Conjugate pairs and pH
Conjugate pairs and pH
- When an acid loses $\text{H}^+$, what's left is its conjugate base.
- pH and the equilibrium constants put numbers on acidity.
- The method to find pH depends on the acid or base.
Practice
The conjugate base of CH₃COOH is:
Removing one H⁺ from the acid leaves its conjugate base; the two differ by a single proton.
Conjugate acid–base pairs
- An acid loses $\text{H}^+$ → its conjugate base; a base gains $\text{H}^+$ → its conjugate acid.
- The two differ by one $\text{H}^+$ — a conjugate pair. E.g. $\text{CH}_3\text{COOH}$ / $\text{CH}_3\text{COO}^-$.
Practice
pH is defined as:
pH = −log[H⁺]; so [H⁺] = 10^(−pH).
Practice
A larger Ka (smaller pKa) means the acid is:
A bigger acid dissociation constant means more dissociation, so a stronger acid.
pH and the constants
- $\text{pH} = -\log[\text{H}^+]$, so $[\text{H}^+] = 10^{-\text{pH}}$.
- weak acid: $K_a = \dfrac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}$, and $\text{p}K_a = -\log K_a$ (larger $K_a$ = stronger acid).
- water: $K_w = [\text{H}^+][\text{OH}^-] = 1.0 \times 10^{-14}$ at 298 K.
Practice
To find [H⁺] for a weak acid, you use:
A weak acid is only partly ionised, so [H⁺] = √(Ka × [HA]); a strong acid uses [H⁺] = concentration.
Calculating pH
- strong acid (fully ionised): $[\text{H}^+]$ = acid concentration, then $-\log$.
- strong alkali: find $[\text{OH}^-]$, then $[\text{H}^+] = K_w / [\text{OH}^-]$.
- weak acid: $[\text{H}^+] = \sqrt{K_a \times [\text{HA}]}$.
You've got it
Key idea
- conjugate pair = acid and base differing by one $\text{H}^+$
- $\text{pH} = -\log[\text{H}^+]$; $K_a$ (bigger = stronger acid); $K_w = 1.0 \times 10^{-14}$
- strong acid: $[\text{H}^+]$ = conc; strong alkali: via $K_w$; weak acid: $\sqrt{K_a[\text{HA}]}$