Weak Acid and Base Equilibria - AP Chemistry

1. A $0.10\,\text{M}$ solution of the weak acid HA is prepared at $25^\circ\text{C}$. The acid dissociation constant is $K_a=1.0\times10^{-5}$ for $\text{HA}(aq)+\text{H}_2\text{O}(l)\rightleftharpoons \text{H}_3\text{O}^+(aq)+\text{A}^-(aq).$ Which inequality is correct at equilibrium?
   Answer: $[\text{HA}] > [\text{A}^-]$ This problem involves weak acid and base equilibria. The weak acid HA establishes equilibrium according to HA + H₂O ⇌ H₃O⁺ + A⁻, with Ka = 1.0×10⁻⁵ indicating limited dissociation. Since Ka is small (much less than 1), the equilibrium lies far to the left, meaning most HA molecules remain undissociated. Therefore, [HA] > [A⁻] at equilibrium, with [HA] staying close to the initial 0.10 M while [A⁻] is much smaller. The incorrect choice A ([A⁻] > [HA]) would require Ka > 1, indicating a strong acid. For weak acids with small Ka values, remember that the undissociated form predominates at equilibrium.
2. A student compares two weak acids at the same initial concentration, $0.10\,\text{M}$: HX with $K_a = 1.0\times 10^{-2}$ and HY with $K_a = 1.0\times 10^{-6}$. In which solution is the ratio $\dfrac{[\text{A}^-]}{[\text{HA}]}$ (conjugate base to acid) larger at equilibrium?
   Answer: HX, because the larger $K_a$ indicates greater ionization. This question tests understanding of weak acid and base equilibria. The ratio $[\text{A}^{-}]$/$[\text{HA}]$ at equilibrium is directly related to the extent of ionization - a larger ratio means more of the acid has ionized to form its conjugate base. For weak acids at the same initial concentration, the acid with the larger $K_a$ will have greater ionization and thus a larger $[\text{A}^{-}]$/$[\text{HA}]$ ratio. Since HX has $K_a = 1.0\times 10^{-2}$ (much larger than HY's $K_a = 1.0\times 10^{-6}$), HX ionizes to a much greater extent, producing more $\text{X}^{-}$ relative to the remaining HX. A common error is thinking smaller $K_a$ means greater ionization, but $K_a$ is the equilibrium constant for ionization - larger $K_a$ means the equilibrium favors products (ions) more. For comparing weak acid ionization, remember that larger $K_a$ always means greater percent ionization and larger $[\text{A}^{-}]$/$[\text{HA}]$ ratio.
3. A student prepares $0.10\,\text{M}$ hydrofluoric acid, HF$(aq)$, in water. The acid establishes the equilibrium $\text{HF}(aq) + \text{H}_2\text{O}(l) \rightleftharpoons \text{H}_3\text{O}^+(aq) + \text{F}^-(aq)$ with $K_a = 6.8\times 10^{-4}$. Which statement about the equilibrium concentrations is correct?
   Answer: $[\text{HF}]_{eq} > [\text{F}^-]_{eq}$ This question tests understanding of weak acid and base equilibria. For weak acids like HF, incomplete dissociation means that at equilibrium, most of the acid remains in its molecular form rather than ionizing to H₃O⁺ and F⁻. The equilibrium constant Ka = 6.8×10⁻⁴ tells us the extent of ionization - since this value is much less than 1, the equilibrium lies far to the left, favoring the reactants. This means [HF]eq will be much greater than [F⁻]eq, making most of the original 0.10 M HF remain undissociated. A common misconception (choice E) is thinking that all the acid dissociates to give [F⁻] = 0.10 M, but this would only occur for a strong acid. When dealing with weak acids, always set up the equilibrium expression first and recognize that Ka < 1 means minimal ionization.
4. A $0.10\,\text{M}$ aqueous solution of ammonia is prepared at $25^\circ\text{C}$. Ammonia reacts with water according to $\text{NH}_3(aq)+\text{H}_2\text{O}(l)\rightleftharpoons \text{NH}_4^+(aq)+\text{OH}^-(aq)$ with $K_b=1.8\times10^{-5}$. Which expression correctly represents $K_b$ for this reaction?
   Answer: $K_b=\dfrac{[\text{NH}_4^+][\text{OH}^-]}{[\text{NH}_3]}$ This question assesses understanding of weak acid and base equilibria. Weak bases like NH3 do not fully accept protons from water, resulting in partial reaction and an equilibrium state. The equilibrium is quantified by Kb = [NH4+][OH-]/[NH3], which dictates the concentrations of ions produced relative to the undissociated base. Water's concentration is omitted from Kb as it is constant, focusing the expression on the base's interaction. A common distractor is choice E, which includes [H2O] in the numerator, but this is erroneous since water is the solvent. For weak species, write the equilibrium first before thinking about pH.
5. A student dissolves $0.10\,\text{mol}$ of nitrous acid, HNO$_2$, in enough water to make $1.0\,\text{L}$ of solution. For $\text{HNO}_2(aq) + \text{H}_2\text{O}(l) \rightleftharpoons \text{H}_3\text{O}^+(aq) + \text{NO}_2^-(aq)$, $K_a = 4.0\times 10^{-4}$. Which relationship between equilibrium concentrations is correct?
   Answer: $[\text{HNO}_2]_{eq} > [\text{NO}_2^-]_{eq}$ This question tests understanding of weak acid and base equilibria. Nitrous acid (HNO₂) is a weak acid with Ka = 4.0×10⁻⁴, which means it only partially ionizes in water according to HNO₂ + H₂O ⇌ H₃O⁺ + NO₂⁻. Since Ka < 1, the equilibrium lies to the left, favoring the reactants, so most HNO₂ molecules remain undissociated. This means [HNO₂]eq > [NO₂⁻]eq at equilibrium. A common misconception (choice C) is thinking all the acid dissociates to give [NO₂⁻] = 0.10 M, but this would only be true for a strong acid. For weak acid equilibria, use the Ka value to determine equilibrium position - when Ka < 1, the undissociated acid concentration exceeds the conjugate base concentration.
6. A solution is made by dissolving $0.20\,\text{mol}$ of ammonia in enough water to make $1.0\,\text{L}$ of solution. Ammonia is a weak base with $K_b=1.8\times10^{-5}$ at $25^\circ\text{C}$. For the equilibrium $\text{NH}_3(\text{aq})+\text{H}_2\text{O}(\ell)\rightleftharpoons \text{NH}_4^+(\text{aq})+\text{OH}^-(\text{aq}),$ which expression correctly represents $K_b$?
   Answer: $K_b=\dfrac{[\text{NH}_4^+][\text{OH}^-]}{[\text{NH}_3]}$ This question tests understanding of weak acid and base equilibria. Ammonia ($\text{NH}_3$) is a weak base that accepts protons from water according to $\text{NH}_3 + \text{H}_2\text{O} \rightleftharpoons \text{NH}_4^+ + \text{OH}^-$. The base dissociation constant $K_b$ represents this equilibrium and equals the product of ion concentrations divided by the base concentration: $K_b = \dfrac{[\text{NH}_4^+][\text{OH}^-]}{[\text{NH}_3]}$. Water is omitted from the expression because it's the solvent and its concentration remains essentially constant. Choice D incorrectly includes $[\text{H}_2\text{O}]$ in the expression, which is a common error when students forget that pure liquids are excluded from equilibrium expressions. For weak base equilibria, write products over reactants, excluding pure liquids like water.
7. A $0.10\,\text{M}$ solution of formic acid, HCOOH(aq), is prepared at $25^\circ\text{C}$. The equilibrium is $\text{HCOOH(aq)}\rightleftharpoons \text{H}^+(\text{aq})+\text{HCOO}^-(\text{aq}),$ with $K_a=1.8\times10^{-4}$. Which statement about the equilibrium position is most accurate?
   Answer: The equilibrium lies far to the left; most HCOOH remains undissociated. This question tests understanding of weak acid and base equilibria. Formic acid (HCOOH) has Ka = 1.8 × $10^{-4}$, which is much less than 1, indicating the equilibrium HCOOH ⇌ H+ + HCOO- lies far to the left. This means most HCOOH molecules remain undissociated at equilibrium, with only a small fraction converting to ions. Choice A incorrectly interprets the small Ka as indicating extensive dissociation, when actually Ka << 1 means minimal dissociation. For weak acid equilibria, remember that Ka < $10^{-3}$ typically means less than 10% dissociation, so the molecular form predominates.
8. A student dissolves benzoic acid, HC$_6$H$_5$COO(aq), to make a $0.10\,\text{M}$ solution at $25^\circ\text{C}$. The acid dissociation constant is $K_a=6.3\times10^{-5}$. Which equilibrium expression is correct for this system?$\text{HC}_6\text{H}_5\text{COO(aq)} \rightleftharpoons \text{H}^+(\text{aq})+\text{C}_6\text{H}_5\text{COO}^-(\text{aq})$
   Answer: $K_a=\dfrac{[\text{H}^+][\text{C}_6\text{H}_5\text{COO}^-]}{[\text{HC}_6\text{H}_5\text{COO}]}$ This question tests understanding of weak acid and base equilibria. For the dissociation of benzoic acid $\text{HC}_6\text{H}_5\text{COO} \rightleftharpoons \text{H}^{+} + \text{C}_6\text{H}_5\text{COO}^{-}$, the acid dissociation constant Ka is defined as the ratio of products to reactants: $K_a = \frac{[\text{H}^{+}][\text{C}_6\text{H}_5\text{COO}^{-}]}{[\text{HC}_6\text{H}_5\text{COO}]}$. This follows the general pattern for equilibrium constants where products appear in the numerator and reactants in the denominator, each raised to their stoichiometric coefficients. Choice A incorrectly inverts the expression, placing reactant over products, which would represent 1/Ka instead. For any weak acid equilibrium, always write Ka with ion products in the numerator and the molecular acid in the denominator.
9. Two separate $0.10\,\text{M}$ solutions are prepared at $25^\circ\text{C}$: Solution 1 contains $\text{HA}(aq)$ with $K_a=1.0\times10^{-5}$, and Solution 2 contains $\text{HB}(aq)$ with $K_a=1.0\times10^{-3}$. Both acids establish equilibrium with water. Which comparison is correct?
   Answer: Solution 2 has a higher $[\text{H}^+]_{eq}$ because it has the larger $K_a$ This question assesses understanding of weak acid and base equilibria. Weak acids exhibit incomplete dissociation, meaning only a portion ionizes to form H+ and the conjugate base. The value of Ka determines the strength: a larger Ka, like for HB ($1.0×10^{-3}$), leads to greater dissociation and higher [H+] compared to a smaller Ka for HA ($1.0×10^{-5}$). Thus, the equilibrium concentrations are governed by Ka, with stronger weak acids producing more H+ at the same initial concentration. Choice A is a tempting distractor, suggesting a smaller Ka yields higher [H+], but this is incorrect as smaller Ka means weaker acid and less ionization. For weak species, write the equilibrium first before thinking about pH.
10. A $0.10\,\text{M}$ solution of the weak acid HA is prepared. The acid dissociation constant is $K_a = 1.0\times 10^{-6}$ for $\text{HA}(aq)+\text{H}_2\text{O}(l)\rightleftharpoons \text{H}_3\text{O}^+(aq)+\text{A}^-(aq)$. Which statement about the equilibrium concentrations is correct?
    Answer: $[\text{HA}]_{eq} \approx 0.10\,\text{M}$ This question tests understanding of weak acid and base equilibria. For a weak acid HA with Ka = 1.0×10⁻⁶, the very small equilibrium constant indicates minimal ionization, meaning the equilibrium HA + H₂O ⇌ H₃O⁺ + A⁻ lies far to the left. Since only a tiny fraction of HA molecules ionize, the equilibrium concentration [HA]eq remains very close to the initial concentration of 0.10 M. The concentrations of H₃O⁺ and A⁻ will be much smaller, approximately equal to each other but far less than 0.10 M. A common error (choice E) is thinking the conjugate base concentration equals the initial acid concentration, which would only occur for complete dissociation. For very weak acids (Ka << 1), assume [HA]eq ≈ initial concentration as a first approximation.
11. A $0.10\,\text{M}$ solution of the weak base B is prepared. The base-ionization constant is $K_b = 1.0\times 10^{-3}$ for $\text{B}(aq)+\text{H}_2\text{O}(l)\rightleftharpoons \text{BH}^+(aq)+\text{OH}^-(aq)$. Which is the correct expression for $K_b$?
    Answer: $K_b = \dfrac{[\text{BH}^+][\text{OH}^-]}{[\text{B}]}$ This question tests understanding of weak acid and base equilibria. For the weak base B undergoing the equilibrium B + H₂O ⇌ BH⁺ + OH⁻, the base ionization constant Kb follows the standard form of products over reactants. Since water is the solvent and its concentration remains essentially constant, it is excluded from the equilibrium expression, giving Kb = [BH⁺][OH⁻]/[B]. This follows the general rule that equilibrium constants equal the concentration of products divided by reactants, each raised to their stoichiometric coefficients. A common mistake (choice D) is including water in the expression, but pure liquids are never included in equilibrium constants. When writing equilibrium expressions for weak bases, always exclude the water even though it appears in the balanced equation.
12. A student dissolves ammonia in water to make $0.10\,\text{M}$ NH$_3(aq)$. The base ionizes according to $\text{NH}_3(aq) + \text{H}_2\text{O}(l) \rightleftharpoons \text{NH}_4^+(aq) + \text{OH}^-(aq)$ with $K_b = 1.8\times 10^{-5}$. Which statement is correct at equilibrium?
    Answer: $[\text{OH}^-] < [\text{NH}_3]$ This question assesses understanding of weak acid and base equilibria. Weak bases like NH3 do not ionize completely, resulting in an equilibrium where NH3 partially accepts protons from water to form NH4+ and OH- ions. The base dissociation constant Kb=1.8×$10^{-5}$ shows that the equilibrium favors reactants, so [NH3] remains much higher than [OH-] or [NH4+]. This incomplete ionization governs the concentrations, with [OH-] < [NH3] at equilibrium due to the small Kb. A common misconception is that [NH4+] = [NH3] at equilibrium, but this would only hold if half ionized, which isn't true for weak bases. For weak species, write the equilibrium first before thinking about pH.
13. A $0.10\,\text{M}$ solution of the weak base $\text{B}$ is prepared at $25^\circ\text{C}$. The base reacts with water according to $\text{B}(aq)+\text{H}_2\text{O}(l)\rightleftharpoons \text{BH}^+(aq)+\text{OH}^-(aq)$ with $K_b=4.0\times10^{-4}$. Compared with pure water at the same temperature, which change occurs as equilibrium is established?
    Answer: $[\text{OH}^-]$ increases because the reaction produces $\text{OH}^-$ This question assesses understanding of weak acid and base equilibria. Weak bases like B partially react with water, not fully, leading to an equilibrium where OH- is produced. Kb = [BH+][OH-]/[B] describes this equilibrium, with a value like 4.0×$10^{-4}$ indicating moderate dissociation and increased [OH-] compared to pure water. This governs concentrations, raising [OH-] and thus making the solution basic. Choice A is a common misconception, stating [H+] increases because BH+ is an acid, but actually [H+] decreases due to higher [OH-]. For weak species, write the equilibrium first before thinking about pH.
14. A $0.20\,\text{M}$ solution of ammonia, $\text{NH}_3(aq)$, is prepared. The base-ionization equilibrium is $\text{NH}_3(aq)+\text{H}_2\text{O}(l)\rightleftharpoons \text{NH}_4^+(aq)+\text{OH}^-(aq)$ with $K_b = 1.8\times 10^{-5}$. Which expression is the correct equilibrium-constant expression for $K_b$?
    Answer: $K_b = \dfrac{[\text{NH}_4^+][\text{OH}^-]}{[\text{NH}_3]}$ This question tests understanding of weak acid and base equilibria. For the weak base ammonia, the equilibrium NH₃ + H₂O ⇌ NH₄⁺ + OH⁻ is described by the base ionization constant Kb. The equilibrium constant expression is always products over reactants, with each species raised to its stoichiometric coefficient. Since water is the solvent and its concentration remains essentially constant, it is not included in the Kb expression, giving Kb = [NH₄⁺][OH⁻]/[NH₃]. A common error (choice A) is including water in the expression, but pure liquids and solids are never included in equilibrium constant expressions. For weak base equilibria, remember to write the balanced equation first, then form the Kb expression excluding pure liquids.
15. A student prepares $0.10\,\text{M}$ hydrofluoric acid, HF$(aq)$, at $25^\circ\text{C}$. The acid establishes the equilibrium $\text{HF}(aq) \rightleftharpoons \text{H}^+(aq) + \text{F}^-(aq)$ with $K_a = 6.8\times 10^{-4}$. Which statement about the equilibrium concentrations is correct?
    Answer: $[\text{F}^-] < [\text{HF}]$ This question assesses understanding of weak acid and base equilibria. Weak acids like HF do not dissociate completely in water, leading to an equilibrium where only a small fraction ionizes to produce H+ and F- ions. The equilibrium constant Ka quantifies this incomplete dissociation, with a value of 6.8×$10^{-4}$ indicating that the position of equilibrium favors the reactants, resulting in [HF] being much larger than [F-] or [H+]. Consequently, at equilibrium, the concentration of the undissociated acid governs the system, with [F-] < [HF] because the dissociation is limited. A tempting distractor is that [F-] = [HF] at equilibrium, but this is incorrect as it would imply complete dissociation or a Ka of 1, which is not the case for weak acids. For weak species, write the equilibrium first before thinking about pH.
16. A $0.10\,\text{M}$ solution of a weak base, $\text{BOH}(aq)$, is prepared and establishes $\text{BOH}(aq) \rightleftharpoons \text{B}^+(aq) + \text{OH}^-(aq)$ with $K_b = 2.5 \times 10^{-6}$. Which is the correct expression for $K_b$?
    Answer: $K_b = \dfrac{[\text{B}^+][\text{OH}^-]}{[\text{BOH}]}$ This question assesses understanding of weak acid and base equilibria. Weak bases like BOH dissociate incompletely, establishing an equilibrium expressed by $K_b = \dfrac{[\text{B}^+][\text{OH}^-]}{[\text{BOH}]}$. This $K_b = 2.5 \times 10^{-6}$ governs the small extent of ionization and resulting concentrations. Water is not in the expression as solvent. A tempting distractor is $K_b = \dfrac{[\text{BOH}]}{[\text{B}^+][\text{OH}^-]}$, but this inverts the dissociation constant. For weak species, write the equilibrium first before thinking about pH.
17. A $0.10\,\text{M}$ solution of methylamine, $\text{CH}_3\text{NH}_2(\text{aq})$, is prepared at $25^\circ\text{C}$. For the equilibrium $\text{CH}_3\text{NH}_2(\text{aq})+\text{H}_2\text{O}(\ell)\rightleftharpoons \text{CH}_3\text{NH}_3^+(\text{aq})+\text{OH}^-(\text{aq}),$ $K_b = 4.4 \times 10^{-4}$. Which statement must be true at equilibrium?
    Answer: $[\text{CH}_3\text{NH}_2] > [\text{CH}_3\text{NH}_3^+]$ This question tests understanding of weak acid and base equilibria. Methylamine ($\text{CH}_3\text{NH}_2$) is a weak base with $K_b = 4.4 \times 10^{-4}$, establishing equilibrium $\text{CH}_3\text{NH}_2 + \text{H}_2\text{O} \rightleftharpoons \text{CH}_3\text{NH}_3^+ + \text{OH}^-$. Since $K_b$ is small (much less than 1), the equilibrium lies far to the left, meaning most methylamine molecules remain unreacted. Therefore, at equilibrium $[\text{CH}_3\text{NH}_2] > [\text{CH}_3\text{NH}_3^+]$. Choice B incorrectly assumes complete reaction to produce $[\text{OH}^-] = 0.10\,\text{M}$, which would only occur for a strong base like NaOH. For weak bases, use the $K_b$ expression to recognize that small $K_b$ means little reaction, so the molecular form predominates.
18. A $0.30\,\text{M}$ solution of the weak base B is prepared at $25^\circ\text{C}$. The base-ionization constant is $K_b=1.0\times10^{-6}$ for $\text{B}(aq)+\text{H}_2\text{O}(l)\rightleftharpoons \text{BH}^+(aq)+\text{OH}^-(aq).$ Which equilibrium expression is correct for this reaction?
    Answer: $K_b=\dfrac{[\text{BH}^+][\text{OH}^-]}{[\text{B}]}$ This problem involves weak acid and base equilibria. For the weak base B, the equilibrium B + H₂O ⇌ BH⁺ + OH⁻ is described by the base ionization constant Kb. The equilibrium expression follows the standard form: Kb = [products]/[reactants], which gives Kb = [BH⁺][OH⁻]/[B]. Water doesn't appear because its concentration is essentially constant in dilute solutions and is incorporated into Kb. The incorrect choice A inverts the expression, placing reactants over products, which would give 1/Kb instead of Kb. When writing equilibrium expressions, always place products in the numerator and reactants in the denominator.
19. Two separate $0.10\,\text{M}$ solutions are prepared at $25^\circ\text{C}$: one contains the weak acid HCN with $K_a=6.2\times10^{-10}$ and the other contains the weak acid HOCl with $K_a=3.0\times10^{-8}$. After equilibrium is established in each solution, which statement is correct?
    Answer: The HOCl solution has the larger $[\text{H}_3\text{O}^+]$ because its $K_a$ is larger This problem involves weak acid and base equilibria. Both HCN and HOCl are weak acids that partially dissociate, with the extent controlled by their Ka values. HOCl has Ka = 3.0×10⁻⁸, which is larger than HCN's Ka = 6.2×10⁻¹⁰, meaning HOCl dissociates more extensively. Greater dissociation produces more H₃O⁺ ions, so the HOCl solution has the larger [H₃O⁺]. The incorrect choice A reverses this relationship, incorrectly claiming that smaller Ka leads to larger [H₃O⁺]. For weak acid comparisons, always remember that larger Ka means stronger acid, more dissociation, and higher [H₃O⁺] concentration.
20. A $0.10\,\text{M}$ solution of a weak monoprotic acid, HA(aq), has $K_a=1.0\times10^{-5}$ at $25^\circ\text{C}$. Which statement about the equilibrium mixture for $\text{HA(aq)} \rightleftharpoons \text{H}^{+}(\text{aq}) + \text{A}^{-}(\text{aq})$ is correct?
    Answer: $[\text{A}^{-}] = [\text{H}^{+}]$ (ignoring the small contribution of water autoionization). This question tests understanding of weak acid and base equilibria. For the weak acid HA dissociating according to $\text{HA} \rightleftharpoons \text{H}^{+} + \text{A}^{-}$, the stoichiometry shows that one $\text{H}^{+}$ and one $\text{A}^{-}$ are produced for each HA that dissociates. Since the solution initially contains no $\text{H}^{+}$ or $\text{A}^{-}$ (ignoring water's small contribution), and they are produced in a 1:1 ratio, $[\text{H}^{+}] = [\text{A}^{-}]$ at equilibrium. Choice B incorrectly assumes complete dissociation ($[\text{A}^{-}] = 0.10\, \text{M}$), which would only occur for a strong acid. For monoprotic weak acids in pure water, always remember that $[\text{H}^{+}] = [\text{A}^{-}]$ because they are produced in equal amounts from the dissociation.

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