Lecture+h - Ch 11 Application of aqueous equilibria Buffer...

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Ch 11 – Application of aqueous equilibria Buffer solutions and the common ion effect The relative stability of neutral and ionic species in aqueous solutions is generally related to the size of the associated K a or K b : - high K a or K b tends to increase the stability of the ionized species -low±K a or K b tends to decrease the stability of the ionized species HCl H + + Cl (strong acid, stable ions are produced, “non-reactive”) HNO 2 (weak acid, unstable ions are produced, “reactive”) H + + NO 2 Let’s examine a solution prepared by mixing 0.50 L of 1.0 M HF with 0.50 L of 1.0 M NaF: The salt will dissociate: NaF Na + (aq) + F (aq) The acid will ionize, according to K a : HF(aq) H + (aq) + F (aq) This is another example of the common-ion effect . not reactive reactive
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Ch 11 – Application of aqueous equilibria Question: what’s the pH of this solution? [F ] 0 = (1.0 × 0.5)/1.0 = 0.50 M [HF] 0 = (1.0 × 0.5)/1.0 = 0.50 M HF(aq) + H 2 O(l) H 3 O + (aq) + F (aq) K a = 7.2 × 10 4 I C E 0.50 N/A 0 0.50 x N/A +x +x 0.50 x x 0.50 + x At equilibrium, we have: from NaF [H 3 O + ] [F ] [HF] = K a If we rearrange a bit and take log of both sides: K a [F ] [HF] = [H 3 O + ] log both sides pH = pK a + log [F ] [HF] For a weak acid/salt solution, this expression can be generalized to the Henderson-Hasselbalch equation : pH = pK a + log [A ] 0 [HA] 0 (HH)
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Ch 11 – Application of aqueous equilibria Some things to note about the Henderson-Hasselbalch equation: - It is assuming the 5% approximation can be made. - It is derived from an acid/water reaction, so a K a value and the associated HA must be involved in the reaction - It is mostly applied when the equilibrium involves the common-ion effect. What is the pH of the solution we have been investigating? pH = pK a + log [A ] 0 [HA] 0 For us: pK a = log(7.2 × 10 4 ) = 3.14 [F ] 0 = 0.50 M [HF] 0 = 0.50 M Therefore: pH = 3.14 + log 0.50 0.50 = 3.14
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Application of aqueous equilibria Question: what happens to the pH if we add 0.01 mols of solid KOH to the 1 L of the NaF/HF solution? First, let’s see what the pH would be if we added 0.01 mols of KOH to just water: KOH(s) K + (aq) + OH (aq) (strong base) After the complete dissociation, [OH ] = 0.01 M. pH = log(10 14 /0.01) = 12 If we sprinkle 0.01 mols of KOH into just water, the pH is 12! Alright. Now let’s sprinkle that 0.01 mols of KOH into the NaF/HF solution. Here are the species floating around in the solution once we do this: K + , Na + , HF, H 2 O, F , H + , OH Speaking in terms of the pH, K + and Na + are non-reactive since they will not be completing any reactions that affect either [H + ] eq or [OH ] eq . In the solution, [H
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This note was uploaded on 11/04/2009 for the course CHEM Chem 1C taught by Professor Farmer during the Spring '09 term at UCL.

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Lecture+h - Ch 11 Application of aqueous equilibria Buffer...

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