the distinction between strong and weak acids and bases using the extent of ionisation, the acid ionisation constant Ka and base ionisation constant Kb, and the relationship between the strength of an acid and the strength of its conjugate base

What this dot point is asking

VCAA wants you to define strong and weak acids and bases in terms of the extent of ionisation in water, to use the acid ionisation constant $K_a$ and base ionisation constant $K_b$ to compare strengths, and to state the inverse relationship: the stronger the acid, the weaker its conjugate base.

The answer

What "strong" and "weak" mean

A strong acid ionises essentially completely in water. Almost every $HA$ molecule donates its proton.

The arrow points one way because the reverse (a chloride ion grabbing a proton off hydronium) is negligible at equilibrium.

A weak acid ionises only partially. Most $HA$ molecules stay protonated.

The arrows go both ways because both forward and reverse processes happen significantly. At equilibrium only a few percent (or less) of the acid is ionised.

The same dichotomy applies to bases. $NaOH$ is strong (fully dissociated to give $OH^-$). Ammonia $NH_3$ is weak (partially protonated by water).

Strong is not the same as concentrated

Strong vs weak describes the extent of ionisation. Concentrated vs dilute describes the amount per litre. These are independent. A weak acid can be concentrated (glacial ethanoic acid is $17\ mol\ L^{-1}$ and still weak). A strong acid can be dilute ($10^{-4}\ mol\ L^{-1}$ HCl is strong and dilute).

The acid ionisation constant Ka

For the general weak-acid ionisation $HA + H_2O \rightleftharpoons H_3O^+ + A^-$, the equilibrium constant (with water activity absorbed into $K_a$) is:

$K_a$ measures how far to the right the ionisation lies. Bigger $K_a$ = more ionisation = stronger acid.

Many VCAA questions give $K_a$ in the question or in the data book. You usually do not need to derive it.

The corresponding $pK_a = -\log_{10}(K_a)$. Smaller $pK_a$ = stronger acid. $HF$ has $pK_a = 3.17$, $CH_3COOH$ has $pK_a = 4.74$, so $HF$ is the stronger acid.

The base ionisation constant Kb

For a weak base $B + H_2O \rightleftharpoons BH^+ + OH^-$:

Bigger $K_b$ = stronger base. $NH_3$ has $K_b = 1.8 \times 10^{-5}$, so it has the same "strength" as ethanoic acid does on the acid side.

The Ka, Kb and Kw relationship

For any conjugate acid-base pair in water at 25 deg C:

so:

This formalises the stronger the acid, the weaker its conjugate base. A very strong acid (small $pK_a$) has a very weak conjugate base (large $pK_b$). $HCl$ has effectively no basic character on $Cl^-$. The conjugate base of a very weak acid like water itself is $OH^-$, which is a strong base.

Why the weak acid pH is higher than the strong acid pH at the same concentration

Two solutions of $0.10\ mol\ L^{-1}$ acid:

$HCl$: strong, fully ionised, $[H_3O^+] = 0.10\ mol\ L^{-1}$, $pH = 1.00$.

$CH_3COOH$: weak, only about $1.3\%$ ionised, $[H_3O^+] \approx 1.3 \times 10^{-3}\ mol\ L^{-1}$, $pH \approx 2.87$.

Same concentration, very different pH. The weak acid pH is higher because most of the acid never lets its proton go.

Worked example

Methylamine ($CH_3NH_2$) has $K_b = 4.4 \times 10^{-4}\ mol\ L^{-1}$ at 25 deg C. Calculate (a) the $K_a$ of its conjugate acid $CH_3NH_3^+$, and (b) state whether methylamine is a stronger or weaker base than ammonia ($K_b = 1.8 \times 10^{-5}$).

(a) $K_a(CH_3NH_3^+) = K_w / K_b = (1.0 \times 10^{-14}) / (4.4 \times 10^{-4}) = 2.3 \times 10^{-11}\ mol\ L^{-1}$.

(b) $K_b$ for methylamine is larger than $K_b$ for ammonia (by a factor of about 24), so methylamine is the stronger base. The methyl group is electron-donating and pushes electron density onto the nitrogen lone pair, making it more available to accept a proton.

Common traps

Confusing $K_a$ with $[H^+]$

$K_a$ is a constant for a given acid at a given temperature. $[H^+]$ depends on both $K_a$ and the concentration of the acid.

Using a one-way arrow for a weak acid

Weak acids must be drawn with equilibrium arrows ($\rightleftharpoons$).

Forgetting that $K_w = K_a \times K_b$

A common SAC trap: given $K_a$, find $K_b$ of the conjugate. Divide $K_w$ by $K_a$.

Calling something a strong base because it has a high pH

$1\ mol\ L^{-1}\ NH_3$ has pH about 11.6, similar to dilute NaOH, but ammonia is still classified as weak because the extent of ionisation, not the absolute pH, is the criterion.

Listing $HF$ as a strong acid

All other hydrogen halides ($HCl$, $HBr$, $HI$) are strong. $HF$ is weak (the $H-F$ bond is unusually strong).

In one sentence

Strong acids and bases ionise completely in water; weak acids and bases ionise only partially with extent measured by $K_a$ or $K_b$; bigger $K_a$ means stronger acid and (because $K_a \times K_b = K_w$) weaker conjugate base.

Examples in context

Example 1. Lactic acid in Mornington Peninsula chardonnay malolactic conversion. Wineries on the Mornington Peninsula encourage malolactic fermentation, in which $\text{Oenococcus oeni}$ bacteria convert the sharp diprotic malic acid ($pK_{a1} = 3.40$, $pK_{a2} = 5.11$) into the softer monoprotic lactic acid ($pK_a = 3.86$). Lactic acid is weaker than malic, so the wine's pH drifts up from $3.0$ to about $3.4$, reducing perceived acidity. Calculating $[\text{H}^+]$ for $0.020 \, \text{mol/L}$ lactic acid using $K_a = 1.38 \times 10^{-4}$: $[\text{H}^+] = \sqrt{K_a \cdot c} = \sqrt{1.38 \times 10^{-4} \times 0.020} = 1.66 \times 10^{-3} \, \text{mol/L}$, pH $= 2.78$ (sub-buffered). Winemakers monitor pH to predict perceived sourness in the bottle.

Example 2. Ammonia release in Murray-Darling fish farms. Australian Native Fish Farms near Mildura monitor ammonia/ammonium equilibria in tank water. Ammonia is a weak base: $\text{NH}_3 + \text{H}_2 \text{O} \rightleftharpoons \text{NH}_4^+ + \text{OH}^-$, $K_b = 1.80 \times 10^{-5}$. At pH $7.5$, almost all dissolved nitrogen sits as the non-toxic $\text{NH}_4^+$. At pH $9.0$, however, the toxic neutral $\text{NH}_3$ accounts for $\sim 30\%$ of the total because the ratio $[\text{NH}_3] / [\text{NH}_4^+] = K_b / [\text{OH}^-]$. Conjugate-base mathematics: $pK_b = 4.74$, so $pK_a$ of $\text{NH}_4^+ = 14 - 4.74 = 9.26$, meaning $\text{NH}_4^+$ is a weak acid much weaker than acetic acid ($pK_a = 4.76$).

Try this

Q1. Distinguish between a "strong" acid and a "concentrated" acid, giving one example of each. [2 marks]

• Cue. Strong: ionises completely (e.g. $0.1 \, \text{mol/L}$ HCl). Concentrated: high concentration regardless of extent (e.g. $17 \, \text{mol/L}$ acetic acid is concentrated but weak).

Q2. A $0.100 \, \text{mol/L}$ solution of a monoprotic weak acid has pH $= 3.20$. (a) Calculate $[\text{H}^+]$. (b) Determine $K_a$. (c) Estimate the percentage ionisation. [4 marks]

• Cue. (a) $[\text{H}^+] = 10^{-3.20} = 6.31 \times 10^{-4} \, \text{mol/L}$. (b) $K_a \approx (6.31 \times 10^{-4})^2 / 0.100 = 3.98 \times 10^{-6}$. (c) Ionisation $= 6.31 \times 10^{-4} / 0.100 \times 100 = 0.63\%$.

Q3. Ethanoic acid has $K_a = 1.74 \times 10^{-5}$. (a) Calculate $pK_a$. (b) Calculate $K_b$ for its conjugate base, ethanoate. (c) Predict whether the pH of $0.10 \, \text{mol/L}$ $\text{CH}_3 \text{COONa}$ will be greater than, equal to or less than $7$. [2+2+2 marks]

• Cue. (a) $pK_a = 4.76$. (b) $K_b = K_w / K_a = 5.75 \times 10^{-10}$. (c) Greater than 7; ethanoate is a weak base.