Chemistry, Canadian EditionChapter 15: Student Study Guide

Chapter 15: Applications of Aqueous Equilibria

Learning Objectives

Upon completion of this chapter you should be able to

• calculate the pH of a buffered solution

• explain how to prepare a buffered solution of known pH and capacity

• calculate an acid or base concentration from titration data

• use the concepts of Ksp and the common-ion effect to calculate solution concentrations

• calculate the concentrations of species involved in complex-ion formation

Practical Aspects

This chapter builds on the concepts in Chapters 13 and 14. The four main new topics in this chapter are buffers, acid/base titrations, the common ion effect, and complex ion equilibria. These last two topics have everyday applications ranging from treating hazardous wastes to treating a person for heavy-metal poisoning. Acid-base titrations are performed routinely in laboratory-based professions. Buffers are found everywhere in biological systems. Anyone studying to be a chemist, biochemist, biologist, environmental scientist, and anyone intending to work in a medical profession should have a solid understanding of buffers and how they work. You will see buffers again in later courses, guaranteed.

By the time you’re done studying this chapter, you should be able to calculate the pH of an aqueous buffered or non-buffered system, calculate concentrations of components present at various stages of a titration, assess solubility equilibria, and use the common ion effect and complex ion equilibria concepts to control solute concentrations.

15.1buffer solutions

Skills to Master:

  • Using the buffer equation to calculate pH of a buffer solution.
  • Calculating the pH of a buffered solution after.

Key Terms:

  • Buffer – a mixture of a weak conjugate acid/base pair in roughly equal proportions.
  • Buffer components – the weak conjugate acid/base pair.
  • Buffer ratio – ratio of weak acid to conjugate base.
  • Buffer capacity – the amount of weak acid and conjugate base present in the buffer.

Key Concepts:

  • A buffer resists changes in pH because the conjugate acid/base pair can react with either added base or acid.
  • A buffer has both a weak acid and its conjugate base as major components in the solution. (Note: The weak acid and conjugate base are typically within a 10-fold concentration range of each other.)
  • Buffer components are often abbreviated “HA” and “A-” to describe the conjugate acid/base pair.
  • The concentration of the total buffer components is the sum of the concentration of HA and A-.

A buffer works because whatever strong acid or base is added to it will react with one of the buffer components essentially 100%, until that entire buffer component has been consumed. For example, if a strong acid is added, it will react with the weak base component:

H3O+ (aq) + A- (aq) →H2O (l) + HA (aq)

Notice that this reaction is the reverse of a weak acid dissociation: Keq = 1/Ka. The value of this reaction’s equilibrium constant will be huge. The reaction will occur essentially 100%.

Key Concepts:

  • A strong acid (H3O+ ) mixing with a weak base (A-) will react essentially 100%.
  • A strong base (OH-) mixing with a weak acid (HA) will react essentially 100%.

A buffer is like a sponge for acids and bases. It can be tailored to your needs:

  • A big sponge: 2 mol HA & 2 mol A-

  • A small sponge: 0.00010 mol HA & 0.00010 mol A-

  • A sponge that specializes in absorbing added acid: 0.1 mol HA & 0.5 mol A-

  • A sponge that specializes in absorbing added base: 0.5 mol HA & 0.1 mol A-

To put the above “sponge” analogies into chemistry terms,

  • the big sponge has a much greater buffer capacity than the small sponge – it can absorb 20,000 times more added acid or base than the small sponge.
  • The big sponge has a buffer ratio of 1:1 A-/HA. So does the little sponge. The sponge specializing in absorbing acid has a buffer ratio of 5:1 A-/HA. What’s the buffer ratio of the sponge that specializes in absorbing added base?

Exercise 1: Use the pictures shown to answer the questions. Color key: black = A-, white = H+, grey = oxygen. Only major species are shown. Water molecules have been omitted for clarity. a) Identify whether or not each is a buffer. b) For the boxes that contain buffers, what is the buffer ratio for each? c) Which buffer is capable of absorbing the most H3O+?

Strategy and Solution:

a)A buffer contains both a weak acid and its conjugate base within a roughly 10-fold concentration range of each other. In these systems, we see that “HA” is being used to describe a weak acid and “A-” is that acid’s conjugate base. Solution A is a buffer because both weak acid and its conjugate base are present in measurable quantities: Solution A contains 1 HA and 2 A-. Solution B is not a buffer. It contains the weak acid, HA, but no amount of conjugate base is shown. Solution C is not a buffer because it contains the conjugate base of HA, but it does not contain HA. Furthermore, Solution C contains excess OH-, which would not be present if a buffer were present because the buffer would consume all of the OH-. Solution D is a buffer because it contains 3 HA and 3 A- (the conjugate acid/base pair is present.)

b)Solutions A and D contain buffers, so only these two solutions must be addressed in this question. The buffer ratio of Solution A is expressed as a ratio of A-/HA is 2:1 and the buffer ratio of Solution D is 1:1.

c)A- will react with any H3O+ that is added to the container. If we add H3O+ to Solutions A and D, we would see that Solution A can absorb 2 A-s while Solution D can absorb 3 A-s. Solution D therefore has a larger capacity to absorb H3O+ than Solution A. Notice that to assess the buffer’s capacity to absorb added H3O+, we must assess the initial number of moles of weak base present, not the buffer ratio.

Exercise 2: Let’s say that you have a buffer composed of 0.50 mol of some acid HA and 0.45 mol of its conjugate base, A-. How many moles of strong acid (H3O+ ) could be added to the buffer before the pH would change appreciably?

Strategy and Solution: The strong acid would react in a 1:1 ratio with A-. As long as there is A- present, there won’t be an excess of H3O+ in the solution. You could therefore add 0.45 mol of H3O+ before a large change in pH would result.

Exercise 3: Consider the buffer in Exercise 2. How many moles of each solute would be present after 0.10 mol of NaOH was added to it? Is the solution still a buffer after the addition of this NaOH?

Strategy: Take an inventory of the chemicals after they react. The OH- will react with the HA in a 1:1 ratio to completion:

OH- (aq) + HA (aq) → A- (aq) + H2O (l)

OH- will be the limiting reagent, since we’re starting with 0.10 moles of OH- and 0.50 moles of HA. All of the OH- will be consumed, 0.10 moles of HA will be consumed (leaving 0.40 moles HA), and 0.10 moles of A- will be formed (making a total of 0.10 mol + 0.45 mol = 0.55 mol A-).

Solution: The amount of each solute remaining in solution is 0.55 mol A- and 0.40 mol HA. There is still an appreciable amount of the conjugate acid/base pair, so the mixture is still a buffer.

Try It #1: Consider the buffer in Exercise 2 (0.50 mol HA and 0.45 mol A-). How many moles of each solute would be present after the addition of 0.55 moles of HCl? Is the mixture still a buffer?

Exercise 4: Which of these mixtures will make a buffer solution?

Mixture A: 45.0 mL of 0.10 M HOCl + 20.0 mL of 0.10 M NaOH

Mixture B: 45.0 mL of 0.10 M NaOH + 20.0 mL of 0.10 M HOCl

Strategy: Take an inventory of the chemicals in each mixture.

Mixture A:

HOCl: 0.0450 L x 0.10 mol/L = 0.0045 mol HOCl to start

NaOH:0.0200 L x 0.10 mol/L = 0.0020 mol OH- to start (strong base)

These chemicals will react completely according to the reaction:

OH- (aq) + HOCl (aq) ↔ OCl- (aq) + H2O (l)

OH- is the limiting reagent with 0.0020 mol initially present. It will be completely consumed in the reaction.

Inventory after reaction:

HOCl: 0.0045 mol to start – 0.0020 mol reacted with OH- = 0.0025 mol remaining.

OCl-:0 mol to start + 0.0020 mol generated.

OH-:0 moles remaining. It was completely consumed by the HOCl.

Solution: Mixture A is a buffer because the mixture will contain a weak acid and its conjugate base as major components: 0.0025 mol HOCl and 0.0020 mol OCl-.

Mixture B:

HOCl: 0.0200 L x 0.10 mol/L = 0.0020 mol HOCl to start

NaOH:0.0450 L x 0.10 mol/L = 0.0045 mol OH- to start

These chemicals will react completely: OH- (aq) + HOCl (aq) → OCl- (aq) + H2O (l)

HOCl is the limiting reagent with 0.0020 mol initially present. It will be completely consumed.

Solution: Since the HOCl is completely consumed, Mixture B cannot be a buffer. We need both HOCl and ClO- as major components for the solution to be a buffer.

Try It #2: Which solution is a buffer? Mixture A: 0.050 mol acetic acid and 0.038 mol of acetate ion, Mixture B: 1.00 mol of HCl and 2.00 moles of KCN, Mixture C: 25.0 mL of 0.250 M NaOH and 30.0 mL of 0.500 M H2S, Mixture D: 10 mL of 1.0 M NaOH and 10 mL of 0.70 M NH4Cl.

The Buffer Equation

Useful Relationship:

/ This is the Henderson-Hasselbalch equation, which is used to determine the pH of a buffer solution. The equation is derived from the Ka expression. (The derivation is provided in the text.)

Key Concepts:

  • HA and A-always refer to the components of the conjugate weak acid/base pair.
  • The equation always utilizes pKa of the acid component of the buffer system. (This is apparent when you look at the derivation of the Henderson-Hasselbalch equation.) Don’t use pKb in this equation.
  • pH = pKa when the buffer ratio is 1:1. When the mole ratio of the weak acid to its conjugate base is 1:1, the log term drops out because log(1) = 0.

Exercise 5: In Exercise 4, we saw that Mixture A was a buffer solution: 45 mL of 0.10 M HOCl mixed with 20 mL of 0.10 M NaOH produced 0.0025 mol of HOCl and 0.0020 mol of OCl-. What is the pH of this buffer solution at 25C?

Strategy: This is a buffer solution, so use the Henderson-Hasselbalch equation to find pH. The pKa of HOCl from Appendix E of the text is 7.40. The total volume of solution is 65 mL – divide moles by this volume to find concentration units.

It makes sense that the buffer pH is slightly less than 7.40 because there is more of the HA component than the A- component.

Helpful Hint

  • Here’s a shortcut: Notice from Exercise 5 that since the buffer components are in the same container, the total volume for both components is the same. The [HA] and [A-] terms can really be based on the number of moles of each. You don’t have to convert them to concentrations for this calculation.

Buffer Action

Exercise 6: Consider the HOCl/ OCl- buffer we’ve been working with: 0.0025 mol of HOCl and 0.0020 mol of OCl- in 65.0 mL of solution. Determine the pH of the solution if 5.1x10-4 mol of KOH is added to it.

Strategy: Take an inventory of the mixture. KOH (a strong base) will react with the HOCl in a 1:1 ratio. Since the KOH is the limiting reagent, it will be completely consumed. In the process, 5.1x10-4 moles of HOCl will be consumed and 5.1x10-4 moles of OCl - will be generated:

Moles of HOCl: 0.0025 – 5.1x10-4 = 0.00199

Moles of OCl-: 0.0020 + 5.1x10-4 = 0.00251

Both the weak acid and its conjugate base are still present as major components, so we still have a buffer. Use the Henderson-Hasselbalch equation to calculate pH:

Solution: The pH of the solution is 7.50. This number seems reasonable for two reasons: 1) the pH did not change significantly (we have a buffer), and 2) the pH is slightly higher than the pKa of the acid component (the buffer ratio is weighted slightly with extra OCl-).

The true power of a buffer solution cannot fully be appreciated until its working ability is compared to a non-buffered environment. Exercise 7 illustrates this.

Exercise 7: Calculate the pH of a solution resulting from the addition of 5.1x10-4 mol of KOH to 65 mL of water. (Notice that these numbers are identical to those in the previous exercise, but this is a non-buffered solution.)

Strategy: The pH can be found from pOH. We need [OH-] first.

[OH-] = 5.1x10-4 mol / 0.065 L = 7.84615x10-3 M

pOH = 2.1053 so pH = 11.89 (2 sig figs)

Solution: The pH of the non-buffered water changed from 7.00 to 11.89. Remember that a 1-unit pH change signifies a 10-fold change in [H3O+], so this represents an enormous change in [H3O+].

Exercise 8: Consider the original HOCl/OCl- buffer again (0.0025 mol of HOCl and 0.0020 mol of OCl- in 65 mL). What volume of 0.0050 M HCl would we need to add to this solution to lower the pH by 0.30 pH units?

Strategy: The pH of the original buffer solution was 7.30 so we want the pH to decrease to 7.00. The question simplifies to “how many moles of HCl should be added?” The HCl and the ClO- will react in a 1:1 ratio to decrease the amount of ClO- and increase the amount of HClO:

so so

Solving for x, we get x = 7.19x10-4 = moles of HCl to add

7.19x10-4 mol x 1 L/0.0050 mol = 0.144 L = 140 mL (2 sig figs)

Solution: 140 mL (or 0.14 L) of 0.0050 M HCl would need to be added to make the pH change to 7.00.

15.2capacity and preparation of buffer solutions

Skill to Master:

  • Calculating the capacity of a buffered solution.
Buffer Capacity

Key Term:

  • Buffer capacity – (first defined in section 15.1, but here is an alternate definition) - the amount of added strong acid or base that the buffer is capable of absorbing without being destroyed.

Key Concept:

  • It is customary to say that a buffer is destroyed if its ratio exceeds 1:10 or 10:1.

Exercise 9: Consider the HOCl buffer one last time (0.0025 mol of HOCl and 0.0020 mol of OCl- in 65 mL). Is the buffer capacity destroyed by adding 30.0 mL of 0.100 M NaOH to it? What is the pH of the resulting solution?

Strategy: Take an inventory of the components. NaOH (a strong base) will react with the HOCl in a 1:1 ratio. We originally had 0.0025 moles of HOCl. The amount of NaOH added is:

0.0300 L x 0.100mol/L = 0.00300 mol OH-

In this scenario, HOCl is the limiting reagent, so it will be completely consumed. We no longer have a buffer. 0.0025 mol of OH- will react with the 0.0025 mole of HOCl and we’ll be left with 0.0005 mol of OH- in a total volume of 95.0 mL (65.0 mL + 30.0 mL). The pH of the solution will be dependent upon the [OH-]:

[OH-] = 0.0005 mol/0.0950 L = 5.26x10-3 M; so pOH = 2.28 and pH = 11.72

Solution: The buffer is extinguished. The pH of the solution is 11.7. (1 sig fig)

Try It #3: Repeat Exercise 9, but use a buffer with a larger concentration of buffer components: 65.0 mL of solution containing 0.025 mol of HOCl and 0.020 mol of OCl-. Would the buffer be extinguished by adding the 30.00 mL of 0.100 M NaOH to it?

Buffer Preparation

In order to prepare a buffer solution, two questions must be answered:

1. What is the desired pH of the buffer?

The desired pH of the buffer will determine which buffer components to use. Typically, a chemist wants the buffer to react equally well to added acid or base, so a buffer ratio of 1:1 is ideal.

Using Henderson-Hasselbalch: , we see that the pH of the buffer will equal the pKa of the acid component in the buffer (log of 1 = 0). This allows us to choose the conjugate weak acid/ base pair of the buffer system. Choose a weak acid that has a pKa approximately equal to the desired pH of the buffer. A combination of this weak acid and its conjugate base will make up the buffer.

2. What is the overall desired buffer concentration?

The buffer capacity will be dictated by the concentration of the buffer components. We need to know this to determine how much weak acid and conjugate base to use. There are three ways a buffer can be made to achieve this mixture of HA and A-. These are general recipes that must be fine-tuned for the details of the actual desired buffer.

1.Mix approximately equimolar amounts of the acid/base conjugate pair. The substance that is charged will be a salt of some type. Non-reactive counter ions typically used are Na+, K+, Cl-.

2.Mix a given amount of HA with approximately ½ the number of moles of strong base.

3.Mix a given amount of A- with approximately ½ the number of moles of strong acid.

Exercise 10: Select a weak conjugate acid/base pair to use in making: a) a pH 8.60 buffer, b) a pH 6.00 buffer, c) a pH 4.30 buffer. Assume all buffers are at 25C.

Strategy: For each, use Appendix E from the text to find an acid with a pKa near the desired pH value. That acid and its conjugate base will constitute the buffer.

Solution: a) pH 8.60: HBrO (pKa = 8.55) and BrO-

b) pH 6.00: HONH3+ (pKa=5.94) and HONH2

c) pH 4.30: Benzoic acid (pKa = 4.20) and benzoate ion

Exercise 11: Make 500.0 mL of a pH 8.75 buffer with a total buffer concentration of 0.250 M. The chemicals available to use in making the buffer are NaBrO and 1.00 M HCl (aq).

Strategy: Given that we have NaBrO and HCl to work with, all of the buffer components will have to originate from the conjugate base portion of the buffer, BrO-. Thus, we need:

0.5000 L x 0.250 mol/L = 0.125 mol of BrO- or NaBrO to start.

NaBrO is an ionic compound, so it will be a solid at room temperature:

0.125 mol x 118.893 g/mol = 14.86 g

The amount of HCl to add will be dictated by the buffer ratio. Whatever HCl is added will react with BrO- (decreasing its concentration) and form HBrO (increasing its concentration). This amount we can call “x.” Determine the number of moles of each buffer component needed to establish the desired pH using the Henderson-Hasselbalch equation.

If we add 0.0589 moles of HCl to the NaBrO, we’ll generate the proper A-/HA ratio. The HCl solution is 1.00 M.

0.0589 mol x 1.00 L/1.00 mol = 0.0589 L = 58.9 mL of 1.00 M HCl.

Solution: Mix 14.86 g of NaBrO with 58.9 mL of 1.00 M HCl in a 500.00 mL volumetric flask and dilute to the mark with water.

Exercise 12: Prepare the same buffer as in Exercise 11, starting with NaBrO(s) and 2.00 M HBrO.

Strategy: In Exercise 11, we saw that the total moles of buffer components had to be 0.125 moles, or we needed 0.0589 moles of HBrO and 0.0661 moles of BrO- (0.125 – 0.0589 = 0.0661 moles).

We need:

HBrO:0.0589 mol x 1 L/2.00 mol = 0.02945 L = 29.5 mL of 2.00 M HBrO

NaBrO:0.0661 mol x 118.893 g/mol = 7.86 g of NaBrO

Solution: Mix 7.86 g of NaBrO with 29.5 mL of 2.00 M HBrO in a 500.00 mL volumetric flask and dilute to the mark with water.

Try It #4: Prepare 250.0 mL of a pH 6.00 buffer with a total buffer concentration of 0.100 M, given these chemicals to work with: solid HONH3Cl and solid NaOH.

15.3acid-base titrations

Skills to Master:

  • Calculating the concentration of a standard base solution.
  • Calculating the pH at the stoichiometric point of a titration.
  • Recognizing qualitative features of a titration curve.
  • Constructing a titration curve for a polyprotic acid.
  • Choosing an appropriate indicator for a particular acid–base titration.

Key Terms:

  • Titration (acid/base) - the gradual addition of an acid to a base (or vice versa) to determine the exact amount of material required to react in stoichiometric proportions.
  • Stoichiometric point – point in titration at which the exact stoichiometric amount of titrant has been added to react with the chemical being titrated. This is also called the equivalence point.
  • Indicator (acid/base) – chemical which changes colors to signal the endpoint of a titration.
  • Titration curve – plot of pH vs. volume of titrant added for a titration.

Acid/base titrations are often used to determine the concentration of some acid (or base) solution by reacting it with a stoichiometric amount of base (or acid). The endpoint can be observed by color change. Scenarios 1 and 2 shown below illustrate that pH calculations for acid-base titrations depend upon the types of chemicals involved.