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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

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).

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).

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

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

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

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

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 (moldm3mol\,dm^{-3}).

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.

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=log10[H+(aq)]pH = -\log_{10}[H^+(aq)]

[H+(aq)]=10pH[H^+(aq)] = 10^{-pH}

Kw=[H+][OH]=1.00×1014 at 298 KK_w = [H^+][OH^-] = 1.00 \times 10^{-14} \text{ at } 298\text{ K}

pOH=log10[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 0.01moldm30.01\,mol\,dm^{-3} solution of HCl(aq)HCl(aq) at 298 K298\text{ K}.

Solution:

pH=log10(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, 0.01moldm30.01\,mol\,dm^{-3}.

Problem 2:

Compare the electrical conductivity of 0.10moldm3H2SO4(aq)0.10\,mol\,dm^{-3}\,H_2SO_4(aq) and 0.10moldm3CH3COOH(aq)0.10\,mol\,dm^{-3}\,CH_3COOH(aq).

Solution:

The conductivity of 0.10moldm3H2SO40.10\,mol\,dm^{-3}\,H_2SO_4 is significantly higher than that of 0.10moldm3CH3COOH0.10\,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 1.0×103moldm31.0 \times 10^{-3}\,mol\,dm^{-3}, what is its pHpH at 298 K298\text{ K}?

Solution:

pOH=log10(1.0×103)=3.0pOH = -\log_{10}(1.0 \times 10^{-3}) = 3.0 pH=14.03.0=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.

Strong and weak acids and bases - Revision Notes & Key Formulas | IB Grade 11 Chemistry