Topic 14

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Constants and conversions

Quantity	Value
ionic product of water, $K_\text{w}$	1.00 × 10⁻¹⁴ mol² dm⁻⁶ at 298 K; it rises with temperature
$\text{p}K_\text{w}$	14.00 at 298 K
pH of pure water	7.00 at 298 K; neutral at any temperature means [H⁺] = [OH⁻]
ethanoic acid, $K_\text{a}$	1.74 × 10⁻⁵ mol dm⁻³, $\text{p}K_\text{a}$ 4.76
ammonium ion, $\text{p}K_\text{a}$	9.25
indicators (pH range)	methyl orange 3.1 to 4.4; methyl red 4.2 to 6.3; bromothymol blue 6.0 to 7.6; phenolphthalein 8.3 to 10.0

To find	Use	Page
pH from [H⁺]	pH = −lg [H⁺]	pH and hydrogen ion concentration
[H⁺] from pH	[H⁺] = $10^{-\text{pH}}$	pH and hydrogen ion concentration
[H⁺] of a strong acid	(number of H⁺) × c	Strong and weak acids
$K_\text{a}$	[H⁺][A⁻] ÷ [HA]	The acid dissociation constant
$\text{p}K_\text{a}$	$-\lg K_\text{a}$ ; $K_\text{a} = 10^{-\text{p}K_\text{a}}$	The acid dissociation constant
[H⁺] of a weak acid	√( $K_\text{a}$ × c)	pH of a weak acid
pH of a weak acid	½ ( $\text{p}K_\text{a}$ − lg c)	pH of a weak acid
$K_\text{a}$ from a pH	[H⁺]² ÷ c	Ka from a measured pH
ionic product of water	$K_\text{w}$ = [H⁺][OH⁻]; $\text{p}K_\text{w}$ = pH + pOH	Water and strong bases
pH of a strong base	[H⁺] = $K_\text{w}$ ÷ [OH⁻]	Water and strong bases
pH of a buffer	[H⁺] = $K_\text{a}$ × [HA] ÷ [A⁻]	pH of a buffer
buffer ratio needed	[A⁻] ÷ [HA] = $K_\text{a}$ ÷ [H⁺]	Making a buffer of a given pH
$K_\text{a}$ from a curve	pH at half $V_\text{eq}$ = $\text{p}K_\text{a}$	Buffers on the titration curve

Brønsted–Lowry acid: a proton donor. Brønsted–Lowry base: a proton acceptor.
Conjugate acid-base pair: an acid and the base formed when it loses one H⁺.
pH: the negative logarithm to base 10 of the hydrogen ion concentration in mol dm⁻³.
Strong acid: completely dissociated in aqueous solution. Weak acid: partially dissociated.
Acid dissociation constant, $K_\text{a}$ : the equilibrium constant for the dissociation of a weak acid in water.
Ionic product of water, $K_\text{w}$ : $K_\text{w} = [\ce{H+}][\ce{OH-}]$ .
$\text{p}K_\text{a}$ and $\text{p}K_\text{w}$ : $-\lg K_\text{a}$ and $-\lg K_\text{w}$ .
Equivalence point: the volume of titrant that has reacted exactly in the ratio of the balanced equation.
Buffer solution: a solution that resists changes in pH when small amounts of acid or alkali are added.

pH of a strong acid: [H⁺] = number of H⁺ × c, then pH = −lg [H⁺].
pH of a weak acid: state [H⁺] = [A⁻] and [HA] = initial c, then [H⁺] = √( $K_\text{a}$ × c).
$K_\text{a}$ from a mass and a pH: mass → n → c; [H⁺] = $10^{-\text{pH}}$ ; $K_\text{a}$ = [H⁺]² ÷ c.
pH of a strong base: [OH⁻] = number of OH⁻ × c; [H⁺] = $K_\text{w}$ ÷ [OH⁻].
pH of a buffer: moles of acid and base after any reaction → ratio → [H⁺] = $K_\text{a}$ × ratio.
Choose an indicator: its pH range must lie inside the vertical section.
$K_\text{a}$ from a curve: find $V_\text{eq}$ from the middle of the steep part, halve it, read the pH, $K_\text{a}$ = $10^{-\text{pH}}$ .

Combination	Start pH	Equivalence pH	Vertical section	Indicator
strong acid + strong base	1.00	7.00	about 3 to 11	any
weak acid + strong base	2.88	8.73	about 6.5 to 11	phenolphthalein
strong acid + weak base	11.12	5.28	about 7.6 to 3	methyl orange or methyl red
weak acid + weak base	2.88	7.00	none	none: pH meter

0.100 mol dm⁻³ each, 25.0 cm³ in the flask. Diprotic acids and bases have a second equivalence point at twice the volume of the first.

Calling a concentrated acid "strong" or a dilute acid "weak".
Using [H⁺] = c for a weak acid, or forgetting the square root.
Forgetting to double [H⁺] for H₂SO₄, or [OH⁻] for Ba(OH)₂.
Using pH = −lg [OH⁻] for a base (that is the pOH).
Taking pH 7 as neutral at every temperature.
Pairing the two reactants as a conjugate pair.
Using the amount of NaOH as the amount of salt, without subtracting it from the acid.
Inverting the ratio [A⁻] ÷ [HA].
Reading the pH at $V_\text{eq}$ instead of half of $V_\text{eq}$ for $K_\text{a}$ .
Choosing an indicator by colour, or because it changes "near pH 7".

Try these with no notes, then check the answers.

Identify the two conjugate acid-base pairs in $\ce{HCO3- + H2O <=> H2CO3 + OH-}$ .
Calculate the pH of 0.0250 mol dm⁻³ sulfuric acid, treating both protons as fully released.
Calculate the pH of 0.150 mol dm⁻³ ethanoic acid. $K_\text{a}$ = 1.74 × 10⁻⁵ mol dm⁻³.
Calculate the pH of 0.0200 mol dm⁻³ barium hydroxide at 298 K.
A 0.0400 mol dm⁻³ solution of a weak acid has pH 2.95. Calculate $K_\text{a}$ .
Explain why phenolphthalein is unsuitable for the titration of ammonia with hydrochloric acid.
Calculate the pH of a buffer containing 0.100 mol dm⁻³ ethanoic acid and 0.300 mol dm⁻³ sodium ethanoate.
A weak acid is titrated with 0.100 mol dm⁻³ NaOH. The equivalence volume is 18.4 cm³, and the pH after 9.20 cm³ is 3.77. Calculate $K_\text{a}$ and explain your method.

Answers cover these until you have tried the questions

HCO₃⁻ gains H⁺ to become H₂CO₃: pair HCO₃⁻ (base) / H₂CO₃ (acid). H₂O loses H⁺ to become OH⁻: pair H₂O (acid) / OH⁻ (base).
[H⁺] = 2 × 0.0250 = 0.0500 mol dm⁻³. pH = −lg 0.0500 = 1.30
[H⁺] = √(1.74 × 10⁻⁵ × 0.150) = √(2.61 × 10⁻⁶) = 1.62 × 10⁻³ mol dm⁻³. pH = 2.79 (assuming [H⁺] = [A⁻] and [HA] = 0.150).
[OH⁻] = 2 × 0.0200 = 0.0400 mol dm⁻³. [H⁺] = 1.00 × 10⁻¹⁴ ÷ 0.0400 = 2.5 × 10⁻¹³ mol dm⁻³. pH = 12.60
[H⁺] = $10^{-2.95}$ = 1.12 × 10⁻³ mol dm⁻³. $K_\text{a}$ = (1.12 × 10⁻³)² ÷ 0.0400 = 1.26 × 10⁻⁶ ÷ 0.0400 = 3.1 × 10⁻⁵, so 3.1 × 10⁻⁵ mol dm⁻³.
The vertical section for ammonia with hydrochloric acid is about pH 7.6 to 3.0. Phenolphthalein changes colour at pH 8.3 to 10.0, which is outside (above) the vertical section, so the colour change happens before the equivalence point and is not sharp. Methyl orange or methyl red is suitable.
[H⁺] = 1.74 × 10⁻⁵ × (0.100 ÷ 0.300) = 5.80 × 10⁻⁶ mol dm⁻³. pH = 5.24
Half of 18.4 cm³ is 9.20 cm³. At half-neutralisation [HA] = [A⁻], so pH = $\text{p}K_\text{a}$ = 3.77. $K_\text{a}$ = $10^{-3.77}$ = 1.7 × 10⁻⁴ mol dm⁻³.