Presentation on theme: "Applications of Aqueous Equilibria"— Presentation transcript:
1Applications of Aqueous Equilibria AP ChemistryChapter 15
2This chapter covers More acid-base equilibria Solubility of salts Formation of complex ions
3Common ion Some solutions contain not only a weak acid… …but also the salt of the weak acid:Acetic acidSodium acetateThese form a solution with powerful results
4Suppose a solution contains HF (Ka = 7.2x10-4) and NaF NaF (s) + H2O Na+ (aq) + F-(aq)HF (aq) H+ (aq) + F-(aq)F- is called the “common ion”Thinking in terms of LeChatelier’s principle, does the presence of a common ion have any effect?With more F-, the equilibrium will not shift as dramatically in the direction of the products.The result will be fewer H+ ions and a higher pH.This is the Common Ion Effect.
515. 1 The equilibrium concentration of H+ in a 1. 0 M HF solution is 2 15.1 The equilibrium concentration of H+ in a 1.0 M HF solution is 2.7x10-2 M, and the percent dissociation of HF is 2.7%. Calculate [H+] and the percent dissociation of HF in a solution containing 1.0M HF (Ka=7.2x10-4) and 1.0 M NaF.HF (aq) H+ (aq) + F-(aq)-x x x1.0-x x xKa=7.2x10-4 = x(1.0+x)1.0-xx = [H+] = 7.2x10-47.2x10-4 100 = 0.072%1.0
6Buffered SolutionsA buffered solution is one that resists a change in its pH when either H+ or OH- is addedOur blood contains a buffering system of HCO3- and H2CO3If it weren’t so, we would die!Our blood must stay in a very narrow pH range.
72 common buffersA solution of a weak acid plus the soluble salt of the weak acidSodium acetateAcetic acidA solution of a weak base plus the soluble salt of the weak baseAmmoniaAmmonium chloride
8Using LeChatalier’s Principle, we can determine what happens to an acid when added to a buffer. HC2H3O H+ + C2H3O2-Adding a strong acid, reacts with the anion and shifts the reaction to the formation of the weak acid.
9Using LeChatalier’s Principle, we can determine what happens to an acid when added to a buffer. HC2H3O H+ + C2H3O2-Adding a strong base, reacts with the H+ forming water and shifts the reaction to the formation of more H+.
1015. 2 a buffered solution contains 0. 50 M acetic acid (Ka=1 15.2 a buffered solution contains 0.50 M acetic acid (Ka=1.8x10-5) and 0.50 M sodium acetate. Calculate the pH of this solution.HC2H3O H+ + C2H3O2--x x x.50-x x xKa = 1.8x10-5 = x (.50+x).50-xx = [H+] = 1.8x10-5pH = 4.74
11Notice, we first had to take care of the reaction between the strong base and the weak acid, then we worked the equilibrium problem!15.3 Calculate the change in pH that occurs when mol solid NaOH is added to 1.0 L of the buffered solution in # Compare the pH change with that which occurs when mol solid NaOH is added to 1.0 L of water.HC2H3O2 + OH- H2O + C2H3O2-HC2H3O H C2H3O2--x x x.49-x x xKa = 1.8x10-5 = x (.51+x).49-xx = [H+] = 1.7x10-5pH = 4.76pH = .02
12Just How Does Buffering Work? As we add OH-, the weak acid is the best source of H+ ions….OH- + HA H2O + A-The OH- ions are not allowed to accumulate and are thus replaced by A- ion
13HA H+ + A- The equilibrium expression is… Ka = [H+][A-] [HA] [H+] = Ka [HA][A-]…and can be written as:So the pH is determined by this ratio. When OH- is added, HA is converted to A-A change in [HA]/[A-] is usually very small and [H+] and pH remain relatively constant.If [H+] is added then H+ + A- HA and free [H+] do not accumulate.If [HA] and [A-] are large compared with [H+] which is added, little pH change occurs
14Calculate [H+] in 0.10 M HF (Ka = 7.2x10-4) and 0.30 M NaF. [H+] = Ka [HA][A-][H+] = 7.2x10-4 [.10][.30][H+] = 2.0x10-4 MAnother way to handle this is to take the –log of both sides of the 1st equation. Trust me, you’ll like this.-log[H+] = -log Ka [HA][A-] pH = pKa + log [A-][HA]pH = pKa + log [base][acid]If this base/acid ratio = 1, the better the buffer and, as a bonus, pH = pKa!
1515. 4 Calculate the pH of a solution containing 0 15.4 Calculate the pH of a solution containing 0.75 M lactic acid (HC3H5O3, Ka=1.4x10-4) and 0.25 M sodium lactate.pH = pKa + log [A-][HA]pH = -log(1.4x10-4) + log [.25][.75]pH = 3.38
1615. 5 A buffered solution contains 0. 25 M NH3 (Kb=1. 8x10-5) and 0 15.5 A buffered solution contains 0.25 M NH3 (Kb=1.8x10-5) and 0.40 M NH4Cl. Calculate the pH of this solution.NH NH OH-pH = pKa + log [A-][HA]pH = -log(5.6x10-10) + log [.25][.40]pH = 9.05Ka = Kw = =5.6x10-10Kb x10-5
1715. 6 Calculate the pH of the solution that results when 0 15.6 Calculate the pH of the solution that results when 0.10 mol gaseous HCl is added to 1.0 L of the buffered solution form example 15.5NH NH OH-pH = pKa + log [A-][HA]pH = -log(5.6x10-10) + log [.15][.50]pH = 8.73Ka = Kw = =5.6x10-10Kb x10-5
18Buffer CapacityThe amount of protons or hydroxide ions the buffer can absorb without significant change in pH.It is determined by the base/acid ratio.
1915. 7 Calculate the change in pH that occurs when 0 15.7 Calculate the change in pH that occurs when mol gaseous HCl is added to 1.0 L of each of the following solutions:SolutionA: M HC2H3O2 and 5.00 M C2H3O2-pH = pKa + log [5.00][5.00]pH = -log 1.8x= initiallyC2H3O2- + H HC2H3O2pH = pKa + log [4.99][5.01]pH == 4.74
20…there is very little effect on the pH Solution B: M HC2H3O2 and .050 M C2H3O2pH = pKa + log [.050][.050]pH = -log 1.8x= initiallyC2H3O2- + H HC2H3O2pH = pKa + log [.04][.06]pH == 4.56As the concentrations of [HA] and [A-] are smaller, the addition of acid or base has a greater effect on pH.When the concentrations of [HA] and [A-] are large and the addition of acid or base is small……there is very little effect on the pH
21Best Buffer To produce the most effective buffer… [HA] = [A-] pKa of weak acid should be as close as possible to the desired pH.A pH of 4.00 is wantedpH = pKa + log [A-][HA]4.00 = pKa + 0Ka = 1.0x10-4
2215. 8 A chemist needs a solution buffered at pH 4 15.8 A chemist needs a solution buffered at pH 4.30 and can choose from the following acids (and their sodium salts):Chloroacetic acid (Ka=1.35x10-3)Propanoic acid (Ka=1.3x10-5)Benzoic acid (Ka=6.4x10-5)Hypochlorous acid (Ka=3.5x10-8)Calculate the ratio [HA]/[A-] required for each system to yield a pH of Which system will work best?5.0x10-5 = [HA]1.35x [A-].00375.0x10-5 = [HA]1.35x [A-]3.85.0x10-5 = [HA]6.4x [A-].785.0x10-5 = [HA]3.5x [A-]14004.30=-log[H+]so[H+]=5.0x10-5[H+] = Ka [HA][A-]5.0x10-5 = [HA]Ka [A-]