Presentation on theme: "Applications of Aqueous Equilibria AP Chemistry Chapter 15."— Presentation transcript:
Applications of Aqueous Equilibria AP Chemistry Chapter 15
This chapter covers More acid-base equilibria Solubility of salts Formation of complex ions
Common ion Some solutions contain not only a weak acid… …but also the salt of the weak acid: Acetic acid Sodium acetate These form a solution with powerful results
Suppose a solution contains HF (Ka = 7.2x10 -4 ) and NaF NaF (s) + H 2 O 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? The result will be fewer H + ions and a higher pH. With more F-, the equilibrium will not shift as dramatically in the direction of the products. This is the Common Ion Effect.
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 (K a =7.2x10 -4 ) and 1.0 M NaF. HF (aq) H + (aq) + F - (aq) x x x 1.0-x x 1.0+x K a =7.2x10 -4 = x(1.0+x) 1.0-x x = [H + ] = 7.2x x10 -4 100 = 0.072% 1.0
Buffered Solutions A buffered solution is one that resists a change in its pH when either H + or OH - is added Our blood contains a buffering system of HCO 3 - and H 2 CO 3 If it weren’t so, we would die! Our blood must stay in a very narrow pH range.
2 common buffers A solution of a weak acid plus the soluble salt of the weak acid Sodium acetate Acetic acid A solution of a weak base plus the soluble salt of the weak base Ammonia Ammonium chloride
Using LeChatalier’s Principle, we can determine what happens to an acid when added to a buffer. HC 2 H 3 O 2 H + + C 2 H 3 O 2 - Adding a strong acid, reacts with the anion and shifts the reaction to the formation of the weak acid.
HC 2 H 3 O 2 H + + C 2 H 3 O 2 - Adding a strong base, reacts with the H + forming water and shifts the reaction to the formation of more H +. Using LeChatalier’s Principle, we can determine what happens to an acid when added to a buffer.
15.2 a buffered solution contains 0.50 M acetic acid (K a =1.8x10 -5 ) and 0.50 M sodium acetate. Calculate the pH of this solution. HC 2 H 3 O 2 H + + C 2 H 3 O 2 - x +x +x.50-x x.50+x Ka = 1.8x10 -5 = x (.50+x).50-x x = [H + ] = 1.8x10 -5 pH = 4.74
15.3 Calculate the change in pH that occurs when mol solid NaOH is added to 1.0 L of the buffered solution in #15.2. Compare the pH change with that which occurs when mol solid NaOH is added to 1.0 L of water. -x +x +x.49-x x.51+x Ka = 1.8x10 -5 = x (.51+x).49-x x = [H + ] = 1.7x10 -5 pH = 4.76 HC 2 H 3 O 2 + OH - H 2 O + C 2 H 3 O pH =.02 Notice, we first had to take care of the reaction between the strong base and the weak acid, then we worked the equilibrium problem! HC 2 H 3 O 2 H + + C 2 H 3 O
Just How Does Buffering Work? As we add OH -, the weak acid is the best source of H + ions…. OH - + HA H 2 O + A - The OH - ions are not allowed to accumulate and are thus replaced by A - ion
The equilibrium expression is… HA H + + A - Ka = [H + ][A - ] [HA] …and can be written as: [H + ] = Ka [HA] [A - ] 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
Calculate [H + ] in 0.10 M HF (K a = 7.2x10 -4 ) and 0.30 M NaF. [H + ] = Ka [HA] [A - ] [H + ] = 7.2x10 -4 [.10] [.30] [H + ] = 2.0x10 -4 M Another way to handle this is to take the – log of both sides of the 1 st 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!
15.4 Calculate the pH of a solution containing 0.75 M lactic acid (HC 3 H 5 O 3, K a =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
15.5 A buffered solution contains 0.25 M NH 3 (K b =1.8x10 -5 ) and 0.40 M NH 4 Cl. Calculate the pH of this solution. pH = pKa + log [A - ] [HA] pH = -log(5.6x ) + log [.25] [.40] pH = 9.05 NH 3 NH OH - Ka = Kw = =5.6x Kb 1.8x10 -5
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 15.5 NH 3 NH OH - Ka = Kw = =5.6x Kb 1.8x10 -5 pH = pKa + log [A - ] [HA] pH = -log(5.6x ) + log [.15] [.50] pH = 8.73
Buffer Capacity The amount of protons or hydroxide ions the buffer can absorb without significant change in pH. It is determined by the base/acid ratio.
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: 5.00 M HC 2 H 3 O 2 and 5.00 M C 2 H 3 O 2 - pH = pKa + log [5.00] [5.00] pH = -log 1.8x = 4.74 initially C 2 H 3 O H + HC 2 H 3 O 2 pH = pKa + log [4.99] [5.01] pH = = 4.74
Solution B:.050 M HC 2 H 3 O 2 and.050 M C 2 H 3 O 2 pH = pKa + log [.050] [.050] pH = -log 1.8x = 4.74 initially C 2 H 3 O H + HC 2 H 3 O 2 pH = pKa + log [.04] [.06] pH = = 4.56 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 pHAs the concentrations of [HA] and [A - ] are smaller, the addition of acid or base has a greater effect on pH.
Best Buffer To produce the most effective buffer… [HA] = [A - ] pKa of weak acid should be as close as possible to the desired pH. pH = pKa + log [A - ] [HA] 4.00 = pKa + 0 A pH of 4.00 is wanted Ka = 1.0x10 -4
15.8 A chemist needs a solution buffered at pH 4.30 and can choose from the following acids (and their sodium salts): a.Chloroacetic acid (K a =1.35x10 -3 ) b.Propanoic acid (K a =1.3x10 -5 ) c.Benzoic acid (K a =6.4x10 -5 ) d.Hypochlorous acid (K a =3.5x10 -8 ) Calculate the ratio [HA]/[A - ] required for each system to yield a pH of Which system will work best? [H + ] = Ka [HA] [A - ] 4.30=-log[H + ] so [H + ]=5.0x x10 -5 = [HA] Ka [A - ] 5.0x10 -5 = [HA] 1.35x10 -3 [A - ] x10 -5 = [HA] 1.35x10 -5 [A - ] x10 -5 = [HA] 6.4x10 -5 [A - ] x10 -5 = [HA] 3.5x10 -8 [A - ] 1400