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pH of Weak Acids AP Chemistry Unit 9 Chapter 14

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Strengths of Acids and Bases “Strength” refers to how much an acid or base ionizes in a solution. STRONGWEAK Ionize completely (~100%) Example: HCl H + + Cl – NaOH Na + + OH – Ionize partially (usually <10%) Example: HF ↔ H + + F – NH 3 + H 2 O ↔ NH 4 + + OH –

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Relative Dissociation of Acids

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Concentration vs- Strength HA ↔ H + + A – Ka = [H+] [A-], therefore larger Ka values indicate stronger acids. [HA]

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What is the pH of a 0.25M HF solution? The Ka = at 25 o C = 7.1 x 10 -4 HF is a weak acid, so we can’t directly calculate the pH from the molarity. We need to set up a table as we did for other equilibrium problems HF(aq) ⇄ H + (aq)+F-(aq) Initial: 0.25M0 M0 M Equilibrium: 0.25-x xx Ka = [H+][F-] = 7.1 x 10 -4 [HF] 7.1 x 10-4 = x 2 (0.25-x)

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7.1 x 10 -4 = x 2 (0.25-x) and x 2 + 7.1 x 10 -4 x -1.775 x 10 -4 = 0 Solving using the quadratic equation: x = 1.30 x 10 -2 or –1.37×10 -2. The negative solution is physically impossible because x was set to be the [H+], so x = 1.30 x 10 -2. Substituting in this x into the original formula for Ka checks, so: pH=-log(1.30 x 10 -2 ) = 1.89

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An easier method than using the quadratic equation is to make the assumption that the dissociation of HF is so insignificant that the initial concentration is not significantly changed at equilibrium. Therefore, we assume that (0.25 – x) = 0.25 So, 7.1 x 10 -4 = x 2 0.25 1.78 x 10 -4 = x 2 therefore x = 1.33 x 10 -2 pH = – log (1.33 x 10 -2 ) = 1.88 Considering we need to round to 2 significant figures, our assumption is valid!

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The 5% Rule We can ignore the (-x) in the denominator if the acid dissociates less than 5% Otherwise, we must use the quadratic equation. % Ionization = [A-] [HA] initial Usually a very low Ka value means we can ignore the (-x) X 100

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