Be able to write the chemical equation for a chemical species acting as an acid: HX(aq) H + (aq) + X - (aq) Example Write the equation showing the bicarbonate

Be able to write the chemical equation for a chemical species acting as an acid:

HX(aq) H+(aq) + X-(aq)

Example Write the equation showing the bicarbonate ion acting as an acid.

HCO3-(aq) H+(aq) + CO3

2-(aq)

Be able to write the chemical equation for a chemical species acting as a base:

X-(aq) + H2O(l) OH-(aq) + HX(aq)

Example Write the equation showing the bicarbonate ion acting as a base.

HCO3-(aq) + H2O(l) OH-(aq) + H2CO3(aq)

The Common Ion EffectWhat is the pH of a solution made by adding 0.30 mol of acetic acid to enough water to make 250 mL of solution? Ka is 1.8 x 10-5 at 25°C.

CH3COOH(aq) CH3COO-(aq) + H+(aq)

initial 0.30/0.250 M 0 M 0 M

change - x M +x M +x M

equilibrium 1.2 - x M x M x M

Ka = [CH3COO-][H+] = x2 = 1.8 x 10-5

[CH3COOH] (1.2 - x)

assume x<<1.2 x2 = 2.16 x 10-5 x = 0.0046 M (<<1.2) pH = -log(0.0046) = 2.33

verify your assumption

The Common Ion EffectWhat is the pH of a solution made by adding 0.30 mol of acetic acid and 0.30 mol of sodium acetate to enough water to make 250 mL of solution?

CH3COOH CH3COO-(aq) + H+(aq)

initial 0.30/0.250 M 0.30/0.250 M 0 M

change - x M +x M +x M

equilibrium 1.2 - x M 1.2 + x M x M

Ka = [CH3COO-][H+] = (1.2 + x)x = 1.8 x 10-5

[CH3COOH] (1.2 - x) (assume x<<1.2)

x = 1.8 x 10-5 M (x<<1.2) pH = -log(1.8 x 10-5) = 4.74


The Common Ion EffectAdding acetate ion shifted the equilibrium to the left, decreasing [H+] and increasing pH.


Common Ion Effect

The extent of ionization of any weak electrolyte is decreased by adding to the solution a strong electrolyte that has an ion in common with the weak electrolyte.

adding C2H3O2-

The Common Ion Effecta) Find the solubility of calcium phosphate in water at

25°C.

b) Find the solubility of calcium phosphate in 0.100M sodium phosphate at 25°C.


25°C.

What information do you need?

formula for calcium phosphate

source: your head (I hope)

the equation for the equilibrium

source: your head

the equilibrium constant

source: Appendix D


25°C.

Ca3(PO4)2(s) 3Ca2+(aq) + 2PO43-(aq)

Ksp = 2.0 x 10-

29

I present

C lose x mol +3x M +2x M

E present 3x M 2x M

2.0 x 10-29 = 108x5 x = 7.1 x 10-7 M

The Common Ion Effectb) Find the solubility of calcium phosphate in 0.100M

sodium phosphate at 25°C.

Ca3(PO4)2(s) 3Ca2+(aq) + 2PO43-(aq)

Ksp = 2.0 x 10-

29

I present 0.100 M

C lose x mol +3x M +2xM

E present 3x M (0.100+2x) M

Assume 2x << .100

2.0 x 10-29 = .27x3 x = 4.2 x 10-10 M

Check: 8.4 x 10-10 << .1 (100,000,000 times smaller)


25°C.

7.1 x 10-7 M

b) Find the solubility of calcium phosphate in 0.100M sodium phosphate at 25°C.

4.2 x 10-10 M

The presence of phosphate from sodium phosphate decreased the solubility of the calcium phosphate by a factor of a 1000.

Buffered SolutionsWhat is the pH of our 1.2M acetic acid solution if 10.0 mL of 1.0 M NaOH is added to it?


If NaOH is added to the solution, it reacts with some of the acetic acid:

CH3COOH(aq) + NaOH(aq) CH3COONa(aq) + H2O(l)

This has the effect of decreasing the acetic acid concentration and increasing the acetate ion concentration. PLUS, addition of the NaOH changes the volume of the system.



If NaOH is added to the solution, it reacts with some of the acetic acid and produces acetate ion:

CH3COOH(aq) + NaOH(aq) CH3COONa(aq) + H2O(l)

This is a stoichiometry problem.



initial: (0.30 - 0.010)/0.260 M 0.010/0.260 M 0 M

change: - x M + x M +x M

equilibrium: 1.115 - x M 0.0385 + x M x M

Ka = [CH3COO-][H+] = (.0385 + x)x = 1.8 x 10-5

[CH3COOH] (1.115 - x) (assume x<<0.0385)

x2 + .0385x - 2.01x10-5 = 0 x = 5.2 x 10-4 M (x<<0.0385) pH = -log(5.2 x 10-4) = 3.29


Buffered SolutionsWhat is the pH of our acetic acid/sodium acetate solution if 10.0 mL of 1.0 M NaOH is added to it?


initial: (0.30-.010)/0.260 M (0.30+.010)/0.260 M 0 M

change: - x M +x M +x M

equilibrium: 1.115 - x M 1.192 + x M x M

Ka = [CH3COO-][H+] = (1.192 + x)x = 1.8 x 10-5

[CH3COOH] (1.115 - x) (assume x<<1.115, 1.192)

x = 1.7 x 10-5 M (x<<1.115 and 1.192) pH = -log(1.7 x 10-5) = 4.77


Buffered Solutions

pH pH after addition pH after addition of 10.0 mL of 10.0 mL

of 1.0 M NaOH of 1.0 M HCl

250 mL of

1.2 M HAc 2.33 3.29 1.47

250 mL of 1.2 M HAc/Ac-

buffer soln 4.74 4.77 4.71

250 mL of

DI water 7.00 12.59 1.41

HAc is short for acetic acid.

Ac- is short for acetate.

Buffered Solutions pH pH after addition pH after addition

of 10 mL 1.0 M NaOH of 10 mL 1.0 M HCl

250 mL of

1.2 M HAc 2.33 3.29 1.47

250 mL of

1.2 M HAc/Ac 4.74 4.77 4.71

250 mL of

DI water 7.00 12.59 1.41

• The pH of the HAc solution ranged 1.47 - 3.20 with the addition of 10.0 mL of the acid or base.

Buffered Solutions pH pH after addition pH after addition

of 10 mL 1.0 M NaOH of 10 mL 1.0 M HCl

250 mL of

1.2 M HAc 2.33 3.29 1.47

250 mL of

1.2 M HAc/Ac 4.74 4.77 4.71

250 mL of

DI water 7.00 12.59 1.41

• When the same amount of acid or base was added to the HAc/Ac solution, its pH ranged 4.71 - 4.77. The addition of the acetate ion buffers the solution against changes in pH.

Buffered Solutions• contain a weak conjugate acid-base pair

such as HAc/Ac-, NH3/NH4+, H2CO3/HCO3

-, HCN/CN-.

• resist changes in pH because they contain an acid to react with OH- and a base to react with H+.

• have the ability to resist changes in pH, known as their buffer capacity, which depends on the concentrations of the members of the pair.

The pH of a buffer solution may be determined by the Henderson-Hasselbalch equation.

Henderson-Hasselbalch Equation

For the general equilibrium

HX(aq) H+(aq) + X-(aq)

the pH is given by

Henderson-Hasselbalch equation

• Let’s recalculate the pH of our 1.2 M HAc/Ac solution.

pH = pKa + log ([base]/[acid])

Ka = 1.8 x 10-5, so pKa = 4.74

pH = 4.74 + log(1.2M/1.2M) = 4.74

• After the addition of 10.0 mL 1.0 M NaOH

pH = 4.74 + log(1.192M/1.115M)

= 4.74 + 0.03 = 4.77

Henderson-Hasselbalch Equation

Buffered SolutionsYou have been asked to prepare 500.0 mL of a buffer with the pH of blood, 7.40. What chemicals and amounts will you use?

Buffers most effectively resist a change in pH in either direction if they contain equal concentrations of both members of the acid-base pair.

Find a weak acid with a Ka ≈ 10-7 (so the pKa will be 7).

Ka3(citric acid) = 4.0 x 10-7 at 25°C. pKa = 6.40

7.40 = 6.40 + log [base]/[acid]

[base]/[acid] = 10

Buffered Solutions

You have been asked to prepare 500.0 mL of a buffer with the pH of blood, 7.40. What chemicals and amounts will you use?

We chose citric acid (abbreviated H3Cit) and found that we need a base-to-acid ratio of 10. For Ka3, the base is Cit3- and the acid is HCit2-.

HCit2-(aq) Cit3-(aq) + H+(aq)

Ka3 = 4.0 x 10-7 at 25°C

pH = pKa3 + log [Cit3-] [HCit2-]

Buffered SolutionsSolution #1:

Put 1.0 mol of Na3Cit and 0.10 mol Na2HCit in a 500-mL volumetric flask and dilute to the mark with DI water.

Solution #2:

Put 0.010 mol of Na3Cit and 0.0010 mol Na2HCit in a 500-mL volumetric flask and dilute to the mark with DI water.

Both solutions will produce the correct pH…what is different about them?

Their BUFFERING CAPACITY. Solution #2 has smaller concentrations of the acid-base pair.

Titration of a Strong Acid with a Strong Base

If we neutralize an acid incrementally and monitor the pH of the solution as we add the base, the resulting data would give us a titration curve, a plot of pH versus volume of titrant added.

Titration of 0.100 M HCl

0.00

2.00

4.00

6.00

8.00

10.00

12.00

14.00

0 20 40 60 80 100

mL 0.100 M NaOH added

pH


For the neutralization of 50.0 mL of 0.100 M HCl with 0.100 M NaOH, the titration curve would be a plot of pH versus volume of NaOH added.

Titration of 0.100 M HCl

0.00

2.00

4.00

6.00

8.00

10.00

12.00

14.00

0 20 40 60 80 100


pH


For the neutralization of 50.0 mL of 0.100 M HCl with 0.100 M NaOH,

here is some useful information:

• 50.0 mL of 0.100 M HCl contain 5.00 mmol H+

• 1.0 mL of 0.100 M NaOH contains 0.10 mmol of OH-

molarity (M)


For the neutralization of 50.0 mL of 0.100 M HCl with 0.100 M NaOH, the [H+] before the equivalence point is the concentration of unreacted HCl.

After 25.0 mL of NaOH have been added, the total volume is (50.0 + 25.0) mL, and

[H+] =(5.00mmol HCl - 2.50mmol NaOH)/75 mL

= 0.033 M

pH = 1.48


For the neutralization of 50.0 mL of 0.100 M HCl with 0.100 M NaOH, the [OH-] after the equivalence point is the concentration of excess NaOH.


[OH-] = (7.50 mmol NaOH - 5.00 mmol HCl)/125 mL

= 0.0200 M

pOH = 1.70 pH = 14.00- 1.70 = 12.30

Titration of a Strong Acid with a Strong BaseTitration of 0.100 M HCl

0.00

2.00

4.00

6.00

8.00

10.00

12.00

14.00

0 20 40 60 80 100


pH The pH at the equivalence

point is 7. This is true for the titration of any strong acid by any strong base…and vice versa.

The initial pH depends solely on the concentration of the acid. Here it is 1.00.

Titration of a Weak Acid with a Strong Base

For the neutralization of 50.0 mL of 0.10 M HAc with 0.100 M NaOH, the [H+] before the equivalence point is calculated from the equilibrium expression using the concentration of unreacted HAc.


HAc Ac-(aq) + H+(aq)

(5.00-2.50)/75.0 M 2.50/75.0 M

We can use the Henderson-Hasselbalch equation:

Ka = 1.8 x 10-5 pKa = 4.74

[Ac-] = [HAc] = 2.50/75.0 M [base]/[acid] = 1.00

pH = 4.74 + log 1.00 = 4.74

Titration of a Weak Acid with a Strong BaseTitration of 0.100 M Acetic Acid

0.00

2.00

4.00

6.00

8.00

10.00

12.00

14.00

0 20 40 60 80 100


pH

The initial pH depends on the concentration of the weak acid and its Ka.


For the neutralization of 50.0 mL of 0.100 M HAc with 0.100 M NaOH, the [H+] at the equivalence point is calculated from the equilibrium expression for the base.

After 50.0 mL of NaOH have been added, the total volume is (50.0 + 50.0) mL, and all of the HAc has been changed to Ac-.

Ac-(aq) + H2O(l) HAc(aq) + OH-(aq)

initial: 5.00/100.0 M 0 M 0 M

change: - x M + x M + x M

equilibrium: 0.0500 - x M x M x M

Kb = Kw/Ka = 1.0 x 10-14/(1.8 x 10-5) = 5.6 x 10-10

[OH-][HAc] = 5.6 x 10-10 ≈ x2 x = [OH-] = 5.3 x 10-6

[Ac-] 0.0500 pOH = 5.28 pH = 14.00- 5.28 = 8.72


0.00

2.00

4.00

6.00

8.00

10.00

12.00

14.00

0 20 40 60 80 100


pH

The pH at the equivalence point is 8.72.


For the neutralization of 50.0 mL of 0.100 M HAc with 0.100 M NaOH, the [OH-] after the equivalence point is the concentration of excess NaOH.

This is exactly the same as the post-equivalence section of the strong acid-strong base titration curve.After 75.0 mL of NaOH have been added, the total volume is (50.0 + 75.0) mL, and

[OH-] = (7.50 mmol NaOH - 5.00 mmol HCl)/125 mL = 0.0200 M

pOH = 1.70 pH = 14.00- 1.70 = 12.30


0.00

2.00

4.00

6.00

8.00

10.00

12.00

14.00

0 20 40 60 80 100


pH

This part of the curve is identical to that of the HCl/NaOH curve, since in both cases the NaOH volume added is the same.

Titration of a Weak Base with a Strong Acid

The titration curve for a weak base with a strong acid is calculated similarly to that of the weak acid/strong base.

You must be careful to write the equilibrium expression correctly and use the correct Kb.

For the titration of aqueous ammonia, use

NH3(aq) + H2O(l) NH4+(aq) + OH-(aq)

Kb = 1.8 x 10-5 at 25°C

Titration of a Weak Base with a Strong AcidTitration of 0.100 M Ammonium Hydroxide

0.00

2.00

4.00

6.00

8.00

10.00

12.00

0 20 40 60 80 100

mL 0.100 M HCl added

pH

Region of Maximum Buffer Capacity

Buffering is best when [base] = [acid].

But when [base] = [acid],

pH = pKa

This occurs when the volume of NaOH added is ½ the NaOH volume needed to reach the equivalence point.

Region of Maximum Buffer CapacityTitration of 0.100 M Acetic Acid

0.00

2.00

4.00

6.00

8.00

10.00

12.00

14.00

0 20 40 60 80 100


pH

pH = pKa = 4.72. Maximum buffering capacity is here.

Titration of a Polyprotic Acid with a Strong Base

0.00

1.00

2.00

3.00

4.00

5.00

6.00

7.00

8.00

9.00

10.00

11.00

12.00

13.00

0 10 20 30 40 50 60 70 80 90 100

pH

volume of NaOH added

2nd equivalence point pH = 9.9

HA- and A2-

H2A and HA-

1st equivalence point pH = 4.5

H2A is completely

neutralized at 2nd eq pt.

Titration of a Polyprotic Acid with a Strong Base

0.00

1.00

2.00

3.00

4.00

5.00

6.00

7.00

8.00

9.00

10.00

11.00

12.00

13.00

0 10 20 30 40 50 60 70 80 90 100

pH

volume of NaOH added

halfway between 40 and 80 mL

halfway between 0 and 40 mL

pKa1= 2.0

pKa2= 7.2

H2A and HA-

HA- and A2-

Not All Equivalence Points are Obvious

Documents

Be able to write the chemical equation for a chemical species acting as an acid: HX(aq) H + (aq) + X - (aq) Example Write the equation showing the bicarbonate