Acids and bases are equilibria between ions in water. The Arrhenius, Brønsted–Lowry and Lewis concepts; the ionic product of water and the pH scale; ionisation constants $K_{a}$ and $K_{b}$ and the degree of ionisation; polybasic acids, what makes an acid strong, and the common-ion effect.
A $0.1\ \mathrm{mol\,L^{-1}}$ solution of hydrochloric acid and a $0.1\ \mathrm{mol\,L^{-1}}$ solution of acetic acid (ethanoic acid, the acid of vinegar) contain the same amount of acid per litre, and both can neutralise exactly the same amount of sodium hydroxide. Yet the hydrochloric acid stings a cut, attacks marble violently and conducts electricity brilliantly, while the acetic acid fizzes gently with marble and lights a conductivity bulb only dimly. The difference is not how much acid is present but how much of it is ionised — and that is an equilibrium question.
In this section the ideas of Sections 6.1 and 6.2 are applied to ions in water. You will compare three definitions of acids and bases, see that water itself is in equilibrium with $\mathrm{H^{+}}$ and $\mathrm{OH^{-}}$ ions, and use that equilibrium to build the pH scale. Then you will calculate the pH of strong and weak acids and bases with the ionisation constants $K_{a}$ and $K_{b}$, look at acids that can lose more than one proton, explain why some acids are stronger than others, and see how adding a common ion holds back the ionisation of a weak acid.
Electrolytes and Three Concepts of Acids and Bases
Michael Faraday called substances whose aqueous solutions conduct electricity electrolytes. Sodium chloride is completely ionised in water and conducts well; acetic acid is only partly ionised, and its solution conducts poorly.
📘 Definition 6.7 — Strong and weak electrolytes
A strong electrolyte is almost completely ionised in aqueous solution (strong acids such as $\mathrm{HCl}$, $\mathrm{HNO_{3}}$, $\mathrm{H_{2}SO_{4}}$ (first step); strong bases such as $\mathrm{NaOH}$, $\mathrm{KOH}$; nearly all salts). A weak electrolyte is only partly ionised; in its solution the un-ionised molecules are in equilibrium with the ions, for example $\mathrm{CH_{3}COOH(aq)\rightleftharpoons H^{+}(aq)+CH_{3}COO^{-}(aq)}$. The equilibria between ions and molecules in solution are called ionic equilibria.
📘 Definition 6.8 — Arrhenius acids and bases
An Arrhenius acid produces hydrogen ions, $\mathrm{H^{+}(aq)}$, in water: $\mathrm{HX(aq)\rightarrow H^{+}(aq)+X^{-}(aq)}$. An Arrhenius base produces hydroxide ions: $\mathrm{MOH(aq)\rightarrow M^{+}(aq)+OH^{-}(aq)}$. A bare proton cannot exist in water; it binds to a water molecule as the hydronium ion $\mathrm{H_{3}O^{+}}$, and $\mathrm{H^{+}(aq)}$ and $\mathrm{H_{3}O^{+}(aq)}$ are used to mean the same thing.
The Arrhenius idea works only in water and cannot explain why ammonia, which contains no $\mathrm{OH}$ group, is a base. In 1923 Johannes Brønsted and Thomas Lowry independently proposed a broader definition based on the transfer of a proton.
📘 Definition 6.9 — Brønsted–Lowry acids, bases and conjugate pairs
A Brønsted–Lowry acid is a proton ($\mathrm{H^{+}}$) donor; a Brønsted–Lowry base is a proton acceptor. When an acid loses a proton, what remains is its conjugate base; when a base gains a proton it becomes its conjugate acid. An acid and its conjugate base differ by exactly one proton and form a conjugate acid–base pair:
$$\mathrm{NH_{3}(aq)+H_{2}O(l)\rightleftharpoons NH_{4}^{+}(aq)+OH^{-}(aq)}$$
Here $\mathrm{H_{2}O}$ (acid) / $\mathrm{OH^{-}}$ (its conjugate base) and $\mathrm{NH_{3}}$ (base) / $\mathrm{NH_{4}^{+}}$ (its conjugate acid) are the two pairs. A species that can act both as an acid and as a base, such as water or $\mathrm{HCO_{3}^{-}}$, is called amphoteric (amphiprotic).
The stronger an acid, the weaker its conjugate base: $\mathrm{HCl}$ gives up its proton completely, so $\mathrm{Cl^{-}}$ has almost no tendency to take one back; acetic acid holds its proton more firmly, so the acetate ion is a noticeably basic species. Some substances act as acids without having a proton to give at all. Boron trifluoride reacts with ammonia, and $\mathrm{AlCl_{3}}$ with chloride ions, by accepting an electron pair. Gilbert Lewis defined acids and bases in terms of electron pairs.
📘 Definition 6.10 — Lewis acids and bases
A Lewis acid is an electron-pair acceptor; a Lewis base is an electron-pair donor. Lewis bases have a lone pair ($\mathrm{NH_{3}}$, $\mathrm{H_{2}O}$, $\mathrm{OH^{-}}$, $\mathrm{F^{-}}$); Lewis acids are electron-deficient molecules ($\mathrm{BF_{3}}$, $\mathrm{AlCl_{3}}$), cations ($\mathrm{H^{+}}$, $\mathrm{Mg^{2+}}$, $\mathrm{Co^{3+}}$) or molecules with a polar multiple bond that can accept a pair. For example $\mathrm{BF_{3}+:NH_{3}\rightarrow F_{3}B{\leftarrow}NH_{3}}$. Every Brønsted base is also a Lewis base; the Lewis definition of acids is much wider.
✏️ Example 1 — Conjugate pairs and Lewis acids
(a) Write the conjugate base of $\mathrm{HF}$, $\mathrm{H_{2}SO_{4}}$, $\mathrm{HCO_{3}^{-}}$ and $\mathrm{NH_{4}^{+}}$, and the conjugate acid of $\mathrm{NH_{2}^{-}}$, $\mathrm{NH_{3}}$, $\mathrm{HCOO^{-}}$ and $\mathrm{H_{2}O}$. (b) In $\mathrm{HCl(aq)+H_{2}O(l)\rightarrow H_{3}O^{+}(aq)+Cl^{-}(aq)}$, identify the two conjugate pairs. (c) Classify as Lewis acid or Lewis base: $\mathrm{OH^{-}}$, $\mathrm{BCl_{3}}$, $\mathrm{F^{-}}$, $\mathrm{H^{+}}$, $\mathrm{NH_{3}}$, $\mathrm{Al^{3+}}$. (d) Show that $\mathrm{HCO_{3}^{-}}$ is amphoteric.
💡 Show solution▶
(a) Conjugate bases (remove one $\mathrm{H^{+}}$): $\mathrm{F^{-}}$, $\mathrm{HSO_{4}^{-}}$, $\mathrm{CO_{3}^{2-}}$, $\mathrm{NH_{3}}$. Conjugate acids (add one $\mathrm{H^{+}}$): $\mathrm{NH_{3}}$, $\mathrm{NH_{4}^{+}}$, $\mathrm{HCOOH}$, $\mathrm{H_{3}O^{+}}$.
(c) Lewis bases (lone-pair donors): $\mathrm{OH^{-}}$, $\mathrm{F^{-}}$, $\mathrm{NH_{3}}$. Lewis acids (pair acceptors): $\mathrm{BCl_{3}}$ (only six electrons on B), $\mathrm{H^{+}}$, $\mathrm{Al^{3+}}$.
(d) As an acid: $\mathrm{HCO_{3}^{-}+OH^{-}\rightarrow CO_{3}^{2-}+H_{2}O}$. As a base: $\mathrm{HCO_{3}^{-}+H_{3}O^{+}\rightarrow H_{2}CO_{3}+H_{2}O}$. It both donates and accepts a proton.
The Ionisation of Water and the pH Scale
Even the purest water conducts electricity very slightly, because water is amphoteric and a tiny fraction of its molecules transfer protons to each other: $\mathrm{H_{2}O(l)+H_{2}O(l)\rightleftharpoons H_{3}O^{+}(aq)+OH^{-}(aq)}$. The concentration of water itself, $1000/18.0=55.5\ \mathrm{mol\,L^{-1}}$, is practically constant and is absorbed into the equilibrium constant.
📐 Rule 6.11 — The ionic product of water
In water and in every dilute aqueous solution,
$$K_{w}=[\mathrm{H^{+}}][\mathrm{OH^{-}}]$$
$$K_{w}=1.0\times10^{-14}\ \mathrm{mol^{2}\,L^{-2}}\ \text{at }298\ \mathrm{K}.$$
In pure water $[\mathrm{H^{+}}]=[\mathrm{OH^{-}}]=1.0\times10^{-7}\ \mathrm{mol\,L^{-1}}$. A solution is acidic if $[\mathrm{H^{+}}]\gt[\mathrm{OH^{-}}]$, neutral if they are equal, and basic if $[\mathrm{H^{+}}]\lt[\mathrm{OH^{-}}]$. The ionisation of water is endothermic, so $K_{w}$ increases with temperature (about $5.5\times10^{-14}$ at $323\ \mathrm{K}$).
Hydrogen-ion concentrations range over many powers of ten — from about $1\ \mathrm{mol\,L^{-1}}$ in battery acid to $10^{-14}\ \mathrm{mol\,L^{-1}}$ in a strong alkali — so the Danish chemist Søren Sørensen introduced (in 1909) a logarithmic scale.
🧰 Tool from mathematics — logarithms to base 10
$\log10^{n}=n$; $\log(ab)=\log a+\log b$; $\log(1/a)=-\log a$. Useful values: $\log2=0.301$, $\log3=0.477$, $\log5=0.699$, $\log1.8=0.255$. If $[\mathrm{H^{+}}]=m\times10^{-n}$, then $\mathrm{pH}=n-\log m$; conversely $[\mathrm{H^{+}}]=10^{-\mathrm{pH}}$ (e.g. $\mathrm{pH}=3.70$ gives $10^{-3.70}=10^{0.30}\times10^{-4}=2.0\times10^{-4}$). Keep as many decimal places in a pH as there are significant figures in $[\mathrm{H^{+}}]$.
📘 Definition 6.11 — pH, pOH and p$K_{w}$
$$\mathrm{pH}=-\log[\mathrm{H^{+}}],\qquad \mathrm{pOH}=-\log[\mathrm{OH^{-}}]$$
(strictly the logarithm of the hydrogen-ion activity, which equals its molar concentration in dilute solution). Taking logarithms of $K_{w}$ gives
$$\mathrm{pH}+\mathrm{pOH}=\mathrm{p}K_{w}=14.00\ \ \text{at }298\ \mathrm{K}.$$
At $298\ \mathrm{K}$: acidic $\mathrm{pH}\lt7$, neutral $\mathrm{pH}=7$, basic $\mathrm{pH}\gt7$. A change of 1 pH unit is a tenfold change in $[\mathrm{H^{+}}]$.
Fluid
gastric juice
lemon juice
black coffee
milk
blood
sea water
milk of magnesia
household ammonia
pH (about)
1.2
2.2
5.0
6.8
7.4
8.1
10.5
11.5
Strong acids and bases are completely ionised, so their $[\mathrm{H^{+}}]$ or $[\mathrm{OH^{-}}]$ follows directly from the concentration and the formula. A very dilute strong acid is the exception: below about $10^{-6}\ \mathrm{mol\,L^{-1}}$ the $\mathrm{H^{+}}$ from water is no longer negligible.
✏️ Example 2 — pH of strong acids and bases
At $298\ \mathrm{K}$, find (a) the pH of $0.010\ \mathrm{mol\,L^{-1}}$ $\mathrm{HCl}$; (b) the pH of $5.0\times10^{-3}\ \mathrm{mol\,L^{-1}}$ $\mathrm{Ba(OH)_{2}}$; (c) $[\mathrm{H^{+}}]$ and $[\mathrm{OH^{-}}]$ in a soft drink of pH $3.76$; (d) the pH of $1.0\times10^{-8}\ \mathrm{mol\,L^{-1}}$ $\mathrm{HCl}$ — and why the answer is not $8$.
(d) An acid cannot have a pH above 7. Here water's own ions matter: $[\mathrm{H^{+}}]=10^{-8}+[\mathrm{OH^{-}}]$ and $[\mathrm{H^{+}}][\mathrm{OH^{-}}]=10^{-14}$. With $[\mathrm{H^{+}}]=h$: $h^{2}-10^{-8}h-10^{-14}=0$, so $h=\dfrac{10^{-8}+\sqrt{10^{-16}+4\times10^{-14}}}{2}=1.05\times10^{-7}$ and $\mathrm{pH}=6.98$ — just on the acidic side of neutral.
Interactive 1 — pH Explorer: Strong and Weak Acids and Bases
Choose a solute and its concentration. The lab solves the ionisation equilibrium exactly (including the ions from water) and places the solution on the pH scale beside some everyday fluids. The bar on the right shows what fraction of the solute is ionised. Press ▶ Play to dilute the solution step by step from $1\ \mathrm{mol\,L^{-1}}$ to $10^{-7}\ \mathrm{mol\,L^{-1}}$ and watch the pH drift towards 7.
0.10 mol L⁻¹
[H⁺] / mol L⁻¹—
[OH⁻] / mol L⁻¹—
pH—
degree of ionisation α—
Weak Acids and Weak Bases: $K_{a}$, $K_{b}$ and the Degree of Ionisation
A weak acid HA sets up the equilibrium $\mathrm{HA(aq)+H_{2}O(l)\rightleftharpoons H_{3}O^{+}(aq)+A^{-}(aq)}$, usually written simply $\mathrm{HA\rightleftharpoons H^{+}+A^{-}}$. Water is the solvent, present in huge excess, so it does not appear in the equilibrium constant.
📘 Definition 6.12 — Ionisation constants and degree of ionisation
For a weak acid and a weak base
$$K_{a}=\frac{[\mathrm{H^{+}}][\mathrm{A^{-}}]}{[\mathrm{HA}]},\qquad K_{b}=\frac{[\mathrm{BH^{+}}][\mathrm{OH^{-}}]}{[\mathrm{B}]}$$
(for $\mathrm{B+H_{2}O\rightleftharpoons BH^{+}+OH^{-}}$), and $\mathrm{p}K_{a}=-\log K_{a}$, $\mathrm{p}K_{b}=-\log K_{b}$. The larger $K_{a}$ (the smaller $\mathrm{p}K_{a}$), the stronger the acid. The degree of ionisation $\alpha$ is the fraction of the dissolved acid (or base) present as ions at equilibrium.
Acid
HF
$\mathrm{CH_{3}COOH}$
HOCl
HCN
$K_{a}$ (298 K)
$3.5\times10^{-4}$
$1.8\times10^{-5}$
$3.0\times10^{-8}$
$4.9\times10^{-10}$
$\mathrm{p}K_{a}$
3.46
4.74
7.52
9.31
For an acid of concentration $c$, the ICE table is: $[\mathrm{HA}]=c(1-\alpha)$, $[\mathrm{H^{+}}]=[\mathrm{A^{-}}]=c\alpha$ (neglecting the tiny contribution of water). Substituting:
📐 Rule 6.12 — Degree of ionisation of a weak electrolyte (Ostwald's dilution law)
$$K_{a}=\frac{c\alpha^{2}}{1-\alpha}$$
When $\alpha$ is small (in practice when $\alpha\lt0.05$), $1-\alpha\approx1$ and
$$\alpha\approx\sqrt{\frac{K_{a}}{c}},\qquad [\mathrm{H^{+}}]=c\alpha\approx\sqrt{K_{a}c}$$
$$\mathrm{pH}\approx\tfrac12(\mathrm{p}K_{a}-\log c).$$
The same formulae hold for a weak base with $K_{b}$ and $[\mathrm{OH^{-}}]$. On dilution ($c$ smaller) the degree of ionisation increases, although $[\mathrm{H^{+}}]$ itself decreases. If the approximation gives $\alpha\gt0.05$, solve the quadratic $c\alpha^{2}+K_{a}\alpha-K_{a}=0$ exactly.
✏️ Example 3 — pH and degree of ionisation of a weak acid
(a) Find the pH and the degree of ionisation of $0.10\ \mathrm{mol\,L^{-1}}$ acetic acid ($K_{a}=1.8\times10^{-5}$). (b) Repeat for $0.0010\ \mathrm{mol\,L^{-1}}$ acetic acid, solving exactly. (c) Compare with $0.10\ \mathrm{mol\,L^{-1}}$ HCl and explain the opening question of this section.
💡 Show solution▶
(a) $\alpha\approx\sqrt{\dfrac{1.8\times10^{-5}}{0.10}}=\sqrt{1.8\times10^{-4}}=1.34\times10^{-2}$ (1.3%, so the approximation is fine). $[\mathrm{H^{+}}]=c\alpha=1.34\times10^{-3}\ \mathrm{mol\,L^{-1}}$, $\mathrm{pH}=3-\log1.34=2.87$.
(b) approx. $\alpha\approx\sqrt{1.8\times10^{-5}/0.0010}=0.134$ — 13%, too large for the approximation.
(b) exact $0.0010\alpha^{2}+1.8\times10^{-5}\alpha-1.8\times10^{-5}=0$ → $\alpha=\dfrac{-1.8\times10^{-5}+\sqrt{3.24\times10^{-10}+7.2\times10^{-8}}}{2\times0.0010}=0.125$. $[\mathrm{H^{+}}]=1.25\times10^{-4}$, $\mathrm{pH}=3.90$. Dilution by a factor of 100 raised $\alpha$ almost tenfold.
(c) $0.10\ \mathrm{mol\,L^{-1}}$ HCl: $[\mathrm{H^{+}}]=0.10$, pH $1.00$ — about 75 times more $\mathrm{H^{+}}$ than the acetic acid. Both neutralise the same amount of NaOH because, as base removes $\mathrm{H^{+}}$, the acetic acid equilibrium shifts forward (Le Chatelier) until all of it has reacted.
For a conjugate pair the two constants are linked. Multiplying $K_{a}$ of $\mathrm{NH_{4}^{+}}$ ($\mathrm{NH_{4}^{+}\rightleftharpoons NH_{3}+H^{+}}$) by $K_{b}$ of $\mathrm{NH_{3}}$ ($\mathrm{NH_{3}+H_{2}O\rightleftharpoons NH_{4}^{+}+OH^{-}}$) cancels $[\mathrm{NH_{3}}]$ and $[\mathrm{NH_{4}^{+}}]$ and leaves $[\mathrm{H^{+}}][\mathrm{OH^{-}}]$: adding the two equations gives the ionisation of water.
📐 Rule 6.13 — Conjugate acid–base pairs
$$K_{a}\times K_{b}=K_{w}$$
$$\mathrm{p}K_{a}+\mathrm{p}K_{b}=\mathrm{p}K_{w}=14.00\ \ (298\ \mathrm{K})$$
for an acid and its own conjugate base. The weaker the acid, the stronger its conjugate base, and vice versa.
✏️ Example 4 — A weak base and its conjugate acid
$K_{b}(\mathrm{NH_{3}})=1.8\times10^{-5}$. (a) Find $[\mathrm{OH^{-}}]$, pOH, pH and the degree of ionisation of $0.20\ \mathrm{mol\,L^{-1}}$ aqueous ammonia. (b) Find $K_{a}$ and $\mathrm{p}K_{a}$ of the ammonium ion. (c) Find $K_{b}$ of the cyanide ion ($K_{a}(\mathrm{HCN})=4.9\times10^{-10}$) and compare it with $K_{b}(\mathrm{NH_{3}})$.
(b) $K_{a}(\mathrm{NH_{4}^{+}})=\dfrac{K_{w}}{K_{b}}=\dfrac{1.0\times10^{-14}}{1.8\times10^{-5}}=5.6\times10^{-10}$; $\mathrm{p}K_{a}=9.26$ (check: $\mathrm{p}K_{b}=4.74$, and $4.74+9.26=14.00$).
(c) $K_{b}(\mathrm{CN^{-}})=\dfrac{1.0\times10^{-14}}{4.9\times10^{-10}}=2.0\times10^{-5}$ — slightly larger than $K_{b}(\mathrm{NH_{3}})$: the cyanide ion, conjugate base of a very weak acid, is a base about as strong as ammonia. Solutions of NaCN are distinctly alkaline (Section 6.4).
Interactive 2 — Dilution and the Degree of Ionisation
Choose the strength of a weak acid ($\mathrm{p}K_{a}$) and its concentration. The red curve is the exact degree of ionisation from $K_{a}=c\alpha^{2}/(1-\alpha)$; the dashed curve is the approximation $\alpha\approx\sqrt{K_{a}/c}$. Where they part company (α above about 5%), the approximation is no longer safe. Press ▶ Play to dilute the acid and watch α rise while $[\mathrm{H^{+}}]$ falls.
4.740.10 mol L⁻¹
α exact—
α ≈ √(Kₐ/c)—
[H⁺] / mol L⁻¹—
pH—
Polybasic Acids, Acid Strength and the Common-Ion Effect
Acids such as $\mathrm{H_{2}S}$, carbonic acid $\mathrm{H_{2}CO_{3}}$ and phosphoric acid $\mathrm{H_{3}PO_{4}}$ have more than one ionisable proton. They are polybasic (polyprotic) acids and ionise in steps, each with its own constant:
$K_{a_{1}}\gt K_{a_{2}}\gt K_{a_{3}}$: removing a proton from a neutral molecule is easier than removing one from a negative ion, which holds it by electrostatic attraction.
Because $K_{a_{2}}\ll K_{a_{1}}$, nearly all the $\mathrm{H^{+}}$ comes from the first step: treat the acid as monoprotic with $K_{a_{1}}$ to find the pH.
For a diprotic acid $\mathrm{H_{2}A}$ alone in water, $[\mathrm{A^{2-}}]\approx K_{a_{2}}$. For the overall reaction $\mathrm{H_{2}A\rightleftharpoons2H^{+}+A^{2-}}$, $K_{a}=K_{a_{1}}K_{a_{2}}$ (Rule 6.4).
What makes one acid stronger than another? For binary acids $\mathrm{H{-}A}$, two factors compete: the strength of the $\mathrm{H{-}A}$ bond and its polarity.
📐 Rule 6.15 — Factors affecting acid strength
Down a group the size of A increases and the $\mathrm{H{-}A}$ bond becomes longer and weaker, so the proton is lost more easily: acid strength $\mathrm{HF\lt HCl\lt HBr\lt HI}$ and $\mathrm{H_{2}O\lt H_{2}S\lt H_{2}Se\lt H_{2}Te}$. Bond strength outweighs electronegativity here.
Across a period the atoms are of similar size, and the electronegativity of A decides: the more polar the $\mathrm{H{-}A}$ bond, the stronger the acid: $\mathrm{CH_{4}\lt NH_{3}\lt H_{2}O\lt HF}$.
Finally, what happens to a weak-acid equilibrium when a salt that supplies one of its ions is added? Adding sodium acetate to acetic acid adds $\mathrm{CH_{3}COO^{-}}$, so $Q\gt K_{a}$ and the equilibrium $\mathrm{CH_{3}COOH\rightleftharpoons H^{+}+CH_{3}COO^{-}}$ shifts backward: $[\mathrm{H^{+}}]$ falls.
📘 Definition 6.13 — Common-ion effect
The common-ion effect is the suppression of the ionisation of a weak electrolyte when a strong electrolyte that supplies one of the same ions is added — for example $\mathrm{CH_{3}COONa}$ added to $\mathrm{CH_{3}COOH}$, $\mathrm{NH_{4}Cl}$ added to $\mathrm{NH_{3}}$, or $\mathrm{HCl}$ added to $\mathrm{H_{2}S}$. It is an application of Le Chatelier's principle.
✏️ Example 5 — A diprotic acid
A saturated solution of $\mathrm{H_{2}S}$ in water is about $0.10\ \mathrm{mol\,L^{-1}}$. Using the constants above, find (a) its pH; (b) $[\mathrm{HS^{-}}]$ and $[\mathrm{S^{2-}}]$; (c) $[\mathrm{S^{2-}}]$ when the same solution is also $0.10\ \mathrm{mol\,L^{-1}}$ in HCl.
💡 Show solution▶
(a) First step only: $[\mathrm{H^{+}}]\approx\sqrt{K_{a_{1}}c}=\sqrt{9.1\times10^{-8}\times0.10}=9.5\times10^{-5}\ \mathrm{mol\,L^{-1}}$, $\mathrm{pH}=4.02$.
(b) $[\mathrm{HS^{-}}]\approx[\mathrm{H^{+}}]=9.5\times10^{-5}\ \mathrm{mol\,L^{-1}}$. In $K_{a_{2}}$ these two cancel: $[\mathrm{S^{2-}}]\approx K_{a_{2}}=1.2\times10^{-13}\ \mathrm{mol\,L^{-1}}$.
(c) Now $[\mathrm{H^{+}}]=0.10$ from the HCl. Use the overall constant: $[\mathrm{S^{2-}}]=\dfrac{K_{a_{1}}K_{a_{2}}[\mathrm{H_{2}S}]}{[\mathrm{H^{+}}]^{2}}=\dfrac{(1.09\times10^{-20})(0.10)}{(0.10)^{2}}=1.1\times10^{-19}\ \mathrm{mol\,L^{-1}}$.
Note The common ion $\mathrm{H^{+}}$ lowers $[\mathrm{S^{2-}}]$ about a millionfold. Qualitative analysis uses this to precipitate only the least soluble sulphides in acid solution (Section 6.4).
✏️ Example 6 — The common-ion effect
(a) Find $[\mathrm{H^{+}}]$, the pH and the degree of ionisation of acetic acid in a solution that is $0.10\ \mathrm{mol\,L^{-1}}$ in $\mathrm{CH_{3}COOH}$ and $0.10\ \mathrm{mol\,L^{-1}}$ in $\mathrm{CH_{3}COONa}$ ($K_{a}=1.8\times10^{-5}$). Compare with Example 3(a). (b) Find the pH of a solution $0.10\ \mathrm{mol\,L^{-1}}$ in $\mathrm{NH_{3}}$ and $0.10\ \mathrm{mol\,L^{-1}}$ in $\mathrm{NH_{4}Cl}$ ($K_{b}=1.8\times10^{-5}$).
💡 Show solution▶
(a) ICE $[\mathrm{CH_{3}COOH}]=0.10-x\approx0.10$, $[\mathrm{CH_{3}COO^{-}}]=0.10+x\approx0.10$, $[\mathrm{H^{+}}]=x$. $K_{a}=\dfrac{x(0.10)}{0.10}$, so $x=1.8\times10^{-5}\ \mathrm{mol\,L^{-1}}$.
(a) result $\mathrm{pH}=4.74$ ($=\mathrm{p}K_{a}$); $\alpha=1.8\times10^{-5}/0.10=1.8\times10^{-4}$ (0.018%), against 1.3% without the salt — the ionisation is suppressed about 75-fold and the pH rises from 2.87 to 4.74.
(b) Likewise $[\mathrm{OH^{-}}]=K_{b}\dfrac{[\mathrm{NH_{3}}]}{[\mathrm{NH_{4}^{+}}]}=1.8\times10^{-5}$, $\mathrm{pOH}=4.74$, $\mathrm{pH}=9.26$. (Mixtures like these resist changes in pH — they are the buffers of Section 6.4.)
Exercises
I
Electrolytes and acid–base concepts.
1 Give two limitations of the Arrhenius concept of acids and bases.
2 Write the conjugate base of $\mathrm{H_{2}PO_{4}^{-}}$, $\mathrm{HS^{-}}$, $\mathrm{H_{2}O}$ and $\mathrm{HNO_{3}}$.
3 Write the conjugate acid of $\mathrm{HPO_{4}^{2-}}$, $\mathrm{CO_{3}^{2-}}$, $\mathrm{OH^{-}}$ and $\mathrm{CH_{3}NH_{2}}$.
4 Identify the conjugate acid–base pairs in $\mathrm{HSO_{4}^{-}(aq)+CO_{3}^{2-}(aq)}$ $\mathrm{\rightleftharpoons SO_{4}^{2-}(aq)+HCO_{3}^{-}(aq)}$.
5 Classify as a Lewis acid or a Lewis base: $\mathrm{H_{2}O}$, $\mathrm{Fe^{3+}}$, $\mathrm{BF_{3}}$, $\mathrm{Cl^{-}}$, $\mathrm{CO_{2}}$ (in its reaction with $\mathrm{OH^{-}}$).
6 Why is $\mathrm{BF_{3}}$ an acid in the Lewis sense but not in the Brønsted–Lowry sense?
7 $0.1\ \mathrm{mol\,L^{-1}}$ solutions of NaCl and of acetic acid are tested with a conductivity cell. Which conducts better, and why?
8 Which is the stronger Brønsted base, $\mathrm{F^{-}}$ or $\mathrm{Cl^{-}}$? Explain using conjugate pairs.
💡 Show solutions▶
1 It applies only to aqueous solutions, and it cannot explain the basicity of substances without $\mathrm{OH}$ groups such as $\mathrm{NH_{3}}$ (or the acidity of $\mathrm{BF_{3}}$, $\mathrm{CO_{2}}$).
5 Bases: $\mathrm{H_{2}O}$, $\mathrm{Cl^{-}}$ (lone pairs). Acids: $\mathrm{Fe^{3+}}$, $\mathrm{BF_{3}}$, $\mathrm{CO_{2}}$ (its C accepts a pair from $\mathrm{OH^{-}}$, giving $\mathrm{HCO_{3}^{-}}$).
6 It has no proton to donate, but its boron atom has only six valence electrons and an empty orbital, so it accepts an electron pair.
7 NaCl, a strong electrolyte, is completely ionised; acetic acid is a weak electrolyte, only about 1% ionised, so its solution has far fewer ions.
8 $\mathrm{F^{-}}$: its conjugate acid HF is weak, while $\mathrm{Cl^{-}}$ is the conjugate base of the strong acid HCl. The weaker the acid, the stronger its conjugate base.
II
$K_{w}$ and the pH of strong acids and bases ($298\ \mathrm{K}$ unless stated).
1 Find the pH of $2.5\times10^{-3}\ \mathrm{mol\,L^{-1}}$ $\mathrm{HNO_{3}}$.
2 Find the pH of $0.020\ \mathrm{mol\,L^{-1}}$ NaOH.
3 Find the pH of $1.0\times10^{-3}\ \mathrm{mol\,L^{-1}}$ $\mathrm{Ca(OH)_{2}}$.
4 $50.0\ \mathrm{mL}$ of $0.10\ \mathrm{mol\,L^{-1}}$ HCl is mixed with $50.0\ \mathrm{mL}$ of $0.060\ \mathrm{mol\,L^{-1}}$ NaOH. Find the pH of the mixture.
5 Find $[\mathrm{H^{+}}]$ and $[\mathrm{OH^{-}}]$ in a sample of rain water of pH $5.30$.
6 Find the pH of $1.0\times10^{-7}\ \mathrm{mol\,L^{-1}}$ HCl.
7 At body temperature ($310\ \mathrm{K}$) $K_{w}=2.4\times10^{-14}$. Find the pH of pure water at $310\ \mathrm{K}$. Is it acidic?
8 A solution of HCl of pH $2.00$ is diluted 100 times with water. What is the new pH?
6 $h^{2}-10^{-7}h-10^{-14}=0$ → $h=\dfrac{10^{-7}+\sqrt{5\times10^{-14}}}{2}=1.62\times10^{-7}$, $\mathrm{pH}=6.79$ (not 7.00: the acid adds to water's own $\mathrm{H^{+}}$).
7 $[\mathrm{H^{+}}]=\sqrt{2.4\times10^{-14}}=1.55\times10^{-7}$, $\mathrm{pH}=6.81$. It is neutral, because $[\mathrm{H^{+}}]=[\mathrm{OH^{-}}]$; "neutral = pH 7" holds only at $298\ \mathrm{K}$.
8 $[\mathrm{H^{+}}]$ falls from $10^{-2}$ to $10^{-4}$: $\mathrm{pH}=4.00$.
III
Weak acids and bases.
1 A $0.10\ \mathrm{mol\,L^{-1}}$ solution of a weak monoprotic acid has pH $4.50$. Find $K_{a}$.
2 Find the pH and degree of ionisation of $0.050\ \mathrm{mol\,L^{-1}}$ HF ($K_{a}=3.5\times10^{-4}$). Is the approximation $\alpha\approx\sqrt{K_{a}/c}$ safe?
3 Find the degree of ionisation and pH of $0.10\ \mathrm{mol\,L^{-1}}$ HCN ($K_{a}=4.9\times10^{-10}$).
4 Find the pH of $0.10\ \mathrm{mol\,L^{-1}}$ methylamine, $\mathrm{CH_{3}NH_{2}}$ ($K_{b}=4.4\times10^{-4}$), solving exactly.
5 Benzoic acid has $K_{a}=6.5\times10^{-5}$. Find its $\mathrm{p}K_{a}$ and $K_{b}$ of the benzoate ion.
6 Acid X has $\mathrm{p}K_{a}=3.46$ and acid Y has $\mathrm{p}K_{a}=4.74$. Which is stronger? By what factor do their $K_{a}$ values differ?
7 A $0.10\ \mathrm{mol\,L^{-1}}$ solution of a monoprotic acid is 1.0% ionised. Find $K_{a}$.
8 Using the approximation, by what factors do $\alpha$ and $[\mathrm{H^{+}}]$ of acetic acid change when $0.10\ \mathrm{mol\,L^{-1}}$ acid is diluted to $0.0010\ \mathrm{mol\,L^{-1}}$?
8 $\alpha\propto1/\sqrt{c}$: $c$ falls 100-fold, so $\alpha$ rises 10-fold; $[\mathrm{H^{+}}]\propto\sqrt{c}$ falls 10-fold. (Example 3 shows the exact factors are slightly smaller.)
IV
Polybasic acids, acid strength and the common-ion effect.
1 Why is $K_{a_{2}}$ of a diprotic acid always smaller than $K_{a_{1}}$?
2 Arrange HF, HCl, HBr and HI in order of increasing acid strength and explain.
3 Arrange $\mathrm{CH_{4}}$, $\mathrm{NH_{3}}$, $\mathrm{H_{2}O}$ and HF in order of increasing acidity and explain.
4 Water saturated with $\mathrm{CO_{2}}$ contains about $0.034\ \mathrm{mol\,L^{-1}}$ $\mathrm{H_{2}CO_{3}}$. Find its pH and $[\mathrm{CO_{3}^{2-}}]$.
5 Find the pH of a solution $0.20\ \mathrm{mol\,L^{-1}}$ in methanoic acid HCOOH ($K_{a}=1.8\times10^{-4}$) and $0.10\ \mathrm{mol\,L^{-1}}$ in HCOONa.
6 Find the pH of a solution $0.10\ \mathrm{mol\,L^{-1}}$ in $\mathrm{NH_{3}}$ and $0.20\ \mathrm{mol\,L^{-1}}$ in $\mathrm{NH_{4}Cl}$ ($K_{b}=1.8\times10^{-5}$).
7 Find $[\mathrm{S^{2-}}]$ in a solution $0.10\ \mathrm{mol\,L^{-1}}$ in $\mathrm{H_{2}S}$ and $0.30\ \mathrm{mol\,L^{-1}}$ in HCl. Why is HCl added in this way in qualitative analysis?
8Applied. Aspirin is a weak acid with $\mathrm{p}K_{a}=3.5$. Using $\dfrac{[\mathrm{A^{-}}]}{[\mathrm{HA}]}=\dfrac{K_{a}}{[\mathrm{H^{+}}]}$, find the ratio of ionised to un-ionised aspirin in the stomach (pH $1.5$) and in the small intestine (pH $7.5$). Un-ionised molecules pass through cell membranes more easily. Where is aspirin absorbed more readily?
💡 Show solutions▶
1 The second proton must be removed from a negatively charged ion, which attracts it electrostatically; the first comes from a neutral molecule.
2 $\mathrm{HF\lt HCl\lt HBr\lt HI}$: down the group the $\mathrm{H{-}X}$ bond gets longer and weaker, so $\mathrm{H^{+}}$ is released more easily.
3 $\mathrm{CH_{4}\lt NH_{3}\lt H_{2}O\lt HF}$: across the period the electronegativity of the atom bonded to H increases, making the bond more polar.
7 $[\mathrm{S^{2-}}]=\dfrac{(1.09\times10^{-20})(0.10)}{(0.30)^{2}}=1.2\times10^{-20}\ \mathrm{mol\,L^{-1}}$. The common ion $\mathrm{H^{+}}$ keeps $[\mathrm{S^{2-}}]$ so low that only the least soluble sulphides precipitate, separating them from the others.
8 Stomach: $10^{1.5-3.5}=10^{-2}$ — about 99% un-ionised. Intestine: $10^{7.5-3.5}=10^{4}$ — almost entirely ionised. Aspirin is absorbed more readily in the stomach (which is also why it can irritate the stomach lining).
🎯 Exam corner BoardJEE MainJEE AdvancedNEET
E1.NEET The conjugate base of $\mathrm{H_{2}PO_{4}^{-}}$ is (A) $\mathrm{H_{3}PO_{4}}$ (B) $\mathrm{HPO_{4}^{2-}}$ (C) $\mathrm{PO_{4}^{3-}}$ (D) $\mathrm{P_{2}O_{5}}$
💡 Show solution▶
Idea A conjugate base has exactly one proton fewer: $\mathrm{H_{2}PO_{4}^{-}\rightarrow HPO_{4}^{2-}+H^{+}}$. (A) is its conjugate acid.
Answer(B).
E2.JEE MainNumerical answer. Assuming complete ionisation in both steps, the pH of $5.0\times10^{-3}\ \mathrm{mol\,L^{-1}}$ $\mathrm{H_{2}SO_{4}}$ at $298\ \mathrm{K}$ is ____.
💡 Show solution▶
Idea Each $\mathrm{H_{2}SO_{4}}$ gives $2\,\mathrm{H^{+}}$: $[\mathrm{H^{+}}]=1.0\times10^{-2}\ \mathrm{mol\,L^{-1}}$.
Answer2.
E3.Board · 3 marks Derive $\alpha\approx\sqrt{K_{a}/c}$ for a weak monoprotic acid, and use it to find the degree of ionisation and pH of $0.050\ \mathrm{mol\,L^{-1}}$ acetic acid ($K_{a}=1.8\times10^{-5}$).
💡 Show solution▶
1 mark At equilibrium $[\mathrm{HA}]=c(1-\alpha)$, $[\mathrm{H^{+}}]=[\mathrm{A^{-}}]=c\alpha$, so $K_{a}=\dfrac{c\alpha^{2}}{1-\alpha}\approx c\alpha^{2}$ for $\alpha\ll1$, giving $\alpha\approx\sqrt{K_{a}/c}$.
1 mark $\alpha=\sqrt{1.8\times10^{-5}/0.050}=\sqrt{3.6\times10^{-4}}=0.019$ (1.9%).
1 mark $[\mathrm{H^{+}}]=c\alpha=9.5\times10^{-4}\ \mathrm{mol\,L^{-1}}$, $\mathrm{pH}=3.02$.
E4.JEE Main A $0.010\ \mathrm{mol\,L^{-1}}$ solution of a weak monoacidic base has $K_{b}=1.0\times10^{-6}$ at $298\ \mathrm{K}$. Its pH is (A) $10$ (B) $11$ (C) $9$ (D) $8$
💡 Show solution▶
Idea $[\mathrm{OH^{-}}]=\sqrt{K_{b}c}=\sqrt{1.0\times10^{-8}}=1.0\times10^{-4}$ (α = 1%, approximation valid); $\mathrm{pOH}=4$.
Answer $\mathrm{pH}=10$: (A).
E5.JEE AdvancedOne or more options correct. At $298\ \mathrm{K}$ unless stated: (A) the pH of $1.0\times10^{-8}\ \mathrm{mol\,L^{-1}}$ HCl is close to $6.98$; (B) $K_{w}$ increases with temperature; (C) the conjugate base of a strong acid is a strong base; (D) HI is a stronger acid than HF because the $\mathrm{H{-}I}$ bond is weaker.
💡 Show solution▶
(A) True (Example 2(d)): water's own $\mathrm{H^{+}}$ dominates.
(B) True: the ionisation of water is endothermic.
(C) False: it is a very weak base ($\mathrm{Cl^{-}}$, $\mathrm{NO_{3}^{-}}$), since $K_{a}K_{b}=K_{w}$.
(D) True: down the group bond strength decides.
Answer(A), (B), (D).
E6.Board · Case study Lemon juice has a pH of about $2.2$, mainly because of citric acid, a weak acid. (i) Define pH and find $[\mathrm{H^{+}}]$ and $[\mathrm{OH^{-}}]$ in lemon juice at $298\ \mathrm{K}$. (2 marks) (ii) At $323\ \mathrm{K}$, $K_{w}=5.5\times10^{-14}$. What is the pH of pure water at this temperature, and is it acidic? (1 mark) (iii) State the effect on the pH of adding sodium acetate to acetic acid, and name the effect. (1 mark)
(ii) $[\mathrm{H^{+}}]=\sqrt{5.5\times10^{-14}}=2.35\times10^{-7}$, $\mathrm{pH}=6.63$. Still neutral: $[\mathrm{H^{+}}]=[\mathrm{OH^{-}}]$.
(iii) The acetate ion suppresses the ionisation of acetic acid, so $[\mathrm{H^{+}}]$ falls and the pH rises — the common-ion effect.
🔬 Research–Teaching NexusResearch-tutored · optional, about 30 minutes
Research question. The average pH of the surface ocean has fallen from about $8.2$ before the industrial revolution to about $8.1$ today. Is a change of $0.1$ really small?
Why it matters. The ocean absorbs roughly a quarter of the carbon dioxide released by burning fuels. Dissolved $\mathrm{CO_{2}}$ forms carbonic acid, a diprotic acid, which lowers the pH and the concentration of carbonate ions that corals, oysters and plankton need to build their shells. Long-term measurements at stations such as Hawaii's ocean time-series and the Bermuda Atlantic site are used by international climate assessments to track this "ocean acidification".
What to do.
Calculate $[\mathrm{H^{+}}]$ at pH $8.2$ and at pH $8.1$, and the percentage increase. Explain why a logarithmic scale makes the change look smaller than it is.
Write the two ionisation steps of carbonic acid and the expressions for $K_{a_{1}}$ and $K_{a_{2}}$.
The ratio $[\mathrm{CO_{3}^{2-}}]/[\mathrm{HCO_{3}^{-}}]$ equals $K_{a_{2}}/[\mathrm{H^{+}}]$. In sea water $K_{a_{2}}\approx1.0\times10^{-9}$ (larger than in pure water because of the dissolved salts). Calculate the ratio at both pH values and the percentage fall.
Use Le Chatelier's principle to explain how adding $\mathrm{CO_{2}}$ to sea water lowers the carbonate concentration.
Find a recent time-series graph of ocean pH (for example from a national oceanographic agency) and estimate the rate of change per decade.
Tools: Definition 6.11, Rule 6.14, Principle 6.10, a calculator, a public ocean-observation data set.
Hand in: Your calculations, a short explanation of why the change matters to shell-building organisms, and the trend you estimated.
Going further. Predict whether the pH of a warm tropical sea or a cold polar sea responds more strongly to the same rise in atmospheric $\mathrm{CO_{2}}$, using what you know about gas solubility (Section 6.1).
Instructor note: $[\mathrm{H^{+}}]=6.3\times10^{-9}$ at pH 8.2 and $7.9\times10^{-9}$ at pH 8.1 — an increase of 26%. $[\mathrm{CO_{3}^{2-}}]/[\mathrm{HCO_{3}^{-}}]=0.158$ and $0.126$, a fall of about 21%. Adding $\mathrm{CO_{2}}$ raises $[\mathrm{H^{+}}]$, which combines with $\mathrm{CO_{3}^{2-}}$ to give $\mathrm{HCO_{3}^{-}}$. Cold water dissolves more $\mathrm{CO_{2}}$, so polar seas acidify faster. Recent trends are about $-0.017$ pH units per decade.
🧪 Self-check quiz
0/5 correct
Q1. Which of these is a Lewis acid but not a Brønsted–Lowry acid?
Q2. The pH of $1.0\times10^{-3}\ \mathrm{mol\,L^{-1}}$ NaOH at $298\ \mathrm{K}$ is
Q3. When a solution of a weak acid is diluted, its degree of ionisation
Q4. For $\mathrm{NH_{3}}$, $\mathrm{p}K_{b}=4.74$ at $298\ \mathrm{K}$. The $\mathrm{p}K_{a}$ of $\mathrm{NH_{4}^{+}}$ is
Q5. Adding solid ammonium chloride to aqueous ammonia