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Reactivity 3. What are the mechanisms of chemical change? - Proton transfer reactions

Grade 11IBChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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A Brønsted–Lowry acid is defined as a proton (H+H^+) donor, while a Brønsted–Lowry base is a proton (H+H^+) acceptor.

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A conjugate acid-base pair consists of two species that differ by a single proton (H+H^+). For every acid HAHA, there is a conjugate base A−A^-.

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Amphiprotic species, such as H2OH_2O, HCO3−HCO_3^-, and HSO4−HSO_4^-, can act as both a Brønsted–Lowry acid and a Brønsted–Lowry base.

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The pHpH scale is a logarithmic measure of the concentration of hydrogen ions in a solution: pH=−log⁡10[H+]pH = -\log_{10}[H^+]. A change of one pHpH unit corresponds to a ten-fold change in [H+][H^+].

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The ionic product constant for water, KwK_w, is given by [H+][OH−][H^+][OH^-]. At 298 K298\text{ K}, Kw=1.0×10−14K_w = 1.0 \times 10^{-14}, which implies pH+pOH=14.00pH + pOH = 14.00.

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Strong acids and bases ionize completely in dilute aqueous solutions (e.g., HClHCl, NaOHNaOH), whereas weak acids and bases (e.g., CH3COOHCH_3COOH, NH3NH_3) only partially ionize, establishing an equilibrium.

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Neutralization reactions are exothermic reactions between an acid and a base to produce a salt and water: acid+base→salt+wateracid + base \rightarrow salt + water.

📐Formulae

pH=−log⁡10[H+(aq)]pH = -\log_{10}[H^+(aq)]

[H+(aq)]=10−pH[H^+(aq)] = 10^{-pH}

pOH=−log⁡10[OH−(aq)]pOH = -\log_{10}[OH^-(aq)]

Kw=[H+][OH−]=1.0×10−14 (at 298 K)K_w = [H^+][OH^-] = 1.0 \times 10^{-14} \text{ (at } 298\text{ K)}

pH+pOH=pKw=14.00 (at 298 K)pH + pOH = pK_w = 14.00 \text{ (at } 298\text{ K)}

Ka=[A−][H3O+][HA]K_a = \frac{[A^-][H_3O^+]}{[HA]}

Kb=[BH+][OH−][B]K_b = \frac{[BH^+][OH^-]}{[B]}

💡Examples

Problem 1:

Identify the conjugate acid-base pairs in the following reaction: NH3(aq)+H2O(l)⇌NH4+(aq)+OH−(aq)NH_3(aq) + H_2O(l) \rightleftharpoons NH_4^+(aq) + OH^-(aq) Calculate the pHpH of a 0.01 mol dm−30.01\text{ mol dm}^{-3} solution of HClHCl.

Solution:

  1. Conjugate pairs: Pair 1: NH3NH_3 (base) and NH4+NH_4^+ (conjugate acid). Pair 2: H2OH_2O (acid) and OH−OH^- (conjugate base).
  2. For HClHCl (a strong acid): [H+]=[HCl]=0.01 mol dm−3=10−2 mol dm−3[H^+] = [HCl] = 0.01\text{ mol dm}^{-3} = 10^{-2}\text{ mol dm}^{-3}. pH=−log⁡10(10−2)=2.0pH = -\log_{10}(10^{-2}) = 2.0

Explanation:

In the reaction, NH3NH_3 accepts a proton to become NH4+NH_4^+, and H2OH_2O donates a proton to become OH−OH^-. Since HClHCl is a strong monoprotic acid, it dissociates fully, making the hydrogen ion concentration equal to the initial concentration of the acid.

Problem 2:

A solution has a hydroxide ion concentration [OH−][OH^-] of 5.0×10−4 mol dm−35.0 \times 10^{-4}\text{ mol dm}^{-3} at 298 K298\text{ K}. Calculate the pHpH of the solution.

Solution:

First, calculate the pOHpOH: pOH=−log⁡10(5.0×10−4)≈3.30pOH = -\log_{10}(5.0 \times 10^{-4}) \approx 3.30 Using the relationship at 298 K298\text{ K}: pH+pOH=14.00pH + pOH = 14.00 pH=14.00−3.30=10.70pH = 14.00 - 3.30 = 10.70

Explanation:

To find the pHpH from the [OH−][OH^-], we first find the pOHpOH using the logarithmic formula. Subtracting the pOHpOH from 1414 (the pKwpK_w at room temperature) gives the pHpH of the basic solution.