Titration and Buffers (5A) - Chem & Physics

1. What is the definition of a buffer in aqueous acid–base chemistry?

   A solution that resists pH change upon small acid/base addition. Buffers maintain stable pH by neutralizing small amounts of added acid or base through reversible reactions involving weak acids/bases and their conjugates.

2. State the Henderson–Hasselbalch equation for a weak acid buffer.

   $\text{pH}=\text{p}K_a+\log\left(\frac{[A^-]}{[HA]}\right)$. Derived from the acid dissociation equilibrium, this equation relates pH to the acid's pKa and the ratio of conjugate base to acid concentrations.

3. What is the relationship between $K_a$ and $\text{p}K_a$?

   $\text{p}K_a=-\log(K_a)$. pKa quantifies acid strength as the negative logarithm of Ka, where lower pKa indicates stronger acids.

4. What is the pH at the equivalence point in a strong acid–strong base titration at $25\,^{\circ}\text{C}$?

   $\text{pH}=7$. Neutralization of strong acid by strong base produces neutral salt solution with pH = 7 at 25°C.

5. Find the pH when $\text{p}K_a=4.76$ and $\frac{[A^-]}{[HA]}=1.0$.

   $\text{pH}=4.76$. Using Henderson-Hasselbalch, when ratio = 1, log(1) = 0, so pH equals pKa directly.

6. What is the relationship between $K_a$ and $K_b$ for a conjugate pair at $25\,^{\circ}\text{C}$?

   $K_aK_b=K_w=1.0\times10^{-14}$. For conjugate acid-base pairs, the product of Ka and Kb equals the water dissociation constant Kw at 25°C.

7. Find the pH when $\text{p}K_a=6.30$ and $\frac{[A^-]}{[HA]}=10$.

   $\text{pH}=7.30$. Henderson-Hasselbalch gives pH = pKa + log(10) = pKa + 1, yielding 7.30.

8. At what condition does a buffer have maximum buffering capacity?

   When $\text{pH}=\text{p}K_a$ and $[A^-]=[HA]$. Maximum capacity occurs when the buffer can equally resist acid or base additions, which is at equal concentrations and pH = pKa.

9. Find the equivalence volume: $25.0\,\text{mL}$ of $0.100\,\text{M}$ HCl titrated by $0.200\,\text{M}$ NaOH.

   $V_{\text{NaOH}}=12.5\,\text{mL}$. Using Ma Va = Mb Vb for 1:1 titration, Vb = (0.100 * 25.0) / 0.200 = 12.5 mL.

10. State the equivalence condition for any titration in terms of moles of titrant and analyte.

    Moles titrant added = stoichiometric moles required for analyte. Equivalence occurs when the titrant provides exactly the moles needed to react completely with the analyte per the stoichiometry.

11. State the formula used to find moles from molarity and volume (with volume in liters).

    $n=MV$. Molarity times volume in liters gives moles, essential for calculating stoichiometry in titrations.

12. Identify the pH relative to $\text{p}K_a$ at the half-equivalence point of a weak acid titration.

    $\text{pH}=\text{p}K_a$. At half-equivalence, [HA] = [A-], so Henderson-Hasselbalch simplifies to pH = pKa.

13. Find $\frac{[A^-]}{[HA]}$ when $\text{pH}=5.00$ and $\text{p}K_a=4.00$.

    $\frac{[A^-]}{[HA]}=10$. From rearranged Henderson-Hasselbalch, ratio = $10^{pH - pKa}$ = $10^{1}$ = 10.

14. What is the relationship between $K_b$ and $\text{p}K_b$?

    $\text{p}K_b=-\log(K_b)$. pKb measures base strength as the negative logarithm of Kb, with lower pKb indicating stronger bases.

15. What is the relationship between $\text{p}K_a$ and $\text{p}K_b$ at $25\,^{\circ}\text{C}$?

    $\text{p}K_a+\text{p}K_b=14$. Taking negative logs of the Ka Kb = Kw relation yields pKa + pKb = 14 at 25°C for conjugate pairs.

16. Identify the primary purpose of an indicator in an acid–base titration.

    To signal the endpoint by a visible color change near equivalence. Indicators change color at a specific pH near the equivalence point to visually indicate when to stop the titration.

17. Identify the dominant species when $\text{pH}<\text{p}K_a$ for a weak acid/conjugate base system.

    The protonated form $HA$ dominates. At pH below pKa, the equilibrium favors the undissociated acid form according to Le Chatelier's principle.

18. In a weak acid–strong base titration, is the pH at equivalence point less than, equal to, or greater than $7$?

    Greater than $7$. The equivalence point forms the conjugate base of the weak acid, which hydrolyzes to produce a basic solution with pH > 7.

19. What two components must be present in an effective buffer solution?

    A weak acid and its conjugate base (or weak base and conjugate acid). Effective buffers require both components to shift equilibrium and absorb added H+ or OH- ions without significant pH change.

20. In a weak base–strong acid titration, is the pH at equivalence point less than, equal to, or greater than $7$?

    Less than $7$. The equivalence point forms the conjugate acid of the weak base, which hydrolyzes to produce an acidic solution with pH < 7.

21. What is the typical effective buffering range for a weak acid buffer relative to $\text{p}K_a$?

    Approximately $\text{p}K_a\pm^1$ pH unit. Buffers are effective within about one pH unit of pKa where the ratio of components allows significant buffering against additions.

22. What is the formula for the equivalence-point volume in a $1{:}1$ acid–base titration?

    $M_aV_a=M_bV_b$. For 1:1 stoichiometry, the product of molarity and volume is equal for acid and base at equivalence.

23. Identify the dominant species when $\text{pH}>\text{p}K_a$ for a weak acid/conjugate base system.

    The deprotonated form $A^-$ dominates. At pH above pKa, the equilibrium shifts to favor the dissociated conjugate base form.

24. What is the pH at the half-equivalence point in a weak acid–strong base titration?

    $\text{pH}=\text{p}K_a$. At half-equivalence, half the acid is neutralized, making [HA] = [A-], so pH = pKa from Henderson-Hasselbalch.

25. Which expression gives the buffer ratio $\frac{[A^-]}{[HA]}$ in terms of pH and $\text{p}K_a$?

    $\frac{[A^-]}{[HA]}=10^{\text{pH}-\text{p}K_a}$. Rearranging the Henderson-Hasselbalch equation isolates the ratio as an exponential function of the pH-pKa difference.

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