Ionic Equilibria AP Chemistry Chapter 19 Mr. Solsman
Goals: • Acid-base buffers, common ion effect, titrations • Slightly soluble salts • Complex ions
I Acid-Base Equilibria • A. Solutions of Acids or Bases Containing a Common Ion • A buffer is something that lessens the impact of an external force.
An acid-base buffer is a solution that lessens changes in [H3O+] resulting from the addition of an acid or base. • When a small amount of H3O+ or OH- is added to an unbuffered solution, the change in pH is much larger than the change in a buffered solution.
Buffers work through the common ion effect. • Suppose we dissolve acetic acid in water and then add sodium acetate as well.
CH3COOH(aq) + H2O(l) CH3COO-(aq) + H3O+(aq) • But adding CH3COONa shifts the equilibrium left (Le Chatelier) in effect decreasing the [H3O+] and lowering the acetic acid dissociation.
The acetate ion is called the common ion because it is common to both solutions. • The common-ion effect occurs when a reactant containing a given ion is added to an equilibrium mixture that already contains that ion and the position of equilibrium shifts away from forming more of it.
Features of a buffer—a buffer consists of high concentrations of the acidic (HA) and basic (A-) components. When small amounts of H3O+ or OH- ions are added to the buffer, they cause a small amount of one buffer component to convert into the other changing the relative concentrations of the two components.
As long as the H3O+ or OH- added is small compared to HA and A-, the added ions have little effect on the pH because they are consumed by one or the other buffer components. • A- consumes H3O+ and HA consumes OH-
Calculate the pH: • (a) Of a buffer solution of 0.50 M acetic acid and 0.50 M sodium acetate. • (b) after adding 0.020 mol solid NaOH to 1.0 L of the buffer in part a. • (c) after adding 0.020 mol HCl to 1.0 L of the buffer in part a.
Calculate the pH of a buffer of 0.50 M HF and 0.45 M F- (a) before and (b) after the addition of 0.40 g NaOH to 1.0 L of the buffer. (Ka of HF = 6.8 x 10-4)
How does a buffer work? • Consider the reaction: • HA + H2O A- + H3O+ • Adding OH- gives: • HA + OH- A- + H2O
We know Ka = [H3O+] [A-] / [HA] • and [H3O+] = Ka [HA] / [A-] • Adding OH- ions causes HA to be converted to A-, the ratio of [HA] / [A-] decreases, but if the original amounts of HA and A- are large, the change is very small. The pH and [H3O+] remain nearly constant.
Similar reasoning applies when protons are added to a buffered solution of a weak acid and a salt of its conjugate base. Because the A- ion has a high affinity for H3O+, the added ions react with A- to form the weak acid: H3O+ + A- HA
There is a net change of A- to HA. Again if [A-] and [HA] are large, little change in pH occurs.
Calculate the pH of a solution containing 0.75 M lactic acid (Ka = 1.4 x 10-4) and 0.25 M sodium lactate.
Henderson-Hasselbalch equation: • For any weak acid: • HA + H2O H3O+ + A- • [H3O+] = Ka [HA] / [A-] • Or pH = pKa + log([Base]/[Acid)]
A buffered solution contains 0.25 M NH3 (Kb = 1.8 x 10-5) and 0.40 M NH4Cl. Find the pH of the solution.
Adding strong acid to a buffered solution • Calculate the pH of the solution that results when 0.10 mol of gaseous HCl is added to 1.0 L of the buffered solution in the previous example.
Buffer capacity • The buffer capacity represents the amount of protons or hydroxide ions the buffer can absorb without significantly changing the pH.
The pH of a buffered solution is determined by the ratio of [A-] to [HA]. • The capacity of a buffered solution is determined by the magnitudes of [HA] and [A-].
Calculate the change in pH that occurs when 0.0100 mol of gaseous HCl is added to each of the following substances: • Solution A: 5.00 M HC2H3O2 and 5.00 M NaC2H3O2 • Solution B: 0.050 M HC2H3O2 and 0.050 M NaC2H3O2
A chemist needs a solution buffered at pH 4.30 and can choose from the following acids and their salts: Which works best? • (a) chloroacetic acid (Ka = 1.35 x 10-3) • (b) propanoic acid (Ka = 1.3 x 10-5) • (c) benzoic acid (Ka = 6.4 x 10-5) • (d) hypochlorous acid (Ka = 3.5 x 10-8)
Describe how you would prepare a “phosphate buffer” with a pH of about 7.40. • Hint: the concentration of the acid component must be roughly equal to the conjugate base component. Or when pH pKa.
How would one prepare a benzoic acid/benzoate buffer with pH 4.25, starting with 5.0 L of 0.050 M sodium benzoate (C6H5COONa) solution and adding the acidic component? (The Ka of benzoic acid is 6.3 x 10-5.)
Summary, Buffered Sol’ns • 1. They contain relatively large concentrations of a weak acid and corresponding weak base. • 2. When H3O+ is added, it reacts essentially to completion with the weak base present. • 3. When OH- is added, it reacts essentially to completion with the weak acid present.
4. The pH is determined by the ratio of the concentrations of the weak acid and base. As long as this ration is constant, the pH will remain virtually constant. This will be true as long as the concentrations of the buffering mtl’s are large compared to the amounts of H3O+ or OH- added.
The pKa of the weak acid to be used in the buffer should be as close as possible to the pH of the solution.
Acid-Base Titration Curves • Two common devices for measuring pH are a pH meter and an acid-base indicator. • An acid-base indicator is a weak organic acid (HIn) that has a different color than its conjugate base (In-).
The color change occurs over a specific and narrow pH range. • The amount of indicator used is small enough that the solution pH is not affected. • The indicator changes color over a 2 pH range.
1. Strong Acid-Strong Base Titrations • A titration curve is a plot of pH versus the volume of added titrant. • There are distinct regions of the curve.
The equivalence point is the point at which the number of moles of OH- added equals the number of moles of H3O+ originally present. This occurs at the nearly vertical portion of the curve where the solutions consists of the anion of the SA and cation of the SB.
These ions do not react with water, so the pH is neutral; pH = 7.00. • Since we are usually titrating with milliliters, an alternative way of looking at molarity is often used: • Molarity = mmol / mL
Strong Acid-Strong Base Titrations • 50.0 mL of 0.200 M HNO3 are titrated with 0.100 M NaOH. Calculate the pH after the additions of 0, 10, 20, 50, 100, 150, and 200 mL of NaOH. Then construct a titration curve and label it properly.
Titrating a strong base with a strong acid is very similar to the last example. • However OH- is in excess before the equivalence point and H+ is in excess after the equivalence point. • The curve is also flipped.
Weak Acid-Strong Base Titrations • 50.0 mL of 0.10 M acetic acid (Ka = 1.8 x 10-5) are titrated with 0.10 M NaOH. Calculate the pH after the additions of 0, 10, 25, 40, 50, 60, and 75 mL samples of NaOH. Then construct a titration curve and label it properly.
Weak Base-Strong Acid Titration • Use the same procedures as in the last problems. Decide which major species are present in solution and decide which reactions run to completion. Choose the dominant equilibrium and calculate the pH!
20.0 mL of 0.10 M triethylamine, (CH3CH2)3N, (Kb = 5.2 x 10-4) are treated with 0.100 M HCl. Calculate the pH after 0.0, 10, 15, 19, 19.95, 20.05, and 25 mL additions of HCl. Construct a titration curve.
The pH Curves for the Titrations of 50.0-mL Samples of 0.10 M Acids with Various Ka Values with 0.10 M NaOH