IB Chemistry Quiz: Apply Proton Transfer Reactions
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Apply Proton Transfer ReactionsQuestion 1 of 20

What is the concentration of hydrogen ions, [H⁺], in a 0.050 mol dm⁻³ solution of sodium hydroxide, NaOH, at 298 K? (Kw = 1.0 × 10⁻¹⁴ mol² dm⁻⁶)

5.0 × 10⁻² mol dm⁻³
1.0 × 10⁻⁷ mol dm⁻³
2.0 × 10⁻¹³ mol dm⁻³
5.0 × 10⁻¹³ mol dm⁻³
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IB Chemistry Quiz

IB Chemistry Quiz: Apply Proton Transfer Reactions

Practice Apply Proton Transfer Reactions in IB Chemistry with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Apply Proton Transfer Reactions, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Chemistry.

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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.

All questions

Question 1

What is the concentration of hydrogen ions, [H⁺], in a 0.050 mol dm⁻³ solution of sodium hydroxide, NaOH, at 298 K? (Kw = 1.0 × 10⁻¹⁴ mol² dm⁻⁶)

  1. 5.0 × 10⁻² mol dm⁻³
  2. 1.0 × 10⁻⁷ mol dm⁻³
  3. 2.0 × 10⁻¹³ mol dm⁻³ (correct answer)
  4. 5.0 × 10⁻¹³ mol dm⁻³
Explanation: Sodium hydroxide is a strong base, so it dissociates completely: [OH⁻] = 0.050 mol dm⁻³. The ionic product constant for water, Kw = [H⁺][OH⁻], can be used to find [H⁺]. Rearranging the formula, [H⁺] = Kw / [OH⁻] = (1.0 × 10⁻¹⁴) / (0.050) = 2.0 × 10⁻¹³ mol dm⁻³.

Question 2

A solution is prepared by dissolving sodium ethanoate (CH₃COONa) in water. What is the net ionic equation for the reaction that occurs and how does it affect the pH?

  1. CH₃COO⁻(aq) + H₂O(l) ⇌ CH₃COOH(aq) + OH⁻(aq); pH increases. (correct answer)
  2. Na⁺(aq) + H₂O(l) ⇌ NaOH(aq) + H⁺(aq); pH decreases.
  3. CH₃COONa(s) ⇌ Na⁺(aq) + CH₃COO⁻(aq); pH remains neutral.
  4. CH₃COO⁻(aq) + H⁺(aq) ⇌ CH₃COOH(aq); pH increases.
Explanation: Sodium ethanoate is the salt of a strong base (NaOH) and a weak acid (CH₃COOH). The Na⁺ ion is a spectator ion. The ethanoate ion, CH₃COO⁻, is the conjugate base of a weak acid and will act as a Brønsted-Lowry base, accepting a proton from water. This process is called hydrolysis. The reaction is CH₃COO⁻(aq) + H₂O(l) ⇌ CH₃COOH(aq) + OH⁻(aq). The production of hydroxide ions (OH⁻) makes the solution basic, thus increasing the pH above 7.

Question 3

A titration is performed by adding a strong base to a strong acid. Which combination of observations is most consistent with this type of titration?

  1. The initial pH is around 1 and the pH at the equivalence point is 7. (correct answer)
  2. The initial pH is around 1 and the pH at the equivalence point is greater than 7.
  3. The initial pH is around 5 and the pH at the equivalence point is 7.
  4. The initial pH is around 13 and the pH at the equivalence point is 7.
Explanation: A strong acid will have a very low initial pH, typically around 1 for a concentration like 0.1 mol dm⁻³. When a strong acid is neutralized by a strong base, the salt formed (e.g., NaCl from HCl + NaOH) does not hydrolyze water, so the pH at the equivalence point is neutral, which is 7 at 298 K. The other options describe titrations involving weak acids/bases or are reversed (base in flask).

Question 4

Which of the following represents the balanced net ionic equation for the reaction between aqueous nitric acid and solid calcium carbonate?

  1. 2H⁺(aq) + CaCO₃(s) → Ca²⁺(aq) + H₂O(l) + CO₂(g) (correct answer)
  2. H⁺(aq) + CO₃²⁻(aq) → HCO₃⁻(aq)
  3. 2HNO₃(aq) + CaCO₃(s) → Ca(NO₃)₂(aq) + H₂CO₃(aq)
  4. 2H⁺(aq) + CO₃²⁻(aq) → H₂O(l) + CO₂(g)
Explanation: The full molecular equation is 2HNO₃(aq) + CaCO₃(s) → Ca(NO₃)₂(aq) + H₂O(l) + CO₂(g). To get the net ionic equation, we write the ions for the strong acid and the soluble salt: 2H⁺(aq) + 2NO₃⁻(aq) + CaCO₃(s) → Ca²⁺(aq) + 2NO₃⁻(aq) + H₂O(l) + CO₂(g). The nitrate ions (NO₃⁻) are spectator ions and are cancelled out. The remaining species give the net ionic equation: 2H⁺(aq) + CaCO₃(s) → Ca²⁺(aq) + H₂O(l) + CO₂(g).

Question 5

A student prepares a 250.0 cm³ solution of 0.100 mol dm⁻³ hydrochloric acid. They then take a 25.00 cm³ sample of this solution and dilute it with deionized water to a final volume of 500.0 cm³. What is the pH of the final diluted solution?

  1. 1.00
  2. 2.30 (correct answer)
  3. 3.00
  4. 3.30
Explanation: First, calculate the moles of HCl in the 25.00 cm³ sample: n = C × V = 0.100 mol dm⁻³ × 0.02500 dm³ = 0.00250 mol. Next, calculate the concentration of the final solution: C = n / V = 0.00250 mol / 0.5000 dm³ = 0.00500 mol dm⁻³. Since HCl is a strong acid, [H⁺] = 0.00500 mol dm⁻³. Finally, calculate the pH: pH = -log₁₀[H⁺] = -log₁₀(0.00500) ≈ 2.30.

Question 6

If the hydrogen ion concentration, [H⁺], in a solution is increased by a factor of 1000, how does the pH value of the solution change?

  1. It increases by a factor of 1000.
  2. It decreases by a factor of 1000.
  3. It increases by 3.
  4. It decreases by 3. (correct answer)
Explanation: The pH scale is logarithmic. pH = -log₁₀[H⁺]. Let the initial concentration be [H⁺]₁ and the final be [H⁺]₂ = 1000 × [H⁺]₁. The change in pH is ΔpH = pH₂ - pH₁ = -log([H⁺]₂) - (-log([H⁺]₁)) = -log(1000 × [H⁺]₁) + log([H⁺]₁) = -(log(1000) + log([H⁺]₁)) + log([H⁺]₁) = -log(1000) - log([H⁺]₁) + log([H⁺]₁) = -log(10³) = -3. Therefore, the pH decreases by 3.

Question 7

A buffer solution is prepared by mixing equal volumes of 0.20 mol dm⁻³ methanoic acid (HCOOH) and 0.30 mol dm⁻³ sodium methanoate (HCOONa). What is the pH of the resulting buffer? (pKa for HCOOH = 3.75)

  1. 3.57
  2. 3.75
  3. 3.93 (correct answer)
  4. 4.12
Explanation: The Henderson-Hasselbalch equation is pH = pKa + log([A⁻]/[HA]). When mixing equal volumes, the concentrations will both halve, but their ratio remains the same. The ratio of concentrations is [HCOONa]/[HCOOH] = 0.30 / 0.20 = 1.5. So, pH = 3.75 + log(1.5) = 3.75 + 0.176 = 3.926 ≈ 3.93.

Question 8

A weak acid is titrated with a strong base, and the pH is monitored. The equivalence point is found to occur at pH = 8.8. Which indicator would be the most suitable to determine the end point of this titration?

  1. Methyl orange (pKa = 3.7)
  2. Bromothymol blue (pKa = 7.1)
  3. Phenolphthalein (pKa = 9.6) (correct answer)
  4. Alizarin yellow (pKa = 11.2)
Explanation: A suitable indicator for a titration has a pKa value close to the pH at the equivalence point. The indicator's color change range (approximately pKa ± 1) should bracket the equivalence point pH. The equivalence point is at pH 8.8. Of the choices, phenolphthalein (pKa = 9.6) is the closest, and its color change range (approx. 8.6-10.6) effectively covers the equivalence point. The other indicators would change color too early or too late.

Question 9

Consider the following equilibrium reaction involving the hydrogen phosphate ion: HPO₄²⁻(aq) + H₂O(l) ⇌ H₃O⁺(aq) + PO₄³⁻(aq). Which statement correctly identifies a Brønsted-Lowry acid and its conjugate base in this system?

  1. HPO₄²⁻ is an acid and its conjugate base is PO₄³⁻. (correct answer)
  2. H₂O is an acid and its conjugate base is PO₄³⁻.
  3. HPO₄²⁻ is an acid and its conjugate base is H₃O⁺.
  4. H₂O is an acid and its conjugate base is HPO₄²⁻.
Explanation: According to the Brønsted-Lowry theory, an acid is a proton (H⁺) donor. In the forward reaction, HPO₄²⁻ donates a proton to H₂O to become PO₄³⁻. Therefore, HPO₄²⁻ is the acid and PO₄³⁻ is its conjugate base, as they differ by one proton.

Question 10

The base dissociation constant, Kb, for ammonia (NH₃) is 1.8 × 10⁻⁵. What is the acid dissociation constant, Ka, for the ammonium ion (NH₄⁺) at 298 K? (Kw = 1.0 × 10⁻¹⁴)

  1. 1.8 × 10⁻⁹
  2. 5.6 × 10⁻¹⁰ (correct answer)
  3. 1.8 × 10⁻⁵
  4. 5.6 × 10⁻⁴
Explanation: For a conjugate acid-base pair, the relationship Ka × Kb = Kw holds. The ammonium ion, NH₄⁺, is the conjugate acid of the base ammonia, NH₃. Therefore, Ka(NH₄⁺) = Kw / Kb(NH₃) = (1.0 × 10⁻¹⁴) / (1.8 × 10⁻⁵) ≈ 5.6 × 10⁻¹⁰.

Question 11

Two buffer solutions, X and Y, are prepared using ethanoic acid and sodium ethanoate. Buffer X contains 0.10 mol dm⁻³ of each component. Buffer Y contains 0.50 mol dm⁻³ of each component. A small amount of strong acid is added to both buffers. Which statement correctly describes the result?

  1. The pH of buffer X changes more than the pH of buffer Y. (correct answer)
  2. The pH of buffer Y changes more than the pH of buffer X.
  3. The pH of both buffers changes by the same amount.
  4. The initial pH of buffer Y is significantly higher than that of buffer X.
Explanation: Both buffers have a [A⁻]/[HA] ratio of 1, so their initial pH will be the same (pH = pKa). However, their buffer capacities are different. Buffer Y has higher concentrations of the acid and conjugate base, meaning it can neutralize more added acid or base before its pH changes significantly. Buffer X has a lower capacity. Therefore, upon adding the same amount of strong acid, the pH of the more dilute buffer (X) will change more than the pH of the more concentrated buffer (Y).

Question 12

A student prepares a buffer solution by mixing 0.10 mol of ammonia (NH3NH_3) with 0.080 mol of ammonium chloride (NH4ClNH_4Cl) in 1.0 L of solution. If 0.020 mol of solid sodium hydroxide (NaOHNaOH) is then added to this buffer, what is the approximate pH of the resulting solution? (KbK_b for NH3NH_3 = 1.8×10−51.8 \times 10^{-5})

  1. 9.1
  2. 9.4
  3. 9.7 (correct answer)
  4. 10.0
Explanation: Initially: [NH₃] = 0.10 M, [NH₄⁺] = 0.080 M. When NaOH is added, it reacts with NH₄⁺: NH₄⁺ + OH⁻ → NH₃ + H₂O. After reaction: [NH₃] = 0.10 + 0.020 = 0.12 M, [NH₄⁺] = 0.080 - 0.020 = 0.060 M. Using Henderson-Hasselbalch: pOH = pKb + log([NH₄⁺]/[NH₃]) = 4.74 + log(0.060/0.12) = 4.74 - 0.30 = 4.44. Therefore pH = 14 - 4.44 = 9.56 ≈ 9.7.

Question 13

In aqueous solution, the diprotic acid H2AH_2A undergoes the following equilibria: H2A⇌H++HA−H_2A \rightleftharpoons H^+ + HA^- (Ka1=1.0×10−3K_{a1} = 1.0 \times 10^{-3}) and HA−⇌H++A2−HA^- \rightleftharpoons H^+ + A^{2-} (Ka2=1.0×10−8K_{a2} = 1.0 \times 10^{-8}). At what pH will the concentration of HA−HA^- be maximized?

  1. 3.0
  2. 5.5 (correct answer)
  3. 8.0
  4. 11.0
Explanation: The concentration of HA⁻ is maximized when d[HA⁻]/d[H⁺] = 0. This occurs at the geometric mean of the two Ka values: pH = ½(pKa1 + pKa2) = ½(3.0 + 8.0) = 5.5. At this pH, the rates of formation (from H₂A) and consumption (to A²⁻) of HA⁻ are balanced optimally.

Question 14

The indicator methyl orange changes color in the pH range 3.1-4.4, with the transition occurring according to: HIn(aq)⇌H+(aq)+In−(aq)HIn(aq) \rightleftharpoons H^+(aq) + In^-(aq) where HInHIn is red and In−In^- is yellow. In a solution where the indicator appears orange (intermediate color), what is the approximate ratio [In−][HIn]\frac{[In^-]}{[HIn]}?

  1. 0.1
  2. 1.0 (correct answer)
  3. 3.2
  4. 10.0
Explanation: When the indicator appears orange (intermediate color), it's at the midpoint of its transition range. The midpoint pH = (3.1 + 4.4)/2 = 3.75. At this pH, which equals the pKa of the indicator, [HIn] = [In⁻] according to the Henderson-Hasselbalch equation, so the ratio [In⁻]/[HIn] = 1.0. This is when we see the intermediate orange color.

Question 15

A solution contains 0.10 M H3PO4H_3PO_4 (phosphoric acid). Given the stepwise dissociation constants Ka1=7.5×10−3K_{a1} = 7.5 \times 10^{-3}, Ka2=6.2×10−8K_{a2} = 6.2 \times 10^{-8}, and Ka3=4.8×10−13K_{a3} = 4.8 \times 10^{-13}, which statement best describes the relative concentrations of phosphate species at equilibrium?

  1. [H3PO4]>[H2PO4−]>[HPO42−]>[PO43−][H_3PO_4] > [H_2PO_4^-] > [HPO_4^{2-}] > [PO_4^{3-}] with significant amounts of all four species
  2. [H3PO4]>>[H2PO4−]>>[HPO42−]>>[PO43−][H_3PO_4] >> [H_2PO_4^-] >> [HPO_4^{2-}] >> [PO_4^{3-}] with minimal ionization overall
  3. [H2PO4−]>[H3PO4]>>[HPO42−]>>[PO43−][H_2PO_4^-] > [H_3PO_4] >> [HPO_4^{2-}] >> [PO_4^{3-}] due to first ionization dominance
  4. [H3PO4]≈[H2PO4−]>>[HPO42−]>[PO43−][H_3PO_4] ≈ [H_2PO_4^-] >> [HPO_4^{2-}] > [PO_4^{3-}] with negligible PO43−PO_4^{3-} (correct answer)
Explanation: When you encounter polyprotic acids like phosphoric acid, you need to analyze each dissociation step separately and compare the relative magnitudes of the KaK_a values to predict species concentrations. For H3PO4H_3PO_4, the first dissociation has Ka1=7.5×10−3K_{a1} = 7.5 \times 10^{-3}, which is relatively large, meaning significant ionization occurs: H3PO4⇌H++H2PO4−H_3PO_4 \rightleftharpoons H^+ + H_2PO_4^-. With a 0.10 M starting concentration and this KaK_a value, you can expect substantial conversion to H2PO4−H_2PO_4^-, making these two species similar in concentration. The second dissociation has Ka2=6.2×10−8K_{a2} = 6.2 \times 10^{-8}, which is much smaller than Ka1K_{a1}. This means H2PO4−H_2PO_4^- ionizes much less readily than H3PO4H_3PO_4, so [HPO42−][HPO_4^{2-}] will be significantly lower than both [H3PO4][H_3PO_4] and [H2PO4−][H_2PO_4^-]. The third dissociation has Ka3=4.8×10−13K_{a3} = 4.8 \times 10^{-13}, which is extremely small, making [PO43−][PO_4^{3-}] negligible. Answer A incorrectly suggests all species exist in significant amounts. Answer B underestimates the first ionization—Ka1K_{a1} is too large for minimal ionization. Answer C overestimates the dominance of H2PO4−H_2PO_4^- over H3PO4H_3PO_4. Answer D correctly shows that H3PO4H_3PO_4 and H2PO4−H_2PO_4^- are comparable, with much lower concentrations of subsequent species. Remember: compare KaK_a values directly—each decrease of about 5 orders of magnitude means dramatically lower concentrations of more deprotonated species.

Question 16

A student prepares a solution by dissolving 0.10 mol of sodium acetate (CH3COONaCH_3COONa) in 1.0 L of water, then adds 0.050 mol of solid acetic acid (CH3COOHCH_3COOH). If the KaK_a for acetic acid is 1.8×10−51.8 \times 10^{-5}, which equation correctly represents the equilibrium expression needed to find the final pH?

  1. Ka=[H+][CH3COO−][CH3COOH]=x(0.10+x)(0.050−x)K_a = \frac{[H^+][CH_3COO^-]}{[CH_3COOH]} = \frac{x(0.10+x)}{(0.050-x)}
  2. Ka=[H+][CH3COO−][CH3COOH]=x(0.10−x)(0.050+x)K_a = \frac{[H^+][CH_3COO^-]}{[CH_3COOH]} = \frac{x(0.10-x)}{(0.050+x)}
  3. Ka=[H+][CH3COO−][CH3COOH]=x(0.10)(0.050)K_a = \frac{[H^+][CH_3COO^-]}{[CH_3COOH]} = \frac{x(0.10)}{(0.050)} (correct answer)
  4. Kw=[H+][OH−]1=x⋅Kw/x1=KwK_w = \frac{[H^+][OH^-]}{1} = \frac{x \cdot K_w/x}{1} = K_w
Explanation: When you encounter a buffer system problem, you're dealing with a weak acid and its conjugate base in equilibrium. This question tests your understanding of how to set up the equilibrium expression when additional weak acid is added to an existing acetate solution. The sodium acetate completely dissociates to give 0.10 M acetate ions (CH3COO−CH_3COO^-), while the added acetic acid establishes the equilibrium: CH3COOH⇌H++CH3COO−CH_3COOH \rightleftharpoons H^+ + CH_3COO^-. Since this is a buffer with relatively high concentrations compared to the expected H+H^+ concentration, you can apply the simplifying assumption that xx (the amount of acid that dissociates) is negligible compared to the initial concentrations. Answer C correctly uses this approximation: Ka=x(0.10)(0.050)K_a = \frac{x(0.10)}{(0.050)}, where x=[H+]x = [H^+], 0.10 represents the initial acetate concentration, and 0.050 represents the initial acetic acid concentration. Answer A incorrectly shows (0.10+x)(0.10+x) and (0.050−x)(0.050-x), suggesting the equilibrium shifts to consume acetic acid and produce more acetate, but the signs are wrong for the established equilibrium direction. Answer B has the opposite error with (0.10−x)(0.10-x) and (0.050+x)(0.050+x), which would apply if we were adding base instead of acid. Answer D completely misapplies the water equilibrium expression, which isn't relevant for calculating pH in a buffer system. Study tip: In buffer problems, when concentrations are much larger than KaK_a, the "x is negligible" approximation almost always applies, simplifying your equilibrium expression significantly.

Question 17

In a titration of 20.0 mL of 0.100 M CH3NH2CH_3NH_2 (methylamine, Kb=4.4×10−4K_b = 4.4 \times 10^{-4}) with 0.100 M HClHCl, at what volume of added HClHCl will the pH equal the pKapK_a of the CH3NH3+CH_3NH_3^+ ion?

  1. 10.0 mL (correct answer)
  2. 15.0 mL
  3. 20.0 mL
  4. 25.0 mL
Explanation: The pH equals pKa when [base] = [conjugate acid], which occurs at the half-equivalence point. Initial moles of CH₃NH₂ = 0.020 L × 0.100 M = 2.0 × 10⁻³ mol. At half-equivalence, 1.0 × 10⁻³ mol of HCl is needed: Volume = (1.0 × 10⁻³ mol)/(0.100 M) = 0.010 L = 10.0 mL. At this point, [CH₃NH₂] = [CH₃NH₃⁺] and pH = pKa.

Question 18

Consider the reaction: HSO4−(aq)+HPO42−(aq)⇌SO42−(aq)+H2PO4−(aq)HSO_4^-(aq) + HPO_4^{2-}(aq) \rightleftharpoons SO_4^{2-}(aq) + H_2PO_4^-(aq). Given that Ka2K_{a2} for H2SO4H_2SO_4 is 1.2×10−21.2 \times 10^{-2} and Ka2K_{a2} for H3PO4H_3PO_4 is 6.2×10−86.2 \times 10^{-8}, what is the equilibrium constant for this proton transfer reaction?

  1. 1.9×1051.9 \times 10^{5} (correct answer)
  2. 5.2×10−65.2 \times 10^{-6}
  3. 1.9×10−51.9 \times 10^{-5}
  4. 7.4×10−107.4 \times 10^{-10}
Explanation: For the reaction HSO₄⁻ + HPO₄²⁻ → SO₄²⁻ + H₂PO₄⁻, the equilibrium constant K = Ka2(H₂SO₄)/Ka2(H₃PO₄) = (1.2 × 10⁻²)/(6.2 × 10⁻⁸) = 1.9 × 10⁵. This makes sense because HSO₄⁻ is a much stronger acid than H₂PO₄⁻, so the equilibrium strongly favors the products.

Question 19

A biochemist studies the amino acid glycine, which has two ionizable groups: −COOH-COOH (pKa1=2.3pK_{a1} = 2.3) and −NH3+-NH_3^+ (pKa2=9.6pK_{a2} = 9.6). At what pH will glycine exist predominantly as a zwitterion (+NH3CH2COO−^+NH_3CH_2COO^-)?

  1. pH 1.0-3.0
  2. pH 11.5-13.0
  3. pH 8.0-11.0
  4. pH 4.5-7.5 (correct answer)
Explanation: When you encounter amino acid ionization problems, focus on the relationship between pH and pK_a values to determine which functional groups are protonated or deprotonated. A zwitterion exists when the amino acid carries both a positive and negative charge simultaneously. For glycine, this means the carboxyl group (−COOH-COOH) must be deprotonated to −COO−-COO^- while the amino group remains protonated as −NH3+-NH_3^+. The carboxyl group deprotonates when pH > pK_{a1} (2.3), and the amino group remains protonated when pH < pK_{a2} (9.6). Therefore, glycine exists as a zwitterion when the pH falls between these two pK_a values: 2.3 < pH < 9.6. Answer choice D (pH 4.5-7.5) is the only range that falls entirely within this window. Answer A (pH 1.0-3.0) is incorrect because at these low pH values, both groups remain protonated, giving glycine a net positive charge (+NH3CH2COOH^+NH_3CH_2COOH). Answer B (pH 11.5-13.0) is wrong because at such high pH, both groups are deprotonated, creating a net negative charge (NH2CH2COO−NH_2CH_2COO^-). Answer C (pH 8.0-11.0) is incorrect because while it includes some zwitterionic range, it extends beyond pK_{a2} = 9.6, where the amino group begins deprotonating. Remember: zwitterions exist between the two pK_a values of diprotic amino acids. Always identify which functional groups should be charged at the given pH by comparing to their respective pK_a values.

Question 20

An aqueous solution of ammonium nitrate, NH₄NO₃, is tested with litmus paper. Which of the following correctly describes the observation and the ion responsible?

  1. The solution is basic because the NO₃⁻ ion hydrolyzes water.
  2. The solution is acidic because the NH₄⁺ ion hydrolyzes water. (correct answer)
  3. The solution is neutral because both ions hydrolyze water equally.
  4. The solution is neutral because neither ion hydrolyzes water.
Explanation: Ammonium nitrate is the salt of a weak base (NH₃) and a strong acid (HNO₃). The nitrate ion (NO₃⁻) is the conjugate base of a strong acid and does not hydrolyze. The ammonium ion (NH₄⁺) is the conjugate acid of a weak base and will hydrolyze water according to the equation: NH₄⁺(aq) + H₂O(l) ⇌ H₃O⁺(aq) + NH₃(aq). The production of H₃O⁺ ions makes the solution acidic.