2.12 · Acids and BasesPractice · 酸与碱

A conjugate pair differs by exactly one $\ce{H+}$. $\ce{H2PO4-}$ gives up an $\ce{H+}$ to become $\ce{HPO4^2-}$: that’s one pair. $\ce{HCO3-}$ takes an $\ce{H+}$ to become $\ce{H2CO3}$: that’s the other. The remaining combinations aren’t one species gaining or losing an $\ce{H+}$.

An acid donates $\ce{H+}$. In C, $\ce{H2O}$ gives an $\ce{H+}$ to $\ce{CO3^2-}$ and becomes $\ce{OH-}$, so it acts as an acid. In A and B, $\ce{H2O}$ accepts an $\ce{H+}$ and becomes $\ce{H3O+}$ (a base); in D, water is a product. Water can be either; it depends on whether it gives or takes $\ce{H+}$ in that reaction.

Let [H⁺] = x, so Ka = $\dfrac{x^2}{0.20 - x}$. Ka is tiny, so x ≪ 0.20 and x² ≈ Ka × c = $2.9 \times 10^{-8} \times 0.20$, giving x = $7.6 \times 10^{-5}$ M (only 0.038% ionized, so the approximation holds). pH = −log($7.6 \times 10^{-5}$) = 4.12.

With Ka = $1.8 \times 10^{-5}$: at 0.10 M, [H⁺] = $0.0013$ M and 1.3% is ionized; at 0.010 M, [H⁺] = $4.2 \times 10^{-4}$ M and 4.2% is ionized. Diluting a weak acid raises the fraction that ionizes, but the solution is still more dilute overall, so [H⁺] falls, just by less than a factor of 10.

KOH is a strong base and dissociates completely: [OH⁻] = $0.0040$ M. pOH = −log($0.0040$) = 2.40, so pH = 14.00 − 2.40 = 11.60. The concentration has 2 significant figures, so the pH gets 2 decimal places.

[H⁺] = $10^{-\text{pH}}$ = $10^{-7.40}$ = $4.0 \times 10^{-8}$ M. The pH has 2 decimal places, so the concentration has 2 significant figures. Blood is slightly basic.

Neutral means [H⁺] = [OH⁻], not pH = 7. At 75 °C a neutral solution has [H⁺] = $\sqrt{K_w}$, so its pH is 12.70 ÷ 2 = 6.35. This solution’s pH of 6.80 is higher than 6.35, so [H⁺] < [OH⁻]: it’s basic. pH 7 means neutral only near 25 °C.

CH₃COOH + NaOH → CH₃COONa + H₂O reacts 1 : 1, so [acetic acid] = 0.5000 × 16.20 ÷ 10.00 = 0.8100 M. 100 mL holds 0.8100 × 0.1000 = 0.08100 mol, which is 0.08100 × 60.05 = 4.864 g.

At the equivalence point the solution is NH₄Cl with [NH₄⁺] = 0.0500 M. NH₄⁺ is a weak acid (Ka = Kw ÷ Kb = $5.6 \times 10^{-10}$), so the solution is acidic, pH ≈ 5.3. Choose an indicator that changes color in the acidic range, such as methyl red (4.4–6.2); phenolphthalein (8.3–10.0) won’t work. The indicator only signals the end point; it doesn’t set the pH.

n(H⁺) = 0.00500 mol and n(OH⁻) = 0.00300 mol, so after neutralization 0.00200 mol of H⁺ is left over. In 0.0800 L, [H⁺] = 0.0250 M and pH = −log 0.0250 = 1.602. You can’t simply average the two pH values.