A weak base takes a proton from water and leaves $\mathrm{OH^-}$ behind. The calculation is the weak-acid one with $K_b$ in place of $K_a$ and $\mathrm{OH^-}$ in place of $\mathrm{H_3O^+}$, plus one last step from pOH to pH.
Key idea
Why it works
Worked examples
Worked example: Ammonia
- $$x = \sqrt{(1.8 \times 10^{-5})(0.15)} = 1.64 \times 10^{-3}\ \text{M} = [\mathrm{OH^-}]$$ Check: 1.1% of 0.15, fine.
- $\text{pOH} = -\log(1.64 \times 10^{-3}) = 2.78$.
- $\text{pH} = 14.00 - 2.78 = 11.22$.
Answer: $\text{pH} = 11.22$.
Worked example: A stronger weak base
- Shortcut: $x = \sqrt{(4.4 \times 10^{-4})(0.050)} = 4.7 \times 10^{-3}$, which is 9.4% of 0.050. Too big.
- Exact: $x^2 + (4.4 \times 10^{-4})x - 2.2 \times 10^{-5} = 0$ gives $x = 4.48 \times 10^{-3}\ \text{M}$.
- $\text{pOH} = 2.35$, $\text{pH} = 11.65$.
Answer: $\text{pH} = 11.65$.
Worked example: When only Ka is given
- They are a conjugate pair: $K_b = \dfrac{K_w}{K_a} = \dfrac{1.0 \times 10^{-14}}{5.6 \times 10^{-10}}$.
- $K_b = 1.8 \times 10^{-5}$.
Answer: $1.8 \times 10^{-5}$.
Worked example: A very weak base
- $$x = \sqrt{(1.7 \times 10^{-9})(0.10)} = 1.3 \times 10^{-5}\ \text{M} = [\mathrm{OH^-}]$$ a tiny fraction of 0.10.
- $\text{pOH} = 4.89$, so $\text{pH} = 9.11$: basic, but far less so than ammonia at the same concentration.
Answer: $\text{pH} = 9.11$.
| Base | Formula | $K_b$ |
|---|---|---|
| ammonia | $\mathrm{NH_3}$ | $1.8 \times 10^{-5}$ |
| methylamine | $\mathrm{CH_3NH_2}$ | $4.4 \times 10^{-4}$ |
| dimethylamine | $\mathrm{(CH_3)_2NH}$ | $5.4 \times 10^{-4}$ |
| trimethylamine | $\mathrm{(CH_3)_3N}$ | $6.3 \times 10^{-5}$ |
| hydrazine | $\mathrm{N_2H_4}$ | $1.3 \times 10^{-6}$ |
| pyridine | $\mathrm{C_5H_5N}$ | $1.7 \times 10^{-9}$ |
Anions of weak acids are weak bases too: acetate, fluoride and cyanide take protons from water. Their $K_b$ values are usually found from the acid's $K_a$, which is the next lesson's main tool. The weak-base calculation is then exactly the one above.
Common mistake