Common Ion Effect Shift in equilibrium position that occurs because of the addition of an ion already involved in the equilibrium reaction. An application.

Common Ion Effect Shift in equilibrium position that occurs because of the addition of an ion already involved in the equilibrium reaction. An application of Le Châtelier’s principle. Copyright © Cengage Learning. All rights reserved

Example HCN(aq) + H2O(l) H3O+(aq) + CN-(aq)
Addition of NaCN will shift the equilibrium to the left because of the addition of CN-, which is already involved in the equilibrium reaction. A solution of HCN and NaCN is less acidic than a solution of HCN alone.

Key Points about Buffered Solutions
Buffered Solution – resists a change in pH, when either hydroxide ion or protons added. They are weak acids or bases containing its salts (a common ion) HF and NaF or NH3 and NH4Cl. After addition of strong acid or base, deal with stoichiometry first, then the equilibrium. Assume that H+ or OH- goes to completion Buffered solutions are weak acids and bases. Copyright © Cengage Learning. All rights reserved

Adding an Acid to a Buffer
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Buffers To play movie you must be in Slide Show Mode
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Solving Problems with Buffered Solutions
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The pH of a Buffered solution I
Major species in Weak acid and its salt (H2C2H3O2, NaC2H3O2 and HNO2, NaNO2) are acetic acid acetate ion , sodium ion and water, but Weak acid and its conjugate base control the pH. Na+ is neither acid nor basic. H2O is either very weak acid or base. If the concentration of acid and its salt is same, the concentration of [H+] is the same as Ka. No change in pH. Problem 15.31

pH Changes in Buffer solutions
If the strong base is added, to the question the major species are weak acid, its conjugate base, Na+ ion OH- ion and H2O. Solution contains relatively large amount of very strong base (OH-), which has greater affinity for proton. The best source is acetic acid. OH-(aq) + HC2H3O2 (aq)→H2O(l) + C2H3O-(aq) OH-(aq) + HNO2 (aq)→H2O(l) + NO2-(aq) Although the acids are weak OH- is such a strong base that reaction above proceed to completion untill all the OH- ions are consumed. It involves two steps. A) The estoichiometry problem B) The equilibrium problem Problem 15.33

Buffering: How Does It Work?

Henderson–Hasselbalch Equation
For a particular buffering system (conjugate acid–base pair), all solutions that have the same ratio [A–] / [HA] will have the same pH. Copyright © Cengage Learning. All rights reserved

EXERCISE! What is the pH of a buffer solution that is 0.45 M acetic acid (HC2H3O2) and 0.85 M sodium acetate (NaC2H3O2)? The Ka for acetic acid is 1.8 × 10–5. pH = 5.02 pH = –logKa + log([C2H3O2–] / [HC2H3O2]) = –log(1.8 × 10–5) + log(0.85 M / 0.45 M) = 5.02 Copyright © Cengage Learning. All rights reserved

Buffered Solution Characteristics
Buffers contain relatively large concentrations of a weak acid and corresponding conjugate base. Added H+ reacts to completion with the weak base. Added OH- reacts to completion with the weak acid. Copyright © Cengage Learning. All rights reserved

Buffered Solution Characteristics
The pH in the buffered solution is determined by the ratio of the concentrations of the weak acid and weak base. As long as this ratio remains virtually constant, the pH will remain virtually constant. This will be the case as long as the concentrations of the buffering materials (HA and A– or B and BH+) are large compared with amounts of H+ or OH– added. Copyright © Cengage Learning. All rights reserved

pH of the buffer solution is determined by the ratio of [A-/[HA]
The amount of protons or hydroxide ions the buffer can absorb without a significant change in pH. pH of the buffer solution is determined by the ratio of [A-/[HA] The capacity of the buffer solution is determined by the magnitudes of [HA] and [A–]. A buffer with large capacity contains large concentrations of the buffering components. Let us suppose a solution of large concentration of acetate ion and small concentration of acetic acid. Addition of proton to form acetic acid will produce a relatively large percentage change in the concentration of acetic acid and so will produce relatively large change in the ratio of [CH3COO-]/CH3COOH] Table 5.1. It will produce large change in pH. If OH- ion are added to remove some acetic acid the percent change in the concentration of acetic acid is large. We want to avoid this situation for the most effective buffering. Copyright © Cengage Learning. All rights reserved

Optimal buffering occurs when [HA] is equal to [A–].
At this condition the ratio [A–] / [HA] is most resistant to change, when H+ or OH– is added to the buffered solution. When choosing the buffering component, we want [A–] / [HA] to equal 1. The pKa of the acid to be used in the buffer should be as close as possible to the desired pH. For example we need a buffered solution with a pH of 4 the most effective buffering will occur when [HA] = [A-]. Ratio = 1 for the most effective buffer 4.0=pKa+log (1)= pKa +0 pKa = 4.00 Best choice of weak acid is one that has pKa = 4.00 or Ka = 1.0 x 10-4 Copyright © Cengage Learning. All rights reserved

Titration Curve Plotting the pH of the solution being analyzed as a function of the amount of titrant added. Equivalence (Stoichiometric) Point – point in the titration when enough titrant has been added to react exactly with the substance in solution being titrated. Mole is inconveniently large therefore we use milli-mole. M= mol solute/L solution = mmol solute/mL solution Number of mole = volume in mL/molarity Calculate t Copyright © Cengage Learning. All rights reserved

Neutralization of a Strong Acid with a Strong Base
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The pH Curve for the Titration of 50. 0 mL of 0. 200 M HNO3 with 0
The pH Curve for the Titration of 50.0 mL of M HNO3 with M NaOH Copyright © Cengage Learning. All rights reserved

The pH Curve for the Titration of 100. 0 mL of 0. 50 M NaOH with 1
The pH Curve for the Titration of mL of 0.50 M NaOH with 1.0 M HCI Copyright © Cengage Learning. All rights reserved

Weak Acid–Strong Base Titration
Step 1: A stoichiometry problem (reaction is assumed to run to completion) then determine concentration of acid remaining and conjugate base formed. Step 2: An equilibrium problem (determine position of weak acid equilibrium and calculate pH). Copyright © Cengage Learning. All rights reserved

CONCEPT CHECK! Consider a solution made by mixing 0.10 mol of HCN (Ka = 6.2 × 10–10) with mol NaOH in 1.0 L of aqueous solution. What are the major species immediately upon mixing (that is, before a reaction)? HCN, Na+, OH–, H2O Major Species: HCN, Na+, OH-, H2O. Copyright © Cengage Learning. All rights reserved

Let’s Think About It… Why isn’t NaOH a major species?
Why aren’t H+ and CN– major species? List all possibilities for the dominant reaction. NaOH is a strong base and completely dissociates. HCN is a weak acid and does not dissociate very much. Copyright © Cengage Learning. All rights reserved

Let’s Think About It… The possibilities for the dominant reaction are:
H2O(l) + H2O(l) H3O+(aq) + OH–(aq) HCN(aq) + H2O(l) H3O+(aq) + CN–(aq) HCN(aq) + OH–(aq) CN–(aq) + H2O(l) Na+(aq) + OH–(aq) NaOH Na+(aq) + H2O(l) NaOH + H+(aq) Copyright © Cengage Learning. All rights reserved

Let’s Think About It… How do we decide which reaction controls the pH?
H2O(l) + H2O(l) H3O+(aq) + OH–(aq) HCN(aq) + H2O(l) H3O+(aq) + CN–(aq) HCN(aq) + OH–(aq) CN–(aq) + H2O(l) In general, the best acid (HCN in this case) will tend to react with the best base (OH- will always be the best base, if present). Discuss how to calculate the K value for this reaction (note -- it is not a Ka or Kb expression). In this case, K = Ka / Kw = 62,000. We can assume the reaction goes to completion. In this problem, OH- is limiting.

Let’s Think About It… HCN(aq) + OH–(aq) CN–(aq) + H2O(l)
What are the major species after this reaction occurs? HCN, CN–, H2O, Na+ After the reaction takes place, the major species are: HCN, CN-, H2O, Na+ Copyright © Cengage Learning. All rights reserved

Let’s Think About It… Now you can treat this situation as before.
List the possibilities for the dominant reaction. Determine which controls the pH. Copyright © Cengage Learning. All rights reserved

CONCEPT CHECK! Calculate the pH of a solution made by mixing 0.20 mol HC2H3O2 (Ka = 1.8 × 10–5) with mol NaOH in 1.0 L of aqueous solution. Major Species: HC2H3O2, Na+, OH-, H2O Possibilities for reactions: 1) HC2H3O2(aq) + H2O  H3O+(aq) + C2H3O2-(aq) 2) H2O + H2O  H3O+(aq) + OH-(aq) 3) HC2H3O2(aq) + OH-(aq)  H2O + C2H3O2-(aq) Reaction 3 is most likely. In general, the best acid (HC2H3O2 in this case) will tend to react with the best base (OH- will always be the best base, if present). Discuss how to calculate the K value for this reaction (note -- it is not a Ka or Kb expression). In this case, K = Ka/ Kw = 1.8 x 109. We can assume the reaction goes to completion. In this problem, OH- is limiting. After the reaction takes place, the major species are: HC2H3O2, C2H3O2-, H2O, Na+ The primary reaction could be: 3) C2H3O2-(aq) + H2O  HC2H3O2(aq) + OH-(aq) We will see that reactions 1 and 3 will give us the same answer (as long as "x" is negligible in calculations, or solved for exactly). In this case, though, Ka (reaction 1) is greater than Kb (reaction 3) and should be used. When students solve this, make sure they include the initial concentration of the conjugate base (they tend to forget and call it zero). pH = 3.99 Copyright © Cengage Learning. All rights reserved

Let’s Think About It… What are the major species in solution?
Na+, OH–, HC2H3O2, H2O Why isn’t NaOH a major species? Why aren’t H+ and C2H3O2– major species? NaOH is a strong base and completely dissociates. HC2H3O2 is a weak acid and does not dissociate very much. Copyright © Cengage Learning. All rights reserved

Let’s Think About It… What are the possibilities for the dominant reaction? H2O(l) + H2O(l) H3O+(aq) + OH–(aq) HC2H3O2(aq) + H2O(l) H3O+(aq) + C2H3O2–(aq) HC2H3O2(aq) + OH–(aq) C2H3O2–(aq) + H2O(l) Na+(aq) + OH–(aq) NaOH(aq) Na+(aq) + H2O(l) NaOH + H+(aq) Which of these reactions really occur? See Slide 12. Copyright © Cengage Learning. All rights reserved

Let’s Think About It… Which reaction controls the pH?
H2O(l) + H2O(l) H3O+(aq) + OH–(aq) HC2H3O2(aq) + H2O(l) H3O+(aq) + C2H3O2–(aq) HC2H3O2(aq) + OH–(aq) C2H3O2–(aq) + H2O(l) How do you know? See Slide 12. Copyright © Cengage Learning. All rights reserved

Let’s Think About It… K = 1.8 × 109 HC2H3O2(aq) + OH– C2H3O2–(aq)
Before 0.20 mol mol Change –0.030 mol –0.030 mol mol After 0.17 mol 0.030 mol K = 1.8 × 109 Copyright © Cengage Learning. All rights reserved

Steps Toward Solving for pH
HC2H3O2(aq) + H2O H3O+ + C2H3O2-(aq) Initial 0.170 M ~0 0.030 M Change –x +x Equilibrium 0.170 – x x x Ka = 1.8 × 10–5 pH = 3.99 Copyright © Cengage Learning. All rights reserved

EXERCISE! Calculate the pH of a mL solution of M acetic acid (HC2H3O2), which has a Ka value of 1.8 × 10–5. pH = 2.87 pH = 2.87 Major Species: HC2H3O2, H2O Dominant reaction: HC2H3O2(aq) + H2O  H3O+(aq) + C2H3O2-(aq) The mL is not necessary to know for this problem. Copyright © Cengage Learning. All rights reserved

CONCEPT CHECK! Calculate the pH of a solution made by mixing mL of a M solution of acetic acid (HC2H3O2), which has a Ka value of 1.8 × 10–5, and 50.0 mL of a 0.10 M NaOH solution. pH = 4.74 pH = 4.74 Major Species: HC2H3O2, Na+, OH-, H2O Dominant reaction: HC2H3O2(aq) + OH-(aq)  H2O + C2H3O2-(aq) In general, the best acid (HC2H3O2 in this case) will tend to react with the best base (OH- will always be the best base, if present). Discuss how to calculate the K value for this reaction (note -- it is not a Ka or Kb expression). In this case, K = Ka/Kw = 1.8 x 109. We can assume the reaction goes to completion. In this problem, OH- is limiting. After the reaction takes place, the major species are: HC2H3O2, C2H3O2-, H2O Primary reaction is: HC2H3O2(aq) + H2O  H3O+(aq) + C2H3O2-(aq) When students solve this, make sure they include the initial concentration of the conjugate base (they tend to forget and call it zero). Copyright © Cengage Learning. All rights reserved

CONCEPT CHECK! Calculate the pH of a solution at the equivalence point when mL of a M solution of acetic acid (HC2H3O2), which has a Ka value of 1.8 × 10–5, is titrated with a 0.10 M NaOH solution. pH = 8.72 pH = 8.72 Major Species: HC2H3O2, Na+, OH-, H2O Dominant reaction: HC2H3O2(aq) + OH-(aq)  H2O + C2H3O2-(aq) In general, the best acid (HC2H3O2 in this case) will tend to react with the best base (OH- will always be the best base, if present). At the equivalence point, there are the same number of moles of HC2H3O2(aq) + OH-(aq) to exactly react due to the 1:1 ratio in the balanced equation. This turns out to be mol of both the acid and base in this case. This means that mL of NaOH had to be added. After the reaction takes place, the major species are: C2H3O2-, H2O, Na+ Primary reaction is: C2H3O2-(aq) + H2O  OH-(aq) + HC2H3O2(aq) Now use an ICE chart to determine the equilibrium concentrations and finally the pH. Take note that the initial concentration of C2H3O2-(aq) is M (0.010 mol / L). Copyright © Cengage Learning. All rights reserved

The pH Curves for the Titrations of 50. 0-mL Samples of 0
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 Copyright © Cengage Learning. All rights reserved

Marks the end point of a titration by changing color.
The equivalence point is not necessarily the same as the end point (but they are ideally as close as possible). Copyright © Cengage Learning. All rights reserved

The Acid and Base Forms of the Indicator Phenolphthalein

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Common Ion Effect Shift in equilibrium position that occurs because of the addition of an ion already involved in the equilibrium reaction. An application.

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Common Ion Effect Shift in equilibrium position that occurs because of the addition of an ion already involved in the equilibrium reaction. An application.

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Presentation on theme: "Common Ion Effect Shift in equilibrium position that occurs because of the addition of an ion already involved in the equilibrium reaction. An application."— Presentation transcript:

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