Why Vinegar Is Far Less Acidic Than Hydrochloric Acid of the Same Molarity

A bottle of vinegar and a bottle of dilute hydrochloric acid can both be labelled $0.1$ M, yet the hydrochloric acid is hundreds of times more acidic. The difference is how completely each acid ionises in water.

Measuring acidity on the pH scale and describing weak acids with an ionisation constant turns that difference into numbers.

This part covers acid-base concepts, the ionic product of water and pH, weak acids and bases, and the link between $K_a$ and $K_b$.

**Arrhenius acids give H${}^{+}$ ions and bases give OH${}^{-}$ ions in water; Bronsted-Lowry acids donate protons and bases accept them; Lewis acids accept an electron pair and Lewis bases donate one.

Worked examples.

- Arrhenius**: HCl gives H${}^{+}$ in water; NaOH gives OH${}^{-}$
- Bronsted-Lowry: in NH${}_3$ + H${}_2$O $\rightleftharpoons$ NH${}_4^{+}$ + OH${}^{-}$, water donates a proton (acid) and ammonia accepts it (base)
- Lewis: in BF${}_3$ + NH${}_3$ $\to$ F${}_3$B–NH${}_3$, BF${}_3$ accepts ammonia's lone pair, so BF${}_3$ is a Lewis acid even though it has no hydrogen

Conjugate pairs differ by one proton:

- HCl + H${}_2$O $\to$ H${}_3$O${}^{+}$ + Cl${}^{-}$: pairs are **HCl/Cl${}^{-}$ and H${}_3$O${}^{+}$/H${}_2$O**
- NH${}_3$ + H${}_2$O $\rightleftharpoons$ NH${}_4^{+}$ + OH${}^{-}$: pairs are **NH${}_4^{+}$/NH${}_3$ and H${}_2$O/OH${}^{-}$

An everyday example. Lemon juice and vinegar** taste sour because they release H${}^{+}$ ions, while soap and baking soda feel slippery because they are bases.

The substance. Water is amphoteric — it acts as a base towards HCl and as an acid towards NH${}_3$.

**Water ionises slightly, with $K_w=[{\text{H}}^{+}][{\text{OH}}^{-}]=1.0\times 10^{-14}$ at $298$ K; pH $=-\log [{\text{H}}^{+}]$, pOH $=-\log [{\text{OH}}^{-}]$, and pH $+$ pOH $=14$.

Strong acids and bases ionise completely, so their ion concentration follows directly from the formula.

Worked example 1 — strong acids.**

- $0.01$ M HCl: $[{\text{H}}^{+}]=10^{-2}$ M, **pH $=2$**
- $0.005$ M H${}_2$SO${}_4$: two H${}^{+}$ per formula, $[{\text{H}}^{+}]=0.01$ M, **pH $=2$

Worked example 2 — strong bases.**

- $0.001$ M NaOH: pOH $=3$, **pH $=11$**
- $0.005$ M Ba(OH)${}_2$: $[{\text{OH}}^{-}]=0.01$ M, pOH $=2$, **pH $=12$

Worked example 3 — a very dilute acid.** $10^{-8}$ M HCl is not pH $8$. Water's own ionisation matters: with $x$ from water, $x(x+10^{-8})=10^{-14}$ gives $x\approx 9.5\times 10^{-8}$, so

$$[{\text{H}}^{+}]\approx 1.05\times 10^{-7}\text{ M}\text{pH}\approx 6.98$$

— very slightly acidic, as it must be.

An everyday example. **Lemon juice has a pH near $2$**, while blood stays close to $7.4$, just on the basic side.

The substance. **$K_w$ increases with temperature**, so pure water at higher temperatures has a pH below $7$ yet is still neutral.

**For a weak acid of concentration $c$ and degree of ionisation $\alpha$, $K_a=\frac{c{\alpha }^2}{1-\alpha }\approx c{\alpha }^2$, so $\alpha \approx \sqrt{\frac{K_a}{c}}$ and $[{\text{H}}^{+}]\approx \sqrt{K_{a}c}$; ionisation increases on dilution (Ostwald's dilution law) and decreases when a common ion is added.

Worked example 1 — acetic acid** ($K_a=1.8\times 10^{-5}$) at $0.1$ M:

$$[{\text{H}}^{+}]=\sqrt{1.8\times 10^{-5}\times 0.1}\approx 1.34\times 10^{-3}\text{ M}\text{pH}\approx 2.87\alpha \approx 1.34\%$$

Worked example 2 — dilution. At $0.001$ M, $\alpha \approx \sqrt{\frac{1.8\times 10^{-5}}{0.001}}\approx 0.134$ — about ten times more ionised, though the solution is less acidic overall.

Worked example 3 — weak base. $0.05$ M ammonia ($K_b=1.8\times 10^{-5}$):

$$[{\text{OH}}^{-}]=\sqrt{1.8\times 10^{-5}\times 0.05}\approx 9.5\times 10^{-4}\text{ M}\text{pOH}\approx 3.02\text{pH}\approx 10.98$$

Worked example 4 — common-ion effect. In $0.1$ M acetic acid with $0.1$ M sodium acetate, the added acetate pushes the ionisation back:

$$[{\text{H}}^{+}]=K_a\frac{[\text{HA}]}{[{\text{A}}^{-}]}=1.8\times 10^{-5}\text{ M}\text{pH}\approx 4.74$$

An everyday example. Vinegar is a dilute solution of acetic acid, which is why it is safe on food while hydrochloric acid of the same molarity is not.

The substance. **The approximation $1-\alpha \approx 1$ holds only when $\alpha$ is small**, roughly under $5\%$; otherwise solve the quadratic.

**For any conjugate acid-base pair, $K_a\times K_b=K_w$, so pK${}_a$ $+$ pK${}_b$ $=14$ at $298$ K; the larger $K_a$ is, the stronger the acid and the weaker its conjugate base.

Worked example 1 — acetate ion.** With $K_a({\text{CH}}_3\text{COOH})=1.8\times 10^{-5}$:

$$K_b({\text{CH}}_3{\text{COO}}^{-})=\frac{1.0\times 10^{-14}}{1.8\times 10^{-5}}\approx 5.6\times 10^{-10}$$

Worked example 2 — ammonium ion. With $K_b({\text{NH}}_3)=1.8\times 10^{-5}$, $K_a({\text{NH}}_4^{+})\approx 5.6\times 10^{-10}$.

Worked example 3 — ranking acids. $K_a$ values: HF $3.5\times 10^{-4}$, CH${}_3$COOH $1.8\times 10^{-5}$, HCN $4.9\times 10^{-10}$.

- Acid strength: HF > CH${}_3$COOH > HCN
- Conjugate base strength: CN${}^{-}$ > CH${}_3$COO${}^{-}$ > F${}^{-}$

The pK${}_a$ values are about $3.46$, $4.74$ and $9.31$ — **a smaller pK${}_a$ means a stronger acid.

An everyday example. Common salt dissolved in water gives a neutral solution** — Cl${}^{-}$ is the conjugate base of the very strong acid HCl, so it is far too weak a base to affect pH.

The substance. Strength and concentration are different ideas — a concentrated weak acid can still be less acidic than a dilute strong one.