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Acids and Bases - Properties of acids and bases

Grade 12IBChemistry

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

🔑Concepts

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Brønsted-Lowry Theory: An acid is a proton (H+H^+) donor and a base is a proton (H+H^+) acceptor.

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Conjugate Acid-Base Pairs: These are two species that differ by a single proton (H+H^+). For every Brønsted-Lowry acid HAHA, there is a conjugate base A−A^-.

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Lewis Theory: A Lewis acid is an electron-pair acceptor (e.g., BF3BF_3, AlCl3AlCl_3), and a Lewis base is an electron-pair donor (e.g., NH3NH_3, H2OH_2O). All Brønsted-Lowry acids are Lewis acids, but not all Lewis acids are Brønsted-Lowry acids.

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Amphiprotic Species: Species that can act as both a Brønsted-Lowry acid and base, such as H2OH_2O, HCO3−HCO_3^-, and H2PO4−H_2PO_4^-.

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Amphoteric Species: A broader term for substances that can react as both an acid and a base (includes Lewis acid/base behavior).

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Properties of Acids: Acids have a pH<7pH < 7 at 298 K298\text{ K}, taste sour, and react with reactive metals to produce H2(g)H_2(g), carbonates to produce CO2(g)CO_2(g), and bases to produce a salt and water.

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Properties of Bases: Bases have a pH>7pH > 7 at 298 K298\text{ K}, taste bitter, and feel slippery/soapy. Alkalis are bases that are soluble in water.

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Neutralization: An exothermic reaction between an acid and a base. The ionic equation for the neutralization of a strong acid and strong base is: H+(aq)+OH−(aq)→H2O(l)H^+(aq) + OH^-(aq) \rightarrow H_2O(l).

📐Formulae

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

[H+]=10−pH[H^+] = 10^{-pH}

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

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

💡Examples

Problem 1:

Identify the conjugate acid-base pairs in the following reaction: CH3COOH(aq)+NH3(aq)⇌CH3COO−(aq)+NH4+(aq)CH_3COOH(aq) + NH_3(aq) \rightleftharpoons CH_3COO^-(aq) + NH_4^+(aq)

Solution:

Pair 1: CH3COOHCH_3COOH (acid) and CH3COO−CH_3COO^- (conjugate base). Pair 2: NH3NH_3 (base) and NH4+NH_4^+ (conjugate acid).

Explanation:

CH3COOHCH_3COOH donates a proton to become CH3COO−CH_3COO^-, making it the acid. NH3NH_3 accepts a proton to become NH4+NH_4^+, making it the base.

Problem 2:

Calculate the pHpH of a 0.01 mol dm−30.01\text{ mol dm}^{-3} solution of HCl(aq)HCl(aq) at 298 K298\text{ K}.

Solution:

pH=−log⁡10(0.01)=2.0pH = -\log_{10}(0.01) = 2.0

Explanation:

HClHCl is a strong acid and dissociates completely: HCl(aq)→H+(aq)+Cl−(aq)HCl(aq) \rightarrow H^+(aq) + Cl^-(aq). Therefore, [H+]=[HCl]=0.01 mol dm−3[H^+] = [HCl] = 0.01\text{ mol dm}^{-3}.

Problem 3:

Write the balanced chemical equation for the reaction between sulfuric acid (H2SO4H_2SO_4) and sodium hydrogen carbonate (NaHCO3NaHCO_3).

Solution:

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

Explanation:

Acids react with metal hydrogen carbonates to produce a salt, carbon dioxide gas, and water.

Properties of acids and bases - Revision Notes & Key Formulas | IB Grade 12 Chemistry