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Acids and Bases - Strong and weak acids and bases

Grade 11IBChemistry

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

🔑Concepts

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Strong acids and bases are defined by their complete dissociation (ionization) in aqueous solution. For example, HCl(aq)→H+(aq)+Cl−(aq)HCl(aq) \rightarrow H^+(aq) + Cl^-(aq).

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Weak acids and bases only partially dissociate in aqueous solution, establishing an equilibrium where the position of equilibrium lies far to the left. For example, CH3COOH(aq)⇌CH3COO−(aq)+H+(aq)CH_3COOH(aq) \rightleftharpoons CH_3COO^-(aq) + H^+(aq).

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Common strong acids include hydrochloric acid (HClHCl), nitric acid (HNO3HNO_3), and sulfuric acid (H2SO4H_2SO_4).

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Common weak acids include ethanoic acid (CH3COOHCH_3COOH), carbonic acid (H2CO3H_2CO_3), and phosphoric acid (H3PO4H_3PO_4).

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Common strong bases include Group 1 hydroxides like LiOHLiOH, NaOHNaOH, and KOHKOH.

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Common weak bases include ammonia (NH3NH_3) and amines like methylamine (CH3NH2CH_3NH_2).

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Strength vs. Concentration: Strength refers to the degree of dissociation, whereas concentration refers to the number of moles of acid or base dissolved per unit volume (mol dm−3mol\,dm^{-3}).

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Experimental distinction: Strong acids have higher electrical conductivity, higher rates of reaction (e.g., with metals or carbonates), and lower pHpH values compared to weak acids of the same concentration.

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The conjugate base of a strong acid is a very weak base, while the conjugate base of a weak acid is a relatively stronger base (though often still weak).

📐Formulae

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

[H+(aq)]=10−pH[H^+(aq)] = 10^{-pH}

Kw=[H+][OH−]=1.00×10−14 at 298 KK_w = [H^+][OH^-] = 1.00 \times 10^{-14} \text{ at } 298\text{ K}

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

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

For a weak acid HA:Ka=[H+][A−][HA]\text{For a weak acid } HA: K_a = \frac{[H^+][A^-]}{[HA]}

💡Examples

Problem 1:

Calculate the pHpH of a 0.01 mol dm−30.01\,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:

Since HClHCl is a strong acid, it dissociates completely. Therefore, the concentration of hydrogen ions [H+][H^+] is equal to the initial concentration of the acid, 0.01 mol dm−30.01\,mol\,dm^{-3}.

Problem 2:

Compare the electrical conductivity of 0.10 mol dm−3 H2SO4(aq)0.10\,mol\,dm^{-3}\,H_2SO_4(aq) and 0.10 mol dm−3 CH3COOH(aq)0.10\,mol\,dm^{-3}\,CH_3COOH(aq).

Solution:

The conductivity of 0.10 mol dm−3 H2SO40.10\,mol\,dm^{-3}\,H_2SO_4 is significantly higher than that of 0.10 mol dm−3 CH3COOH0.10\,mol\,dm^{-3}\,CH_3COOH.

Explanation:

Conductivity depends on the concentration of mobile ions. H2SO4H_2SO_4 is a strong diprotic acid that dissociates fully, producing a high concentration of ions. CH3COOHCH_3COOH is a weak acid and only partially dissociates, resulting in a much lower concentration of ions in solution.

Problem 3:

If a solution of NaOHNaOH has a [OH−][OH^-] of 1.0×10−3 mol dm−31.0 \times 10^{-3}\,mol\,dm^{-3}, what is its pHpH at 298 K298\text{ K}?

Solution:

pOH=−log⁡10(1.0×10−3)=3.0pOH = -\log_{10}(1.0 \times 10^{-3}) = 3.0 pH=14.0−3.0=11.0pH = 14.0 - 3.0 = 11.0

Explanation:

First, calculate the pOHpOH from the hydroxide ion concentration. Then, use the relationship pH+pOH=14pH + pOH = 14 at standard temperature to find the pHpH.