How Your Blood Stays at pH 7.4 While Acids Pour Into It

Common salt dissolves to give a neutral solution, but washing soda makes water slippery and basic, and ammonium chloride makes it slightly acidic. Salts can react with water, and that reaction shifts the pH.

This part covers salt hydrolysis, buffer solutions, solubility product, and the common-ion effect. Take $K_w=1.0\times 10^{-14}$ and pK${}_a$(acetic acid) $=$ pK${}_b$(ammonia) $=4.74$.

**A salt of a strong acid and strong base gives a neutral solution; a salt of a weak acid and strong base is basic because its anion takes H${}^{+}$ from water; a salt of a strong acid and weak base is acidic because its cation gives H${}^{+}$; and a salt of a weak acid and weak base depends on whether $K_a$ or $K_b$ is larger.

Worked example 1 — sodium acetate**, $0.1$ M. The acetate ion hydrolyses with $K_h=\frac{K_w}{K_a}\approx 5.6\times 10^{-10}$:

$$\text{pH}=7+\frac{1}{2}(\text{p}{K}_a+\log c)=7+\frac{1}{2}(4.74-1)\approx 8.87$$

Worked example 2 — ammonium chloride, $0.1$ M:

$$\text{pH}=7-\frac{1}{2}(\text{p}{K}_b+\log c)=7-\frac{1}{2}(4.74-1)\approx 5.13$$

Worked example 3 — ammonium acetate. For a weak acid-weak base salt, $\text{pH}=7+\frac{1}{2}(\text{p}{K}_a-\text{p}{K}_b)$. Here pK${}_a$ $=$ pK${}_b$, so pH $=7.0$.

An everyday example. A solution of washing soda (sodium carbonate) feels slippery and cuts grease, because the carbonate ion makes the water basic.

The substance. A neutral pH for ammonium acetate does not mean no hydrolysis — both ions hydrolyse, to an equal extent.

**A buffer — a weak acid with its salt, or a weak base with its salt — keeps pH nearly constant because it contains a reservoir of species that react with added acid or base; its pH is $\text{pH}=\text{p}{K}_a+\log \frac{[\text{salt}]}{[\text{acid}]}$, or for a basic buffer $\text{pOH}=\text{p}{K}_b+\log \frac{[\text{salt}]}{[\text{base}]}$.

How it works.** In an acetic acid-acetate buffer, added H${}^{+}$ is taken up by acetate ions, and added OH${}^{-}$ is neutralised by acetic acid molecules.

Worked example 1 — acidic buffer. $0.1$ M acetic acid with $0.2$ M sodium acetate:

$$\text{pH}=4.74+\log \frac{0.2}{0.1}=4.74+0.30=5.04$$

Worked example 2 — adding acid. $0.01$ mol of HCl is added to $1$ L of a buffer that is $0.1$ M in both acetic acid and acetate. The acid becomes $0.11$ M and the salt $0.09$ M:

$$\text{pH}=4.74+\log \frac{0.09}{0.11}\approx 4.74-0.09=4.65$$

The same acid in $1$ L of pure water would drop the pH from $7$ to $2$.

Worked example 3 — basic buffer. $0.1$ M ammonia with $0.1$ M ammonium chloride: pOH $=4.74$, so **pH $=9.26$.

An everyday example. Blood is buffered by carbonic acid and hydrogen carbonate ions**, holding its pH close to $7.4$ even as muscles release acids.

The substance. A buffer works best when the salt and acid concentrations are about equal, so that pH is close to pK${}_a$.

**For a sparingly soluble salt in equilibrium with its ions, the solubility product $K_{sp}$ is the product of the ion concentrations raised to their coefficients; it links to molar solubility $s$ — for example $K_{sp}=s^2$ for AB and $4s^3$ for AB${}_2$ — and a precipitate forms when the ionic product exceeds $K_{sp}$.

Worked example 1 — silver chloride.** AgCl(s) $\rightleftharpoons$ Ag${}^{+}$ + Cl${}^{-}$ with $K_{sp}=1.8\times 10^{-10}$:

$$s=\sqrt{1.8\times 10^{-10}}\approx 1.34\times 10^{-5}\text{ M}\text{in grams: }1.34\times 10^{-5}\times 143.5\approx 1.9\times 10^{-3}\text{ g/L}$$

**Worked example 2 — an AB${}_2$ salt.** With $K_{sp}=4.0\times 10^{-12}$:

$$4s^3=4.0\times 10^{-12}\Rightarrow s=1.0\times 10^{-4}\text{ M}$$

Worked example 3 — will it precipitate? Equal volumes of $2\times 10^{-4}$ M AgNO${}_3$ and $2\times 10^{-4}$ M NaCl are mixed, so each ion becomes $1\times 10^{-4}$ M:

$$Q=(1\times 10^{-4}{)}^2=1\times 10^{-8}>1.8\times 10^{-10}\Rightarrow \text{AgCl precipitates}$$

An everyday example. White scale builds up inside kettles in hard-water areas when boiling pushes the ionic product of calcium carbonate above its solubility product.

The substance. **Compare solubilities through $s$, not $K_{sp}$, when salts have different formulas** — a smaller $K_{sp}$ does not always mean lower solubility.

**Adding an ion already present in a salt's equilibrium pushes the equilibrium back towards the solid, lowering the salt's solubility while $K_{sp}$ stays the same — and chemists use this to control which ions precipitate in qualitative analysis.

Worked example 1 — AgCl in salt solution.** In $0.1$ M NaCl, $[{\text{Cl}}^{-}]\approx 0.1$ M:

$$s=\frac{K_{sp}}{0.1}=\frac{1.8\times 10^{-10}}{0.1}=1.8\times 10^{-9}\text{ M}$$

That is roughly **$7400$ times less soluble than in pure water.

Uses in qualitative analysis:

- Group II**: H${}_2$S is passed in acidic solution, where extra H${}^{+}$ suppresses S${}^{2-}$; only sulphides with very low $K_{sp}$, such as CuS and PbS, precipitate
- Group III: NH${}_4$Cl is added with NH${}_4$OH, so the common NH${}_4^{+}$ ion lowers [OH${}^{-}$]; hydroxides of Fe${}^{3+}$, Al${}^{3+}$ and Cr${}^{3+}$ precipitate but Mg(OH)${}_2$ does not
- Group IV: H${}_2$S in basic solution gives enough S${}^{2-}$ to precipitate ZnS, MnS, NiS and CoS

An everyday example. Adding common salt during soap making throws the soap out of solution — the shared Na${}^{+}$ ion lowers its solubility.

The substance. **The common-ion effect changes solubility, never the value of $K_{sp}$** at a fixed temperature.