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3. Acids, Bases, and Buffers

Acids donate $\text{H}^+$; bases accept it. The pH scale compresses enormous concentration ranges into tidy numbers: $$\text{pH} = -\log[\text{H}_3\text{O}^+]$$

Three Definitions

Arrhenius: acids produce $\text{H}^+$ in water; bases produce $\text{OH}^-$.
Brønsted–Lowry: acids are proton donors, bases are proton acceptors — every acid has a conjugate base (e.g., $\text{CH}_3\text{COOH}$ / $\text{CH}_3\text{COO}^-$).
Lewis: acids accept an electron pair; bases donate one.

pH of Strong Acids and Bases

Strong acids and bases dissociate completely.

Example: 0.010 M $\text{HCl}$ gives $[\text{H}_3\text{O}^+] = 0.010$ M, so $\text{pH} = 2.00$.
Example: 0.0050 M $\text{Ba}(\text{OH})_2$ gives $[\text{OH}^-] = 0.010$ M, so $\text{pOH} = 2.00$ and $\text{pH} = 14.00 - 2.00 = 12.00$.

Weak Acids and Ka

Example: For 0.10 M $\text{CH}_3\text{COOH}$ ($K_a = 1.8 \times 10^{-5}$): let $x = [\text{H}_3\text{O}^+]$ at equilibrium. $$\frac{x^2}{0.10 - x} \approx \frac{x^2}{0.10} = 1.8 \times 10^{-5} \;\Rightarrow\; x = 1.34 \times 10^{-3}$$ $\text{pH} = 2.87$. (The $x \ll$ initial approximation is valid here.)

Buffers and the Henderson–Hasselbalch Equation

A buffer — a weak acid plus its conjugate base — resists pH change: $$\text{pH} = \text{p}K_a + \log\frac{[\text{A}^-]}{[\text{HA}]}$$

Example: An acetate buffer with 0.20 M $\text{CH}_3\text{COO}^-$ and 0.30 M $\text{CH}_3\text{COOH}$ ($\text{p}K_a = 4.74$): $$\text{pH} = 4.74 + \log\frac{0.20}{0.30} = 4.74 - 0.176 = 4.56$$

Practice Problems

Find the pH of 0.0025 M HNO₃.
A 0.20 M solution of a weak acid HA has pH 3.10. Estimate K_a.
Calculate the pH of a buffer that is 0.15 M NH₃ and 0.10 M NH₄⁺ (pK_a of NH₄⁺ = 9.25).
Why does adding a small amount of strong acid barely change a buffer's pH?