AP Chemistry Quiz: Weak Acid And Base Equilibria

What this quiz covers

This quiz focuses on Weak Acid And Base Equilibria, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Chemistry.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. The HCN solution has the larger $[\text{H}^+]$ because its $K_a$ is smaller.
2. Both solutions have the same $[\text{H}^+]$ because the initial concentration is the same.
3. The acetic acid solution has the larger $[\text{H}^+]$ because its $K_a$ is larger. (correct answer)
4. Both acids fully dissociate, so $[\text{H}^+]=0.10\,\text{M}$ in each.
5. The HCN solution has the larger $[\text{H}^+]$ because it is a molecular acid.

Explanation: This question tests understanding of weak acid and base equilibria. Both acetic acid and HCN are weak acids that establish equilibria with their conjugate bases and H+ ions. The strength of a weak acid is determined by its Ka value - larger Ka means stronger acid and more dissociation. Since acetic acid has Ka = 1.8 × 10^-5 while HCN has Ka = 6.2 × 10^-10, acetic acid is the stronger acid and will produce more H+ ions at equilibrium. Choice D incorrectly assumes both acids fully dissociate, which only happens with strong acids like HCl or HNO3. For comparing weak acids at the same concentration, the one with larger Ka always produces more H+ and thus has lower pH.

Question 2

1. $[\text{CH}_3\text{COO}^-]_{eq} > [\text{CH}_3\text{COOH}]_{eq}$ because ions are favored in water
2. $[\text{H}^+]_{eq}=[\text{CH}_3\text{COO}^-]_{eq}$ because they are produced in a 1:1 ratio (correct answer)
3. $[\text{CH}_3\text{COOH}]_{eq}=0.10\,\text{M}$ exactly because the acid is weak
4. $[\text{CH}_3\text{COO}^-]_{eq}=0$ because the acid is weak and does not ionize
5. $[\text{H}^+]_{eq}=[\text{CH}_3\text{COOH}]_{eq}$ because each mole of acid contains one acidic H

Explanation: This question assesses understanding of weak acid and base equilibria. Weak acids like acetic acid partially dissociate, forming H+ and CH3COO- in equilibrium. Ka = 1.8×10^{-5} indicates limited ionization, but the reaction stoichiometry ensures [H+] = [CH3COO-] from each dissociation event. This equality governs the concentrations, neglecting water's autoionization. Choice A is a common misconception, suggesting [CH3COO-]{eq} > [CH3COOH]{eq} because ions are favored, but small Ka means the opposite. For weak species, write the equilibrium first before thinking about pH.

Question 3

1. K_a=\dfrac{[\text{HC}_6\text{H}_5\text{COO}]}{[\text{H}^+][\text{C}_6\text{H}_5\text{COO}^-]}}
2. K_a=\dfrac{[\text{H}^+]}{[\text{HC}_6\text{H}_5\text{COO}][\text{C}_6\text{H}_5\text{COO}^-]}}
3. $K_a=\dfrac{[\text{H}^+][\text{C}_6\text{H}_5\text{COO}^-]}{[\text{HC}_6\text{H}_5\text{COO}]}$ (correct answer)
4. K_a=\dfrac{[\text{HC}_6\text{H}_5\text{COO}][\text{H}_2\text{O}]}{[\text{H}^+][\text{C}_6\text{H}_5\text{COO}^-]}}
5. $K_a=\dfrac{[\text{C}_6\text{H}_5\text{COO}^-]}{[\text{HC}_6\text{H}_5\text{COO}]}$

Explanation: 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.

Question 4

1. $[\text{A}^-]_{eq} = [\text{HA}]_{eq}$
2. $[\text{H}_3\text{O}^+]_{eq} = 0.10\,\text{M}$
3. $[\text{HA}]_{eq} \approx 0.10\,\text{M}$ (correct answer)
4. $[\text{HA}]_{eq} = 0$
5. $[\text{A}^-]_{eq} \approx 0.10\,\text{M}$

Explanation: 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.

Question 5

1. HX, because the larger $K_a$ indicates greater ionization. (correct answer)
2. HY, because the smaller $K_a$ indicates greater ionization.
3. HX, because the smaller $K_a$ indicates greater ionization.
4. HY, because the larger $K_a$ indicates greater ionization.
5. The ratio is the same in both because the initial concentrations are equal.

Explanation: 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.

Question 6

1. $K_b = \dfrac{[\text{NH}_4^+][\text{OH}^-]}{[\text{NH}_3][\text{H}_2\text{O}]}$
2. $K_b = \dfrac{[\text{NH}_3]}{[\text{NH}_4^+][\text{OH}^-]}$
3. $K_b = \dfrac{[\text{NH}_4^+][\text{OH}^-]}{[\text{NH}_3]}$ (correct answer)
4. $K_b = \dfrac{[\text{NH}_3][\text{H}_2\text{O}]}{[\text{NH}_4^+][\text{OH}^-]}$
5. $K_b = \dfrac{[\text{OH}^-]}{[\text{NH}_3][\text{NH}_4^+]}$

Explanation: 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.

Question 7

1. Equilibrium lies far to the right, so most CH$_3$COOH is converted to CH$_3$COO$^-$
2. Equilibrium lies far to the left, so most CH$_3$COOH remains undissociated (correct answer)
3. Equilibrium lies to the right because $K_a>1$
4. Equilibrium lies to the left because $K_a$ is a base constant
5. Equilibrium lies equally left and right because the initial concentration is $0.10\,\text{M}$

Explanation: This problem involves weak acid and base equilibria. Acetic acid (CH₃COOH) is a weak acid with Ka = 1.8×10⁻⁵, which is much less than 1. This small Ka value indicates that the equilibrium CH₃COOH + H₂O ⇌ H₃O⁺ + CH₃COO⁻ lies far to the left, meaning the reverse reaction is favored. Consequently, most CH₃COOH molecules remain undissociated at equilibrium, with only a small fraction converting to CH₃COO⁻ and H₃O⁺. The incorrect choice A claims extensive dissociation, which would require Ka >> 1, characteristic of a strong acid. For weak acids, remember that Ka << 1 means the equilibrium strongly favors the undissociated form.

Question 8

1. $[\text{H}_3\text{O}^+]_{\text{HA}} < [\text{H}_3\text{O}^+]_{\text{HB}}$
2. $[\text{H}_3\text{O}^+]_{\text{HA}} = [\text{H}_3\text{O}^+]_{\text{HB}}$
3. $[\text{H}_3\text{O}^+]_{\text{HA}} > [\text{H}_3\text{O}^+]_{\text{HB}}$ (correct answer)
4. $[\text{H}_3\text{O}^+]_{\text{HA}} = 0.10\,\text{M}$ and $[\text{H}_3\text{O}^+]_{\text{HB}} = 0.10\,\text{M}$
5. $[\text{H}_3\text{O}^+]_{\text{HA}} = [\text{OH}^-]_{\text{HB}}$

Explanation: This problem involves weak acid and base equilibria. Both HA and HB are weak acids that partially dissociate according to their Ka values. Since Ka(HA) = 1.0×10⁻³ is larger than Ka(HB) = 1.0×10⁻⁶, acid HA dissociates to a greater extent than HB. A larger Ka means more H₃O⁺ is produced at equilibrium, so the HA solution will have a higher [H₃O⁺] concentration. The incorrect choice B assumes equal [H₃O⁺] concentrations, which would only be true if the acids had identical Ka values. When comparing weak acids at the same initial concentration, always remember that larger Ka means greater dissociation and higher [H₃O⁺].

Question 9

1. $[\text{CH}_3\text{NH}_2] > [\text{CH}_3\text{NH}_3^+]$ (correct answer)
2. $[\text{OH}^-] = 0.10\,\text{M}$
3. $[\text{CH}_3\text{NH}_3^+] > [\text{CH}_3\text{NH}_2]$
4. $[\text{CH}_3\text{NH}_2] = 0$ because it reacts with water.
5. $[\text{H}^+] > [\text{OH}^-]$ because the solution contains an amine.

Explanation: 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.

Question 10

1. The HA solution has a higher pH because $K_a$ is larger.
2. The HA solution has a lower $[\text{H}^+]$ because $K_a$ is larger.
3. The HB solution has a higher $[\text{H}^+]$ because it is less dissociated.
4. The HA solution has a higher $[\text{H}^+]$ because its $K_a$ is larger. (correct answer)
5. Both solutions have the same $[\text{H}^+]$ because both are $0.10\,\text{M}$.

Explanation: This question tests understanding of weak acid and base equilibria. For weak acids at the same initial concentration, the acid with larger Ka dissociates more and produces higher [H+]. Since HA has Ka = 1.0 × 10^-3 (larger) and HB has Ka = 1.0 × 10^-6 (smaller), HA dissociates more extensively. This means the HA solution has higher [H+] and therefore lower pH than the HB solution. Choice A incorrectly states that higher Ka leads to higher pH, confusing the inverse relationship between [H+] and pH. For weak acid comparisons, remember: larger Ka → more dissociation → higher [H+] → lower pH.

Question 11

1. $[\text{F}^{-}] > [\text{HF}]$
2. $[\text{HF}] = [\text{F}^{-}]$
3. $[\text{H}^{+}] = 0.10\,\text{M}$
4. $[\text{HF}] > [\text{F}^{-}]$ (correct answer)
5. $[\text{F}^{-}] = 0.10\,\text{M}$

Explanation: This question tests understanding of weak acid and base equilibria. Hydrofluoric acid (HF) is a weak acid with $K_a = 6.8 \times 10^{-4}$, meaning it only partially dissociates in water according to $\text{HF} \rightleftharpoons \text{H}^{+} + \text{F}^{-}$. Since $K_a$ is relatively small (much less than 1), the equilibrium lies far to the left, meaning most HF molecules remain undissociated. At equilibrium, $[\text{HF}] > [\text{F}^{-}]$ because only a small fraction of the original 0.10 M HF dissociates. Choice B ($[\text{H}^{+}] = 0.10\,\text{M}$) is incorrect because it assumes complete dissociation, which only occurs for strong acids. For weak acids, always set up the equilibrium expression $K_a = \frac{[\text{H}^{+}][\text{F}^{-}]}{[\text{HF}]}$ and recognize that most of the acid remains in molecular form.

Question 12

1. $[\text{H}^+]_{eq}=[\text{A}^-]_{eq}$ (correct answer)
2. $[\text{H}^+]_{eq}=[\text{HA}]_{eq}$
3. $[\text{A}^-]_{eq}=[\text{HA}]_{eq}$
4. $[\text{H}^+]_{eq} > [\text{A}^-]_{eq}$ because $\text{H}^+$ is produced first
5. $[\text{A}^-]_{eq}=0$ because the acid is weak

Explanation: This question assesses understanding of weak acid and base equilibria. Weak acids such as HA undergo incomplete dissociation, producing H+ and A- in equal amounts from the ionization reaction. The equilibrium Ka = [H+][A-]/[HA] governs the concentrations, but stoichiometry ensures [H+] = [A-] neglecting water's contribution. This equality holds because each dissociated HA molecule yields one H+ and one A-. Choice C is a tempting distractor, claiming [A-]{eq} = [HA]{eq}, but this is wrong as weak acids have [HA]{eq} >> [A-]{eq}. For weak species, write the equilibrium first before thinking about pH.

Question 13

1. $[\text{H}^+]$ increases because $\text{BH}^+$ is an acid
2. $[\text{H}^+]$ remains $1.0\times10^{-7}\,\text{M}$ because the base is weak
3. $[\text{OH}^-]$ remains $1.0\times10^{-7}\,\text{M}$ because water controls $[\text{OH}^-]$
4. $[\text{OH}^-]$ decreases because $K_b < 1$
5. $[\text{OH}^-]$ increases because the reaction produces $\text{OH}^-$ (correct answer)

Explanation: 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.

Question 14

1. $K_b=\dfrac{[\text{NH}_3]}{[\text{NH}_4^+][\text{OH}^-]}$
2. $K_b=\dfrac{[\text{NH}_4^+][\text{OH}^-]}{[\text{NH}_3]}$ (correct answer)
3. $K_b=\dfrac{[\text{OH}^-]}{[\text{NH}_4^+][\text{NH}_3]}$
4. $K_b=\dfrac{[\text{NH}_4^+]}{[\text{NH}_3][\text{OH}^-]}$
5. $K_b=\dfrac{[\text{NH}_3][\text{H}_2\text{O}]}{[\text{NH}_4^+][\text{OH}^-]}$

Explanation: 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.

Question 15

1. $K_a=\dfrac{[\text{HNO}_2]}{[\text{H}^+][\text{NO}_2^-]}$
2. $K_a=\dfrac{[\text{H}^+][\text{NO}_2^-]}{[\text{HNO}_2]}$ (correct answer)
3. $K_a=\dfrac{[\text{H}^+]}{[\text{HNO}_2][\text{NO}_2^-]}$
4. $K_a=\dfrac{[\text{NO}_2^-]}{[\text{HNO}_2][\text{H}^+]}$
5. $K_a=\dfrac{[\text{HNO}_2][\text{H}_2\text{O}]}{[\text{H}^+][\text{NO}_2^-]}$

Explanation: This question assesses understanding of weak acid and base equilibria. Weak acids like HNO2 partially dissociate in water, establishing an equilibrium rather than complete ionization. The equilibrium constant Ka is defined as [H+][NO2-]/[HNO2], reflecting the extent of dissociation and controlling the concentrations of products relative to reactants. Since water is the solvent, its concentration is constant and not included in the Ka expression, ensuring the focus on the acid's ionization. A common misconception is choice E, which includes [H2O] in the numerator, but this is wrong because water's concentration is incorporated into Ka. For weak species, write the equilibrium first before thinking about pH.

Question 16

1. $K_a = \dfrac{[\text{HNO}_2]}{[\text{H}^+][\text{NO}_2^-]}$
2. $K_a = \dfrac{[\text{H}^+][\text{NO}_2^-]}{[\text{HNO}_2]}$ (correct answer)
3. $K_a = \dfrac{[\text{H}^+]}{[\text{HNO}_2][\text{NO}_2^-]}$
4. $K_a = \dfrac{[\text{NO}_2^-]}{[\text{HNO}_2][\text{H}^+]}$
5. $K_a = \dfrac{[\text{HNO}_2][\text{NO}_2^-]}{[\text{H}^+]}$

Explanation: This question assesses understanding of weak acid and base equilibria. Weak acids like HNO2 undergo incomplete dissociation in aqueous solution, establishing an equilibrium where HNO2 partially breaks into H+ and NO2- ions. The acid dissociation constant Ka is defined as the ratio of products to reactants, specifically Ka = [H+][NO2-]/[HNO2], reflecting the extent of dissociation and controlling the equilibrium concentrations. This expression ensures that the concentrations satisfy the equilibrium condition, with water not included as it's the solvent. A common misconception is that Ka = [HNO2]/[H+][NO2-], but this inverts the actual definition and would imply a constant for association rather than dissociation. For weak species, write the equilibrium first before thinking about pH.

Question 17

1. $K_b = \dfrac{[\text{BOH}]}{[\text{B}^+][\text{OH}^-]}$
2. $K_b = \dfrac{[\text{B}^+]}{[\text{BOH}][\text{OH}^-]}$
3. $K_b = \dfrac{[\text{B}^+][\text{OH}^-]}{[\text{BOH}]}$ (correct answer)
4. $K_b = \dfrac{[\text{OH}^-]}{[\text{BOH}][\text{B}^+]}$
5. $K_b = \dfrac{[\text{BOH}][\text{OH}^-]}{[\text{B}^+]}$

Explanation: 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.

Question 18

1. $[\text{OH}^-] > [\text{NH}_3]$
2. $[\text{NH}_4^+] = [\text{NH}_3]$ at equilibrium
3. $[\text{NH}_4^+] = 0.10\,\text{M}$ because NH$_3$ is a base
4. $[\text{OH}^-] < [\text{NH}_3]$ (correct answer)
5. $[\text{NH}_3] = 0$ because it reacts completely with water

Explanation: 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.

Question 19

1. $[\text{H}^+] = 0.10\,\text{M}$ because HF is an acid
2. $[\text{F}^-] < [\text{HF}]$ (correct answer)
3. $[\text{F}^-] = [\text{HF}]$ at equilibrium
4. $[\text{H}^+] > [\text{HF}]$
5. $[\text{HF}] = 0$ at equilibrium because it dissociates

Explanation: 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.

Question 20

1. $[\text{H}_3\text{O}^+]$ increases and $[\text{OH}^-]$ decreases.
2. $[\text{H}_3\text{O}^+]$ decreases and $[\text{OH}^-]$ increases. (correct answer)
3. Neither $[\text{H}_3\text{O}^+]$ nor $[\text{OH}^-]$ changes because CH$_3$NH$_2$ is weak.
4. Both $[\text{H}_3\text{O}^+]$ and $[\text{OH}^-]$ increase.
5. Both $[\text{H}_3\text{O}^+]$ and $[\text{OH}^-]$ decrease.

Explanation: This question tests understanding of weak acid and base equilibria. Methylamine (CH₃NH₂) is a weak base that accepts protons from water, producing CH₃NH₃⁺ and OH⁻ ions according to the equilibrium CH₃NH₂ + H₂O ⇌ CH₃NH₃⁺ + OH⁻. As this equilibrium establishes, [OH⁻] increases from its pure water value of 1.0×10⁻⁷ M. Since Kw = [H₃O⁺][OH⁻] must remain constant at a given temperature, an increase in [OH⁻] requires a decrease in [H₃O⁺]. A common error (choice E) is thinking weak bases don't affect ion concentrations, but even weak bases shift the water equilibrium. For weak base problems, remember that adding base increases [OH⁻] and decreases [H₃O⁺] to maintain constant Kw.