Acids and bases

Water is very slightly ionised, and every acid and base calculation is a statement about which way that ionisation has been pushed.

The equilibrium machinery of the fourth lesson applies unchanged to proton transfer, and this is where it earns its keep, because the numbers involved span sixteen orders of magnitude and are therefore always handled as logarithms. Nothing new is assumed here beyond $K$ and the reaction quotient.

Three definitions, each for a different job

Arrhenius, in 1884, defined an acid as a substance that releases hydrogen ions in water and a base as one that releases hydroxide. It is correct as far as it goes and it excludes ammonia, which is plainly a base and contains no hydroxide.

The definition used throughout this lesson is Brønsted and Lowry's, from 1923: an acid is a proton donor and a base is a proton acceptor. The gain is that acidity becomes a relationship rather than a property, since a donor needs an acceptor. Every acid has a conjugate base, what remains when the proton has gone, and the reaction is always a competition between two bases for one proton:

CCOOH + O rightleftharpoons CCO +

Here water is the base. In the presence of hydrogen chloride water is also the base; in the presence of ammonia it is the acid. A substance that can do either is amphiprotic, and water's being so is why it is the reference solvent.

Lewis's definition, from the same year, is broader still: an acid is an electron pair acceptor. That covers boron trifluoride and metal ions, which have no proton to donate, and it is the definition organic chemistry works in. It is not needed below, where every reaction involves a proton moving.

The free proton itself does not exist in solution. It is , or more accurately a shifting cluster of several water molecules, and writing is a convenient abbreviation rather than a species.

Water ionises, slightly

Water transfers a proton to itself, $2 O rightleftharpoons + O$ , and since the solvent is a pure liquid with activity $1$ , the equilibrium constant contains only the two ions:

K_{w} = [] [O] = 1.0 \times 10^{- 14} at 25 °C

In pure water the two concentrations are equal, so each is $1.0 \times 10^{- 7}$ mol dm⁻³. Out of about $55$ moles of water in a cubic decimetre, roughly one molecule in $5 \times 10^{8}$ is ionised at any moment.

The product is the point. $K_{w}$ holds in every aqueous solution, so fixing one ion fixes the other, and an acidic solution still contains hydroxide, just very little of it.

$K_{w}$ is an equilibrium constant, so it depends on temperature, and the ionisation is endothermic ( $Δ H = + 55.8$ kJ mol⁻¹), so van 't Hoff says it rises with temperature. At $60$ °C, $K_{w} = 9.6 \times 10^{- 14}$ , so neutral water has $[] = 3.1 \times 10^{- 7}$ and a pH of $6.51$ . That water is not acidic: it has equal concentrations of both ions, which is what neutral means. Neutrality is pH $7$ only at $25$ °C, and blood at $37$ °C is neutral at about $6.8$ .

The pH scale

Søren Sørensen introduced the logarithmic measure in 1909 while working on beer at the Carlsberg laboratory:

pH = - \log_{10} a () \approx - \log_{10} []

with pOH defined the same way, and taking logarithms of $K_{w}$ gives $pH + pOH = 14.00$ at $25$ °C. The same operator applied to an equilibrium constant gives $p K = - \log_{10} K$ , which is used constantly below.

A strong acid is one that transfers its proton completely, so its concentration is the hydronium concentration and the pH follows immediately.

Example. Find the pH of $0.025$ mol dm⁻³ nitric acid, and of $0.0050$ mol dm⁻³ barium hydroxide.

Nitric acid is strong, so $[] = 0.025$ and $pH = - \log_{10} (0.025) = 1.60$ . Barium hydroxide is strong and supplies two hydroxides per formula unit, so $[O] = 0.010$ , giving $pOH = 2.00$ and $pH = 12.00$ .

Now you. Find the pH of $1.0 \times 10^{- 8}$ mol dm⁻³ hydrochloric acid. Be careful.

Answer

Not $8$ , which would make an acid alkaline. At this dilution the water's own ionisation dominates. Charge balance requires $[] = [C] + [O] = 10^{- 8} + K_{w} / []$ , a quadratic whose root is $[] = 1.05 \times 10^{- 7}$ , giving $pH = 6.98$ . Slightly acidic, as it must be, and the shortcut of reading the concentration off the label fails below about $10^{- 6}$ mol dm⁻³.

$K_{a}$ , $p K_{a}$ , and conjugate pairs

A weak acid is one whose proton transfer to water is an equilibrium with a constant well below one:

K_{a} = \frac{[] []}{[HA]}

Water is omitted, being the solvent. Acetic acid has $K_{a} = 1.75 \times 10^{- 5}$ , so $p K_{a} = 4.76$ ; chloroacetic acid $1.36 \times 10^{- 3}$ , so $2.87$ ; hydrofluoric acid $3.17$ ; ammonium ion $9.25$ . A smaller $p K_{a}$ means a stronger acid, and each unit is a factor of ten.

For the conjugate base, $K_{b}$ describes its reaction with water, and multiplying the two expressions together makes every term cancel except the ions of water:

K_{a} K_{b} = K_{w} or p K_{a} + p K_{b} = 14.00

So a strong acid has a negligibly weak conjugate base, which is why chloride does nothing in solution, and a weak acid has a conjugate base that matters: acetate has $K_{b} = 5.7 \times 10^{- 10}$ , small but not zero, which is the subject of the last section. One table of $K_{a}$ values therefore covers bases too.

Water also imposes a ceiling. Any acid stronger than hands its proton to the solvent completely, so hydrochloric, nitric and perchloric acids are indistinguishable in water at equal concentration. That is the levelling effect, and comparing them requires a less basic solvent such as acetic acid, which is how their true relative strengths are known.

The pH of a weak acid

The calculation is the equilibrium table of the fourth lesson. Let $x$ be the hydronium concentration produced; then the acid is depleted to $C - x$ and the conjugate base is $x$ , so $K_{a} = x^{2} / (C - x)$ . That is a quadratic, and the universal shortcut is to assume $x ≪ C$ and drop it from the denominator, giving $x = \sqrt{K_{a} C}$ .

The shortcut has to be checked rather than trusted. The convention is that it is acceptable when $x$ is under five per cent of $C$ , and the discipline is to compute $x$ and then look, because when it fails it fails quietly.

Example. Find the pH of $0.100$ mol dm⁻³ acetic acid, $K_{a} = 1.75 \times 10^{- 5}$ , and the fraction ionised.

The approximation gives $x = \sqrt{1.75 \times 10^{- 5} \times 0.100} = 1.32 \times 10^{- 3}$ mol dm⁻³, which is $1.3$ per cent of $0.100$ , comfortably within the limit. So $pH = - \log_{10} (1.32 \times 10^{- 3}) = 2.88$ . Solving the quadratic exactly gives $1.314 \times 10^{- 3}$ and the same pH to two decimals. Only about one molecule in eighty has given up its proton, which is what "weak" means quantitatively.

Now you. Find the pH of $0.0010$ mol dm⁻³ acetic acid, using the approximation first and then checking it.

Answer

The approximation gives $x = \sqrt{1.75 \times 10^{- 8}} = 1.32 \times 10^{- 4}$ , which is $13$ per cent of $0.0010$ and so not acceptable. Solving $x^{2} + 1.75 \times 10^{- 5} x - 1.75 \times 10^{- 8} = 0$ gives $x = 1.24 \times 10^{- 4}$ and $pH = 3.91$ , against $3.88$ from the shortcut. Note also that the fraction ionised has risen from $1.3$ to $12$ per cent on dilution, which is the equilibrium shifting toward more particles exactly as the fifth lesson said it would.

What makes one acid stronger than another

Three effects account for most of the variation, and all three are statements about the stability of the conjugate base, since that is what the acid becomes.

Down a group, bond strength dominates. Hydrofluoric acid is weak with $p K_{a} = 3.17$ while hydrochloric, hydrobromic and hydroiodic acids are all strong and increasingly so, because the hydrogen to halogen bond gets longer and weaker down the group. Electronegativity would predict the opposite order, and it loses.

Across a period, and for oxoacids, charge stabilisation dominates. Adding an oxygen to a chlorine oxoacid pulls electron density away from the resulting anion and spreads its negative charge over more atoms, and each addition drops $p K_{a}$ by roughly five units: hypochlorous $7.5$ , chlorous $1.9$ , chloric about $- 1$ , perchloric about $- 8$ . Sixteen orders of magnitude, from three oxygen atoms.

The same logic works at a distance, which is the inductive effect. Replacing one hydrogen of acetic acid with chlorine drops $p K_{a}$ from $4.76$ to $2.87$ , a factor of nearly eighty, although the chlorine is two atoms from the acidic proton. Adding two more chlorines gives trichloroacetic acid at $0.66$ , comparable to a mineral acid. The electronegative atom pulls electron density along the bonds and stabilises the anion.

Salts are not neutral

Dissolve sodium acetate in water and the solution is alkaline, because acetate is a base and takes a proton from water. Dissolve ammonium chloride and the solution is acidic. This is not a new phenomenon: it is the $K_{a} K_{b} = K_{w}$ relation being used from the other end.

Example. Find the pH of $0.100$ mol dm⁻³ ammonium chloride, given $K_{b} = 1.8 \times 10^{- 5}$ for ammonia.

Chloride is the conjugate base of a strong acid and does nothing. The ammonium ion is a weak acid with $K_{a} = K_{w} / K_{b} = 1.0 \times 10^{- 14} / 1.8 \times 10^{- 5} = 5.6 \times 10^{- 10}$ . Then $x = \sqrt{5.6 \times 10^{- 10} \times 0.100} = 7.5 \times 10^{- 6}$ , which is a tiny fraction of $0.100$ , so $pH = 5.13$ .

Now you. Is a solution of sodium acetate acidic, alkaline or neutral, and what is $K_{b}$ for acetate given $K_{a} = 1.75 \times 10^{- 5}$ for acetic acid?

Answer

Alkaline. Sodium does nothing, but acetate accepts a proton from water and releases hydroxide, with $K_{b} = K_{w} / K_{a} = 1.0 \times 10^{- 14} / 1.75 \times 10^{- 5} = 5.7 \times 10^{- 10}$ . For a $0.100$ mol dm⁻³ solution that gives $[O] = \sqrt{5.7 \times 10^{- 11}} = 7.6 \times 10^{- 6}$ , so $pOH = 5.12$ and $pH = 8.88$ .

Polyprotic acids ionise in stages, each with its own constant, and the constants fall steeply because pulling a second proton off an already negative ion is much harder. Carbonic acid has $p K_{a 1} = 6.35$ and $p K_{a 2} = 10.33$ , phosphoric acid $2.15$ , $7.20$ and $12.35$ . Separations of four or five units mean the stages barely overlap, so each can usually be treated on its own.

The mixture that resists

One case has been left out and it is the most useful of all. Every calculation above involved a weak acid alone or its conjugate base alone. Put both in the same solution, in comparable amounts, and something new happens: the pH becomes insensitive to added acid or base, and to dilution.

That is a buffer, and since blood, seawater and every enzyme assay depend on one, it deserves its own treatment. It also turns out to be the key to reading a titration curve, which is the next lesson.