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AP Chemistry Chapter 11

Ionic Equilibria

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When salts, acids, or bases are added to water they ionize. For example:

NaCl(s) → Na+(aq) + Cl-(aq)

Could an equilibrium be established for this reaction? What happens if you add a lot of salt to a small glass of water? What would you be able to see if it was at equilibrium? What would you not see if it was at equilibrium?

How would this be written at equilibrium? NaCl(s) Na+(aq) + Cl-(aq)

Are there any salts that would never come to equilibrium? What about acids or bases? Why?

For acids and bases to set up an equilibrium they must be weak by definition. When they come to equilibrium (in water) we can write equilibrium expressions for them.

IONIZATION (EQUILIBRIUM) CONSTANTS FOR MONOPROTIC ACIDS AND BASES

In general for a weak, monoprotic acid:

HA H+ + A-

so Ka = Why “Ka”?

But acids must be in water, so it can look like this:

HA + H2O H3O+ + A-

Ka = [H2O] usually considered = 1

[H+][A-][HA]

[H3O+][A-]

[HA]

A solution of nicotinic acid was found at equilibrium to contain [HA] = 0.049M, [H3O+] = [A-] = 8.4 x 10-4 M. What is the Ka and pKa?

Starting with a 0.0100 M solution of acetic acid, when added to water 4.2% ionizes. What is the Ka and pKa?

The pH of a 0.15 M solution of chloroscetic acid (ClCH2COOH) is 1.92. What is the value of Ka?

What are the equilibrium concentrations if the Ka of 0.10 M HOCl is 3.5 x 10-8?

Calculate the percent ionization of a 0.10 M acetic acid solution.

If you haven’t caught it by now:

The higher the Ka value (less negative exponent) the lower the pKa value (closer to 0).

Also, the “rule of thumb” for the simplification of the quadratic only works if Ka is “x 10-4” or a larger negative exponent. When compared to the [HA]!

Calculate the [OH-] for a 0.20 M solution of NH3. Kb is 1.8 x 10-5.

Calculate the pH for a 0.20 M solution of NH3. Kb is 1.8 x 10-5.

Calculate the % ionization for a 0.20 M solution of NH3. Kb is 1.8 x 10-5.

IONIZATION (EQUILIBRIUM) CONSTANTS FOR POLYPROTIC ACIDS AND BASES

Polyprotic acids can give off more than one H+, so it is considered to occur in steps, each with their own Ka value. For example:

H3PO4 H+ + H2PO4- Ka1 = 7.5 x 10-3

H2PO4- H+ + HPO4

2- Ka2 = 6.2 x 10-8

HPO42- H+ + PO4

3- Ka3 = 3.6 x 10-13

So if the concentrations were wanted, we would need to solve all three reactions and then add up the results. (Not hard, but definitely time consuming.)

Would anyone like a shortcut?

Ka3 = 3.6 x 10-13 and (remember) Kw = 1 x10-14, so the contribution of the third equation to the total of the acid is only a little larger than water’s own contribution.

Specifically, if we started with 0.10 M H3PO4, the third ionization would only add 0.00000000000000000093 M H+ to the total.

Rule of Thumb: “x 10-9” and larger negative exponents indicate that reaction does not contribute in any noticeable way to the total (the additional amount will be rounded away).

Calculate the concentrations in 0.10 M H2SO4. Ka2 = 1.2 x 10-2

SALTS AND HYDROLYSIS

When salts ionize it is possible one of the ions formed would react with water in a side reaction, but this reaction could change the pH.

Bronstead Lowry definitions of acids and bases give us conjugate acid-base pairs. Strong acids/bases give weak conjugate bases/acids as products, and weak acids/bases give strong conjugate bases/acids.

It is the conjugates of weak acids/bases that are the ones that typically react with water.

For nonexample:NaCl → Na+ + Cl-

Na+ would be the cation of NaOH (a strong base) so it would be a weak acid, and would not react with water.

Cl- would be the anion of HCl (a strong acid) so it would be a weak base, and would not react with water.

Strong base - weak acid example:NaC2H3O2 → Na+ + C2H3O2

-

Na+ would be the cation of NaOH (a strong base) so it would be a weak acid, and would not react with water.

C2H3O2- would be the anion of HC2H3O2 (a weak

acid) so it would be a strong base, and would react with water, like this:

C2H3O2- + H2O HC2H3O2 + OH-

What type of solution is the result? What would be the pH? Why would we use Kb?

Strong acid - weak base example:NH4Cl → NH4

+ + Cl-

NH4+ would be the cation of NH4OH (a weak base)

so it would be a strong acid, and would react with water, like this:

NH4+ + H2O NH3 + H3O+

Cl- would be the anion of HCl (a strong acid) so it would be a weak base, and would not react with water.

What type of solution is the result? What would be the pH? Ka or Kb?

Once the side reaction is known, the amount of acid or base formed can be calculated just like we did earlier.

It might be helpful to know:

Kw = 1 x10-14 = KaKb

so if you know Ka and need Kb or vise-versa you can calculate what you need

Weak acid - weak base:

If the salt of a strong base with weak acid gave a basic solution, and the salt of a strong acid with a weak base gave an acidic solution, then

What happens if they are both weak?if Ka = Kb, then neutralif Ka > Kb, then acidic (acid was stronger than base, but still weak)if Kb > Ka, then basic(base was stronger than acid, but still weak)

Let’s recap:

NaClNa+ + H2O NaOH + H+

Cl- + H2O HCl + OH-

Let’s recap:

NaClNa+ + H2O NaOH + H+

Cl- + H2O HCl + OH-

Strong Base

Let’s recap:

NaClNa+ + H2O NaOH + H+

Cl- + H2O HCl + OH-

Strong Base = No Reaction

Let’s recap:

NaClNa+ + H2O NaOH + H+

Cl- + H2O HCl + OH-

Strong Base = No Reaction

Strong Acid

Let’s recap:

NaClNa+ + H2O NaOH + H+

Cl- + H2O HCl + OH-

Strong Base = No Reaction

Strong Acid = No Reaction

Let’s recap:

NaClNa+ + H2O NaOH + H+

Cl- + H2O HCl + OH-

Strong Base = No Reaction

Strong Acid = No Reaction

Conclusion: Neither Acidic or Basic = Neutral

Let’s recap:

NH4ClNH4

+ + H2O NH4OH + H+

Cl- + H2O HCl + OH-

Let’s recap:

NH4ClNH4

+ + H2O NH4OH + H+

Cl- + H2O HCl + OH-

Weak Base

Let’s recap:

NH4ClNH4

+ + H2O NH4OH + H+

Cl- + H2O HCl + OH-

Weak Base = Yes Reaction

Let’s recap:

NH4ClNH4

+ + H2O NH4OH + H+

Cl- + H2O HCl + OH-

Weak Base = Yes Reaction

Strong Acid

Let’s recap:

NH4ClNH4

+ + H2O NH4OH + H+

Cl- + H2O HCl + OH-

Weak Base = Yes Reaction

Strong Acid = No Reaction

Let’s recap:

NH4ClNH4

+ + H2O NH4OH + H+

Cl- + H2O HCl + OH-

Weak Base = Yes Reaction

Strong Acid = No Reaction

Conclusion: as H+ is produced = acidic

Let’s recap:

NaC2H3O2

Na+ + H2O NaOH + H+

C2H3O2- + H2O HC2H3O2 + OH-

Let’s recap:

NaC2H3O2

Na+ + H2O NaOH + H+

C2H3O2- + H2O HC2H3O2 + OH-

Strong Base

Let’s recap:

NaC2H3O2

Na+ + H2O NaOH + H+

C2H3O2- + H2O HC2H3O2 + OH-

Strong Base = No Reaction

Let’s recap:

NaC2H3O2

Na+ + H2O NaOH + H+

C2H3O2- + H2O HC2H3O2 + OH-

Strong Base = No Reaction

Weak Acid

Let’s recap:

NaC2H3O2

Na+ + H2O NaOH + H+

C2H3O2- + H2O HC2H3O2 + OH-

Strong Base = No Reaction

Weak Acid = Yes Reaction

Let’s recap:

NaC2H3O2

Na+ + H2O NaOH + H+

C2H3O2- + H2O HC2H3O2 + OH-

Strong Base = No Reaction

Weak Acid = Yes Reaction

Conclusion: as OH- is produced = basic

Let’s recap:

NH4C2H3O2 – Your Turn…

Let’s recap:

NH4C2H3O2

NH4+ + H2O NH4OH + H+

C2H3O2- + H2O HC2H3O2 + OH-

Let’s recap:

NH4C2H3O2

NH4+ + H2O NH4OH + H+

C2H3O2- + H2O HC2H3O2 + OH-

Weak Base = Yes Reaction

Weak Acid = Yes Reaction

Let’s recap:

NH4C2H3O2

NH4+ + H2O NH4OH + H+

C2H3O2- + H2O HC2H3O2 + OH-

Weak Base = Yes Reaction

Weak Acid = Yes Reaction

Conclusion: we cannot yet say if it is acidic or basic!

As Kb = 1.8 x 10-5 and Ka = 1.8 x 10-5 it is neutral,otherwise the larger K wins! (larger = small -exponent

More Ionic Equilibria

(Acid-Base Interactions)

Much of life involves acids and bases, including your body. •What is in your stomach?•What must your pancreas produce to avoid dissolving your intestines? •When carbon dioxide is dissolved in water (blood) it forms carbonic acid. What must also be in our blood to avoid dissolving our vessels?•There are people who get sick after drinking a coke (coke contain acids). What must they be lacking?

Much of life, then, involves buffers.

A buffer contains a conjugate acid-base pair with both the acid and base in reasonable concentrations. (By Bronstead-Lowry definition.)

The acidic part would react with added base.

The basic part would react with added acid.

For example: acetic acid and sodium acetateNaC2H3O2 → Na+ + C2H3O2

-

HC2H3O2 H+ + C2H3O2-

If OH- were added, what would react?

OH- + H+ H2O How would pH be affected?

If H+ were added, what would react?

H+ + C2H3O2- HC2H3O2 How would pH be

affected?

Notice that the addition of acetic acid and sodium acetate both give off acetate ion.

NaC2H3O2 → Na+ + C2H3O2-

HC2H3O2 H+ + C2H3O2-

What would having this much acetate ion do to the equilibrium of acetic acid? What would this mean to the amount of acetic acid that would ionize?

When a solution of a weak electrolyte (weak acid or base or not very soluble salt) is altered by adding one of the ions from another source, the ionization of that weak electrolyte is decreased. This is called the common-ion effect.

Why is the common-ion effect just an application of LeChatelier’s Principle?

NaC2H3O2 → Na+ + C2H3O2-

HC2H3O2 H+ + C2H3O2-

Why is it just the acetic acid that is affected?What happens to the pH of the mixture compared to the pH of acetic acid?

Let’s prove it: what is the pH of a 0.10 M acetic acid solution if Ka = 1.8 x 10-5?

Let’s prove it: what is the pH of a 0.10 M acetic acid and 0.20 M sodium acetate solution if Ka = 1.8 x 10-5?

Shortcut anyone?Thanks to the Henderson-Hasselbalch equation…

and by logic…

pH = pKa + log[salt]

[acid]

pOH = pKb + log[salt]

[base]

What is the pH of a 0.10 M acetic acid and 0.20 M sodium acetate solution if Ka = 1.8 x 10-5?Using the H-H equation…

The purpose of a buffer system in our bodies or anywhere else is to minimize the change that adding an acid or base causes to the pH. For example: adding 0.010 mols of NaOH to…

1L water makes pH 7→ 120.1 M acetic acid makes pH 2.89 → 3.78

What about 1L of acetic acid/acetate buffer?Start by finding the initial pH, and then the change…

What is the original pH of a 1.00 L solution that is 0.100 M acetic acid and 0.100 M sodium acetate.

What is the new pH of a 1.00 L solution that is 0.100 M acetic acid and 0.100 M sodium acetate if 0.010 mol solid NaOH is added. (No volume change.)

Buffers are prepared by mixing solutions, but there are several methods to do this. Regardless of the method remember:

When mixing solutions the final volume is the sum of the two solutions’ volume, so the concentrations will need to be recalculated!

Calculate the pH of a buffer solution prepared by mixing 200. mL of 0.10 M NaF and 100. mL of 0.050 M HF. Ka = 7.2 x 10-4

Calculate the grams of NH4Cl that must be used to prepare 500. mL of a buffer solution that is 0.10 M in NH3 and has a pH of 9.15.

Calculate the pH of a solution obtained by mixing 400. mL of a 0.200 M acetic acid solution and 100. mL of a 0.300 M NaOH soln.

What is the concentration of a 25.0 mL of ammonia solution if it took 30.0 mL of 1.0 M HCl to titrate it to the equivalence point?

What will be the pH of the titrated ammonia-HCl mixture at the equivalence point?

What will be the pH of the titrated ammonia-HCl mixture at the half equivalence point?

What was the original pH of the ammonia?

Draw a titration curve for all of this.

Titration Curves and Buffers

Titration Curves graphically represent the results on pH of adding acid to base or base to acid. Points to note:

Equivalence Point - the point where [H+]=[OH-]

Indicators change at the end point, ideally identical to the equivalence point.

Half Equivalence Point - the point at where the pH = pKa, so Ka could easily be calculated

Strong Acid-Strong Base Curves

Where should the endpoint occur? Where is the half equivalent point?

Strong Acid-Strong Base Curve

Weak Acid-Weak Base Curve

Triprotic Acid-Strong Base Curve