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Acids and Bases - Properties of acids and bases

Grade 11IBChemistry

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

🔑Concepts

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The Brønsted-Lowry theory defines an acid as a proton (H+H^+) donor and a base as 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 example, in the reaction of NH3NH_3 with H2OH_2O, NH4+NH_4^+ is the conjugate acid of the base NH3NH_3.

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Amphiprotic species can act as both Brønsted-Lowry acids and bases. Examples include H2OH_2O, HCO3−HCO_3^-, and H2PO4−H_2PO_4^-.

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Acids react with reactive metals (e.g., MgMg, ZnZn) to produce a salt and hydrogen gas: Acid+Metal→Salt+H2(g)Acid + Metal \rightarrow Salt + H_2(g).

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Acids react with bases (metal oxides and hydroxides) in neutralization reactions to produce a salt and water: Acid+Base→Salt+H2O(l)Acid + Base \rightarrow Salt + H_2O(l).

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Acids react with carbonates and hydrogencarbonates to produce a salt, water, and carbon dioxide gas: Acid+Carbonate→Salt+H2O(l)+CO2(g)Acid + Carbonate \rightarrow Salt + H_2O(l) + CO_2(g).

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The pHpH scale is a logarithmic representation of the concentration of hydrogen ions in a solution: pH=−log⁡10[H+]pH = -\log_{10}[H^+].

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Universal indicator and pHpH probes are used to measure the pHpH of a solution, where pH<7pH < 7 is acidic, pH=7pH = 7 is neutral, and pH>7pH > 7 is alkaline at 298 K298\text{ K}.

📐Formulae

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

[H+(aq)]=10−pH[H^+(aq)] = 10^{-pH}

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

pH+pOH=14.00 at 298 KpH + pOH = 14.00 \text{ at } 298\text{ K}

💡Examples

Problem 1:

Identify the conjugate acid-base pairs in the following equilibrium reaction: H2PO4−(aq)+OH−(aq)⇌HPO42−(aq)+H2O(l)H_2PO_4^-(aq) + OH^-(aq) \rightleftharpoons HPO_4^{2-}(aq) + H_2O(l).

Solution:

Pair 1: H2PO4−H_2PO_4^- (acid) and HPO42−HPO_4^{2-} (conjugate base). Pair 2: OH−OH^- (base) and H2OH_2O (conjugate acid).

Explanation:

H2PO4−H_2PO_4^- donates a proton to become HPO42−HPO_4^{2-}, making it the acid. OH−OH^- accepts a proton to become H2OH_2O, making it the base.

Problem 2:

Calculate the pHpH of a solution with a hydrogen ion concentration [H+][H^+] of 2.5×10−4 mol dm−32.5 \times 10^{-4} \text{ mol dm}^{-3}.

Solution:

pH=−log⁡10(2.5×10−4)≈3.60pH = -\log_{10}(2.5 \times 10^{-4}) \approx 3.60

Explanation:

The pHpH is calculated using the negative logarithm of the molar concentration of H+H^+ ions.

Problem 3:

Write the balanced molecular equation for the reaction between sulfuric acid (H2SO4H_2SO_4) and sodium hydrogencarbonate (NaHCO3NaHCO_3).

Solution:

H2SO4(aq)+2NaHCO3(aq)→Na2SO4(aq)+2H2O(l)+2CO2(g)H_2SO_4(aq) + 2NaHCO_3(aq) \rightarrow Na_2SO_4(aq) + 2H_2O(l) + 2CO_2(g)

Explanation:

Acids react with hydrogencarbonates to form a salt, water, and carbon dioxide gas. The coefficients are adjusted to balance the atoms on both sides.

Properties of acids and bases - Revision Notes & Key Formulas | IB Grade 11 Chemistry