange for weak acid -- strong base.

10. Units in Ka/Kb: Concentrations in $K_a$ and $K_b$ expressions are in mol/L. The units of $K_a$ and $K_b$ are mol/L. $K_w$ has units of mol$^2$/L$^2$.

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Practice Problems

Question 1: Strong Acid/Base pH

(a) Calculate the pH of $0.0030\mathrm{ mol/L}$ HNO$_3$.

(b) Calculate the pH of $0.0025\mathrm{ mol/L}$ Ca(OH)$_2$ at $25\degree\mathrm{C}$.

(c) A solution has pH = 3.40. What is $[\mathrm{H}^+]$?

Answer:

(a) HNO$_3$ is a strong monoprotic acid:

$$[\mathrm{H}^+] = 0.0030\mathrm{ mol/L}$$$$\mathrm{pH} = -\log(3.0 \times 10^{-3}) = 3 - \log 3.0 = 3 - 0.477 = 2.52$$

(b) Ca(OH)$_2$ is a strong base giving 2 OH$^-$ per formula unit:

$$[\mathrm{OH}^-] = 2 \times 0.0025 = 0.0050\mathrm{ mol/L}$$$$\mathrm{pOH} = -\log(5.0 \times 10^{-3}) = 3 - \log 5.0 = 3 - 0.699 = 2.30$$$$\mathrm{pH} = 14 - 2.30 = 11.70$$

(c)

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

Question 2: Weak Acid pH and Degree of Dissociation

Hypochlorous acid (HOCl) has $K_a = 3.5 \times 10^{-8}$.

(a) Calculate the pH of $0.050\mathrm{ mol/L}$ HOCl.

(b) Calculate the percentage dissociation of HOCl at this concentration.

Answer:

(a)

$$\mathrm{HOCl} \rightleftharpoons \mathrm{H}^+ + \mathrm{OCl}^-$$$$K_a = \frac{x^2}{0.050} = 3.5 \times 10^{-8}$$$$x^2 = 1.75 \times 10^{-9}$$$$x = 4.18 \times 10^{-5}\mathrm{ mol/L}$$$$\mathrm{pH} = -\log(4.18 \times 10^{-5}) = 4.38$$

(b)

$$\mathrm{Percentage dissociation} = \frac{4.18 \times 10^{-5}}{0.050} \times 100\% = 0.084\%$$

Question 3: Conjugate Pairs and Ka/Kb

(a) The $K_b$ of NH$_3$ is $1.8 \times 10^{-5}$. Calculate $K_a$ for NH$_4^+$.

(b) Is NH$_4^+$ acidic, basic, or neutral in aqueous solution? Explain.

Answer:

(a)

$$K_a(\mathrm{NH}_4^+) = \frac{K_w}{K_b(\mathrm{NH}_3)} = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} = 5.56 \times 10^{-10}$$

(b) NH$_4^+$ is the conjugate acid of the weak base NH$_3$. Since NH$_3$ is a weak base, its conjugate acid NH$_4^+$ will donate protons in water, making the solution acidic. This is confirmed by the relatively large $K_a$ value ($5.56 \times 10^{-10} \gg K_b$ of NH$_4^+$ which would be $K_w/K_a = 1.8 \times 10^{-5}$, but wait -- we already have $K_a$ for NH$_4^+$, so we can see it is an acid). A $0.1\mathrm{ mol/L}$ NH$_4$Cl solution would have pH \lt 7.

Question 4: Buffer Preparation and pH Change

A buffer is prepared by mixing $0.150\mathrm{ mol/L}$ HCOOH ($K_a = 1.8 \times 10^{-4}$, p$K_a = 3.74$) with $0.100\mathrm{ mol/L}$ HCOONa in equal volumes.

(a) Calculate the pH of the buffer.

(b) To $100\mathrm{ mL}$ of this buffer, $5.0\mathrm{ mL}$ of $0.100\mathrm{ mol/L}$ NaOH is added. Calculate the new pH.

Answer:

(a) Equal volumes of $0.150$ and $0.100\mathrm{ mol/L}$ give concentrations of $0.0750$ and $0.0500\mathrm{ mol/L}$ respectively:

$$\mathrm{pH} = 3.74 + \log\frac{0.0500}{0.0750} = 3.74 + \log(0.667) = 3.74 + (-0.176) = 3.56$$

(b) Moles in $100\mathrm{ mL}$ of buffer:

• $n(\mathrm{HCOOH}) = 0.0750 \times 0.100 = 0.00750\mathrm{ mol}$
• $n(\mathrm{HCOO}^-) = 0.0500 \times 0.100 = 0.00500\mathrm{ mol}$

Moles of NaOH added: $n = 0.100 \times 0.0050 = 0.000500\mathrm{ mol}$

After reaction:

• $n(\mathrm{HCOOH}) = 0.00750 - 0.000500 = 0.00700\mathrm{ mol}$
• $n(\mathrm{HCOO}^-) = 0.00500 + 0.000500 = 0.00550\mathrm{ mol}$

$$\mathrm{pH} = 3.74 + \log\frac{0.00550}{0.00700} = 3.74 + \log(0.786) = 3.74 + (-0.105) = 3.64$$

Question 5: Titration Curve Analysis (Paper 2 Style)

$20.0\mathrm{ mL}$ of $0.100\mathrm{ mol/L}$ NH$_3$ ($K_b = 1.8 \times 10^{-5}$) is titrated with $0.100\mathrm{ mol/L}$ HCl.

(a) Calculate the pH at the equivalence point.

(b) State and explain which indicator would be most suitable for this titration.

(c) Calculate the pH when $10.0\mathrm{ mL}$ of HCl has been added (half-equivalence point).

Answer:

(a) At equivalence point, moles of HCl = moles of NH$_3$:

$$n = 0.100 \times 0.0200 = 0.00200\mathrm{ mol}$$

Volume of HCl: $V = 0.00200 / 0.100 = 0.0200\mathrm{ L} = 20.0\mathrm{ mL}$.

Total volume: $20.0 + 20.0 = 40.0\mathrm{ mL} = 0.0400\mathrm{ L}$.

All NH$_3$ is converted to NH$_4^+$:

$$[\mathrm{NH}_4^+] = \frac{0.00200}{0.0400} = 0.0500\mathrm{ mol/L}$$$$K_a(\mathrm{NH}_4^+) = \frac{K_w}{K_b} = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} = 5.56 \times 10^{-10}$$$$[\mathrm{H}^+] = \sqrt{K_a \times [\mathrm{NH}_4^+]} = \sqrt{5.56 \times 10^{-10} \times 0.0500}$$$$[\mathrm{H}^+] = \sqrt{2.78 \times 10^{-11}} = 5.27 \times 10^{-6}\mathrm{ mol/L}$$$$\mathrm{pH} = -\log(5.27 \times 10^{-6}) = 5.28$$

(b) The equivalence point pH is 5.28, which falls within the transition range of methyl orange (3.1--4.4) only partially. A better choice would be bromocresol green (3.8--5.4) or methyl red (4.4--6.2), which has a transition range that includes pH 5.28. From the common indicators listed in the IB syllabus, methyl orange is the closest suitable indicator for a strong acid -- weak base titration.

(c) At the half-equivalence point, $[\mathrm{NH}_3] = [\mathrm{NH}_4^+]$, so:

$$\mathrm{pH} = \mathrm{p}K_a(\mathrm{NH}_4^+) = 14 - \mathrm{p}K_b = 14 - (-\log(1.8 \times 10^{-5}))$$$$\mathrm{pH} = 14 - 4.74 = 9.26$$

Question 6: Polyprotic Acid (Paper 2 Style)

Calculate the pH of a $0.100\mathrm{ mol/L}$ H$_3$PO$_4$ solution. Use the following data: $K_{a1} = 7.5 \times 10^{-3}$, $K_{a2} = 6.2 \times 10^{-8}$, $K_{a3} = 4.2 \times 10^{-13}$.

Answer:

For the first dissociation:

$$\mathrm{H}_3\mathrm{PO}_4 \rightleftharpoons \mathrm{H}^+ + \mathrm{H}_2\mathrm{PO}_4^-$$

Since $K_{a1}$ is not very small compared to $c$, the approximation $c - x \approx c$ may not be valid. Check: $K_{a1}/c = 7.5 \times 10^{-3}/0.100 = 0.075 \gt 0.05$. The $5\%$ rule fails.

Solve the quadratic: $x^2 + K_a x - K_a \cdot c = 0$

$$x^2 + 7.5 \times 10^{-3}x - 7.5 \times 10^{-4} = 0$$$$x = \frac{-7.5 \times 10^{-3} + \sqrt{(7.5 \times 10^{-3})^2 + 4 \times 7.5 \times 10^{-4}}}{2}$$$$x = \frac{-7.5 \times 10^{-3} + \sqrt{5.625 \times 10^{-5} + 3.0 \times 10^{-3}}}{2}$$$$x = \frac{-7.5 \times 10^{-3} + \sqrt{3.056 \times 10^{-3}}}{2}$$$$x = \frac{-7.5 \times 10^{-3} + 5.528 \times 10^{-2}}{2} = \frac{4.778 \times 10^{-2}}{2} = 2.389 \times 10^{-2}$$$$[\mathrm{H}^+] \approx 0.0239\mathrm{ mol/L}$$$$\mathrm{pH} = -\log(0.0239) = 1.62$$

Note: The second and third dissociations contribute negligible H$^+$ since $K_{a2} \ll K_{a1}$.

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