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Acids and Bases - pH curves (HL only)

Grade 12IBChemistry

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

πŸ”‘Concepts

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The equivalence point in a titration is the point where the amount of titrant added is stoichiometrically equal to the amount of analyte present. The pHpH at this point depends on the nature of the salts formed.

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Strong Acid (SA) - Strong Base (SB) titration: The pHpH at the equivalence point is exactly 7.07.0 (at 298K298K) because the resulting salt does not undergo hydrolysis.

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Weak Acid (WA) - Strong Base (SB) titration: The pHpH at the equivalence point is >7> 7 because the conjugate base (Aβˆ’A^-) reacts with water (hydrolysis): Aβˆ’(aq)+H2O(l)ightleftharpoonsHA(aq)+OHβˆ’(aq)A^-(aq) + H_2O(l) ightleftharpoons HA(aq) + OH^-(aq).

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Strong Acid (SA) - Weak Base (WB) titration: The pHpH at the equivalence point is <7< 7 because the conjugate acid (BH+BH^+) reacts with water: BH+(aq)+H2O(l)ightleftharpoonsB(aq)+H3O+(aq)BH^+(aq) + H_2O(l) ightleftharpoons B(aq) + H_3O^+(aq).

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The half-equivalence point is reached when half the volume of titrant required for equivalence has been added. At this point, for a weak acid titration, [HA]=[Aβˆ’][HA] = [A^-], and therefore pH=pKapH = pK_a.

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The buffer region is the range of the pHpH curve (usually for WA-SB or SA-WB) where the pHpH changes very slowly despite the addition of titrant. This occurs when both the weak species and its salt are present in significant concentrations.

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Acid-Base indicators are weak acids or bases that change color over a specific pHpH range. For an indicator to be effective, its pKinpK_{in} should be as close as possible to the pHpH at the equivalence point, or fall within the steep vertical section of the curve.

πŸ“Formulae

pH=βˆ’log⁑10[H+]pH = -\log_{10}[H^+]

pOH=βˆ’log⁑10[OHβˆ’]pOH = -\log_{10}[OH^-]

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

pH=pKa+log⁑10([Aβˆ’][HA])pH = pK_a + \log_{10}\left(\frac{[A^-]}{[HA]}\right) (Henderson-Hasselbalch Equation)

pKa=βˆ’log⁑10KapK_a = -\log_{10}K_a

Kw=KaΓ—KbK_w = K_a \times K_b

πŸ’‘Examples

Problem 1:

Calculate the pHpH at the half-equivalence point when 25.0 cm325.0\,cm^3 of 0.100 mol dmβˆ’30.100\,mol\,dm^{-3} CH3COOHCH_3COOH is titrated with 0.100 mol dmβˆ’30.100\,mol\,dm^{-3} NaOHNaOH. The acid dissociation constant for ethanoic acid is Ka=1.8Γ—10βˆ’5K_a = 1.8 \times 10^{-5} at 298K298K.

Solution:

pH=pKa=βˆ’log⁑10(1.8Γ—10βˆ’5)β‰ˆ4.74pH = pK_a = -\log_{10}(1.8 \times 10^{-5}) \approx 4.74

Explanation:

At the half-equivalence point, exactly half of the CH3COOHCH_3COOH has reacted to form CH3COOβˆ’CH_3COO^-. Thus, [CH3COOH]=[CH3COOβˆ’][CH_3COOH] = [CH_3COO^-]. Substituting these equal values into the Henderson-Hasselbalch equation makes the log term log⁑10(1)=0\log_{10}(1) = 0, leaving pH=pKapH = pK_a.

Problem 2:

Explain why the pHpH at the equivalence point for the titration of NH3(aq)NH_3(aq) with HCl(aq)HCl(aq) is less than 77.

Solution:

The salt formed is NH4ClNH_4Cl. The NH4+NH_4^+ ion is the conjugate acid of a weak base and undergoes hydrolysis: NH4+(aq)+H2O(l)β‡ŒNH3(aq)+H3O+(aq)NH_4^+(aq) + H_2O(l) \rightleftharpoons NH_3(aq) + H_3O^+(aq).

Explanation:

Since the hydrolysis of the ammonium ion produces hydronium ions (H3O+H_3O^+), the concentration of [H+][H^+] increases, resulting in an acidic pHpH at the stoichiometric equivalence point.

pH curves (HL only) - Revision Notes & Key Formulas | IB Grade 12 Chemistry