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Equilibria

Equilibria. An Introduction for Chem Honors Students. Reaction is reversible Both products and reactants are present Forward and reverse reactions take place at the same rate. For the reaction 2A  B. What is an Equilibrium Reaction?. K expression. In general terms

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Equilibria

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  1. Equilibria An Introduction for Chem Honors Students

  2. Reaction is reversible Both products and reactants are present Forward and reverse reactions take place at the same rate For the reaction 2A  B What is an Equilibrium Reaction?

  3. K expression In general terms Keq = [C]c[D]d [A]a[B]b For aA + bB --> cC + dD

  4. Dissociation of Water H2O <---> H1+ + OH1- • H2O + H2O <---> H3O1+ + OH1- equilibrium expression is products over reactants. K = [H3O1+] [OH1-] [H2O] [H2O] • The molarity for the water is a constant at any specific temperature. So,

  5. Dissociation of Water • Equilibrium constants exist then for both acid dissociation and base. (Ka and Kb) • The higher the Ka, the stronger the acid and the higher the Kb, the stronger the base. • Ka and Kb are related by the previous equation. Kw = KaKb

  6. Dissociation of Water • As Ka gets larger the strength of the acid gets higher, but Kb must fall.  Therefore the stronger the acid, the weaker the conjugate base. • It can now be said that the conjugate base (acid) of a weak acid (base) is a weak base (acid) and the conjugate base (acid) of a strong acid (base) is a worthless base (acid).

  7. Conjugate Acid and Base Acid and Conjugate Base explanation of strength. Pair of substances differing only by H+ HF(aq) + H2O(l) <==> H3O+(aq) + F-(aq) acid base conjugate acid conjugate base H3O+(aq) + OH-(aq) <==> H2O(l) + H2O(l) acid base conjugate acid conjugate base

  8. pH in Solutions of Strong Acids and Strong Bases Strong acids • Certain acids are known as strong acids. These are acids that fully ionize when placed in water: • HA + H2O  A- + H3O+ The Reaction goes to completion and thus • Ka = [A-][H3O+]/[HA] = infinity • Some common strong acids are: • HCl, hydrochloric acid • HBr, hyrdobromic acid • HI, hydroiodic acid • H2SO4, sulfuric acid • HNO3, nitric acid • HClO4, perchloric acid

  9. pH in Solutions of Strong Acids and Strong Bases Strong Bases • Certain bases are known as strong bases. These are bases that fully ionize when placed in water. • Some common strong bases are: • LiOH, lithium hydroxide • NaOH, sodium hydroxide • KOH, potassium hydroxide • Ca(OH)2, calcium hydroxide • Sr(OH)2, strontium hydroxide • Ba(OH)2, barium hydroxide • Alkaline earth oxides. • Lime (CaO)

  10. Equilibrium in Solutions of Weak Acids • HA(aq) + H2O(l)  A-(aq) + H3O+(aq) • The equilibrium constant for a weak acid is • Ka = [H3O+][A-]/[HA] • For a weak acid then Ka << 1 • For a strong acid Ka >> 1 • A common way to express the strength of an acid is the pKa, which is similar in form to the pH • pKa = -logKa

  11. Solving Problems • Hydrazoic acid, HN3, is a weak acid. The [H3O+] of a 0.102 M solution of HN3 is 1.39 x 10 – 3 M. Determine the pH of this solution, and calculate the Ka for HN3. • 1. Write the equation: • HN3 H+ + N3-

  12. 2. Now we will fill in an ICE table. I is initial, C is change and E is equilibrium HN3 H+ + N3- • I 0.102 M 0 0 The Molarity of the HN3 is given, and this is always the initial M. The H+ and N3- ions will be 0 to start and will increase as the HN3 dissociates. C -x +x +x The change is how much will increase or decrease as the dissociation starts. It is determined by the stoichiometric ratios. Here that is 1:1:1 so all are the same.

  13. HN3 H+ + N3- I 0.102 M 0 0 C -x +x +x 3. Now look at what will happen at equilibrium. The amount of HN3 will decrease until equilibrium when it will become steady. The amount of H+ and N3- will increase until they too become steady. They become steady because the rate of change forward equals the rate of change backwards. The HN3 changes in to H+ and N3- at the same rate that H+ and N3-turn into HN3. E 0.102 – x xx

  14. HN3 H+ + N3- I 0.102 M 0 0 C -x +x +x E 0.102 – x xx The amount of NH3 that dissociates is much smaller than its molarity, so to avoid doing some complicated math we will pretend it is still 0.102 The problem tells us that the [H30+] = 1.39 x 10 -3 M So, substituting in we get E 0.102 1.39 x 10 -3 1.39 x 10 -3

  15. Now we have HN3  H+ + N3- I 0.102 M 0 0 C -x +x +x E 0.102 1.39 x 10 -3 1.39 x 10 -3 Ka = [H+][N3-] = (1.39 x 10-3)(1.39 x 10-3) [NH3] 0.102 Ka = 1.89 x 10-5

  16. Finding the pH pH = -log [H+] [H+] = 1.39 x 10-3 from the rest of the problem So pH = -log (1.39 x 10-3) pH = 2.9

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