1. Chapter 16 Aqueous Ionic Equilibrium
Chemistry: A Molecular Approach, 1st Ed. Nivaldo Tro
Chapter 16 Aqueous Ionic Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
2008, Prentice Hall

2. Buffers
buffers are solutions that resist changes in pH when an acid or base is added
they act by neutralizing the added acid or base
but just like everything else, there is a limit to what they can do, eventually the pH changes
many buffers are made by mixing a solution of a weak acid with a solution of soluble salt containing its conjugate base anion
Tro, Chemistry: A Molecular Approach

3. Making an Acid Buffer
Tro, Chemistry: A Molecular Approach

4. How Acid Buffers Work HA(aq) + H2O(l)  A−(aq) + H3O+(aq)
buffers work by applying Le Châtelier’s Principle to weak acid equilibrium
buffer solutions contain significant amounts of the weak acid molecules, HA – these molecules react with added base to neutralize it
you can also think of the H3O+ combining with the OH− to make H2O; the H3O+ is then replaced by the shifting equilibrium
the buffer solutions also contain significant amounts of the conjugate base anion, A− - these ions combine with added acid to make more HA and keep the H3O+ constant
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5. Common Ion Effect HA(aq) + H2O(l)  A−(aq) + H3O+(aq)
adding a salt containing the anion, NaA, that is the conjugate base of the acid (the common ion) shifts the position of equilibrium to the left
this causes the pH to be higher than the pH of the acid solution
lowering the H3O+ ion concentration
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6. Common Ion Effect
Tro, Chemistry: A Molecular Approach

7. Ex 16. 1 - What is the pH of a buffer that is 0. 100 M HC2H3O2 and 0
Ex What is the pH of a buffer that is M HC2H3O2 and M NaC2H3O2?
Write the reaction for the acid with water
Construct an ICE table for the reaction
Enter the initial concentrations – assuming the [H3O+] from water is ≈ 0
HC2H3O2 + H2O  C2H3O2 + H3O+
[HA]
[A-]
[H3O+]
initial
0.100
≈ 0
change
equilibrium
Tro, Chemistry: A Molecular Approach

8. Ex 16. 1 - What is the pH of a buffer that is 0. 100 M HC2H3O2 and 0
Ex What is the pH of a buffer that is M HC2H3O2 and M NaC2H3O2?
HC2H3O2 + H2O  C2H3O2 + H3O+
represent the change in the concentrations in terms of x
sum the columns to find the equilibrium concentrations in terms of x
substitute into the equilibrium constant expression
[HA]
[A-]
[H3O+]
initial
0.100
change
equilibrium x +x +x
0.100 x x x
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9. Ex 16. 1 - What is the pH of a buffer that is 0. 100 M HC2H3O2 and 0
Ex What is the pH of a buffer that is M HC2H3O2 and M NaC2H3O2?
Ka for HC2H3O2 = 1.8 x 10-5
determine the value of Ka
since Ka is very small, approximate the [HA]eq = [HA]init and [A−]eq = [A−]init solve for x
[HA]
[A-]
[H3O+]
initial
0.100
≈ 0
change -x +x
equilibrium x
0.100 x x
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10. Ex 16. 1 - What is the pH of a buffer that is 0. 100 M HC2H3O2 and 0
Ex What is the pH of a buffer that is M HC2H3O2 and M NaC2H3O2?
[HA]
[A-]
[H3O+]
initial
0.100
≈ 0
change -x +x
equilibrium
1.8E-5
substitute [H3O+] into the formula for pH and solve
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11. Practice - What is the pH of a buffer that is 0. 14 M HF (pKa = 3
Practice - What is the pH of a buffer that is 0.14 M HF (pKa = 3.15) and M KF?
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12. Practice - What is the pH of a buffer that is 0. 14 M HF (pKa = 3
Practice - What is the pH of a buffer that is 0.14 M HF (pKa = 3.15) and M KF?
Write the reaction for the acid with water
Construct an ICE table for the reaction
Enter the initial concentrations – assuming the [H3O+] from water is ≈ 0
HF + H2O  F + H3O+
[HA]
[A-]
[H3O+]
initial
0.14
0.071
≈ 0
change
equilibrium
Tro, Chemistry: A Molecular Approach

13. Practice - What is the pH of a buffer that is 0. 14 M HF (pKa = 3
Practice - What is the pH of a buffer that is 0.14 M HF (pKa = 3.15) and M KF?
HF + H2O  F + H3O+
represent the change in the concentrations in terms of x
sum the columns to find the equilibrium concentrations in terms of x
substitute into the equilibrium constant expression
[HA]
[A-]
[H3O+]
initial
0.14
0.071
change
equilibrium x +x +x
0.14 x x x
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14. Practice - What is the pH of a buffer that is 0. 14 M HF (pKa = 3
Practice - What is the pH of a buffer that is 0.14 M HF (pKa = 3.15) and M KF?
Ka for HF = 7.0 x 10-4
determine the value of Ka
since Ka is very small, approximate the [HA]eq = [HA]init and [A−]eq = [A−]init solve for x
[HA]
[A-]
[H3O+]
initial
0.14
0.071
≈ 0
change -x +x
equilibrium
0.012
0.100 x
0.14 x x
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15. Practice - What is the pH of a buffer that is 0. 14 M HF (pKa = 3
Practice - What is the pH of a buffer that is 0.14 M HF (pKa = 3.15) and M KF?
Ka for HF = 7.0 x 10-4
check if the approximation is valid by seeing if x < 5% of [HC2H3O2]init
[HA]
[A2-]
[H3O+]
initial
0.14
0.071
≈ 0
change -x +x
equilibrium x
x = 1.4 x 10-3
the approximation is valid
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16. Practice - What is the pH of a buffer that is 0. 14 M HF (pKa = 3
Practice - What is the pH of a buffer that is 0.14 M HF (pKa = 3.15) and M KF?
[HA]
[A2-]
[H3O+]
initial
0.14
0.071
≈ 0
change -x +x
equilibrium
0.072
1.4E-3
substitute x into the equilibrium concentration definitions and solve
0.14 x x x
x = 1.4 x 10-3
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17. Practice - What is the pH of a buffer that is 0. 14 M HF (pKa = 3
Practice - What is the pH of a buffer that is 0.14 M HF (pKa = 3.15) and M KF?
[HA]
[A-]
[H3O+]
initial
0.14
0.071
≈ 0
change -x +x
equilibrium
0.072
1.4E-3
substitute [H3O+] into the formula for pH and solve
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18. Henderson-Hasselbalch Equation
calculating the pH of a buffer solution can be simplified by using an equation derived from the Ka expression called the Henderson-Hasselbalch Equation
the equation calculates the pH of a buffer from the Ka and initial concentrations of the weak acid and salt of the conjugate base
as long as the “x is small” approximation is valid
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19. How Much Does the pH of a Buffer Change When an Acid or Base Is Added?
though buffers do resist change in pH when acid or base are added to them, their pH does change
calculating the new pH after adding acid or base requires breaking the problem into 2 parts
a stoichiometry calculation for the reaction of the added chemical with one of the ingredients of the buffer to reduce its initial concentration and increase the concentration of the other
added acid reacts with the A− to make more HA
added base reacts with the HA to make more A−
an equilibrium calculation of [H3O+] using the new initial values of [HA] and [A−]
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20. Ex What is the pH of a buffer that has mol HC2H3O2 and mol NaC2H3O2 in 1.00 L that has mol NaOH added to it?
If the added chemical is a base, write a reaction for OH− with HA. If the added chemical is an acid, write a reaction for it with A−.
Construct a stoichiometry table for the reaction
HC2H3O2 + OH−  C2H3O2 + H2O HA A-
OH−
mols Before
0.100
mols added -
0.010
mols After
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21. Ex What is the pH of a buffer that has mol HC2H3O2 and mol NaC2H3O2 in 1.00 L and than has mol NaOH added to it?
HC2H3O2 + OH−  C2H3O2 + H2O
Fill in the table – tracking the changes in the number of moles for each component HA A-
OH−
mols Before
0.100
≈ 0
mols added -
0.010
mols After
0.090
0.110
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22. Ex What is the pH of a buffer that has mol HC2H3O2 and mol NaC2H3O2 in 1.00 L that has mol NaOH added to it?
Write the reaction for the acid with water
Construct an ICE table for the reaction
Enter the initial concentrations – assuming the [H3O+] from water is ≈ 0, and using the new molarities of the [HA] and [A−]
HC2H3O2 + H2O  C2H3O2 + H3O+
[HA]
[A-]
[H3O+]
initial
0.090
0.110
≈ 0
change
equilibrium
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23. Ex What is the pH of a buffer that has mol HC2H3O2 and mol NaC2H3O2 in 1.00 L that has mol NaOH added to it?
HC2H3O2 + H2O  C2H3O2 + H3O+
represent the change in the concentrations in terms of x
sum the columns to find the equilibrium concentrations in terms of x
substitute into the equilibrium constant expression
[HA]
[A-]
[H3O+]
initial
0.090
0.110
change
equilibrium x +x +x
0.090 x x x
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24. Ex What is the pH of a buffer that has mol HC2H3O2 and mol NaC2H3O2 in 1.00 L that has mol NaOH added to it?
Ka for HC2H3O2 = 1.8 x 10-5
determine the value of Ka
since Ka is very small, approximate the [HA]eq = [HA]init and [A−]eq = [A−]init solve for x
[HA]
[A-]
[H3O+]
initial
0.100
≈ 0
change -x +x
equilibrium
0.090
0.110 x
0.090 x x
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25. the approximation is valid
Ex What is the pH of a buffer that has mol HC2H3O2 and mol NaC2H3O2 in 1.00 L that has mol NaOH added to it?
Ka for HC2H3O2 = 1.8 x 10-5
check if the approximation is valid by seeing if x < 5% of [HC2H3O2]init
[HA]
[A-]
[H3O+]
initial
0.090
0.110
≈ 0
change -x +x
equilibrium x
x = 1.47 x 10-5
the approximation is valid
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26. Ex What is the pH of a buffer that has mol HC2H3O2 and mol NaC2H3O2 in 1.00 L that has mol NaOH added to it?
[HA]
[A-]
[H3O+]
initial
0.090
0.110
≈ 0
change -x +x
equilibrium
1.5E-5
substitute x into the equilibrium concentration definitions and solve
0.090 x x x
x = 1.47 x 10-5
Tro, Chemistry: A Molecular Approach

27. Ex What is the pH of a buffer that has mol HC2H3O2 and mol NaC2H3O2 in 1.00 L that has mol NaOH added to it?
[HA]
[A-]
[H3O+]
initial
0.090
0.110
≈ 0
change -x +x
equilibrium
1.5E-5
substitute [H3O+] into the formula for pH and solve
Tro, Chemistry: A Molecular Approach

28. Ex What is the pH of a buffer that has mol HC2H3O2 and mol NaC2H3O2 in 1.00 L that has mol NaOH added to it?
Ka for HC2H3O2 = 1.8 x 10-5
check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated Ka to the given Ka
[HA]
[A-]
[H3O+]
initial
0.090
0.110
≈ 0
change -x +x
equilibrium
1.5E-5
the values match
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29. Ex What is the pH of a buffer that has mol HC2H3O2 and mol NaC2H3O2 in 1.00 L that has mol NaOH added to it?
find the pKa from the given Ka
Assume the [HA] and [A-] equilibrium concentrations are the same as the initial
HC2H3O2 + H2O  C2H3O2 + H3O+
Ka for HC2H3O2 = 1.8 x 10-5
[HA]
[A-]
[H3O+]
initial
0.090
0.110
≈ 0
change -x +x
equilibrium x
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30. Ex 16. 3 – Compare the effect on pH of adding 0
Ex 16.3 – Compare the effect on pH of adding mol NaOH to a buffer that has mol HC2H3O2 and mol NaC2H3O2 in 1.00 L to adding mol NaOH to 1.00 L of pure water?
HC2H3O2 + H2O  C2H3O2 + H3O+
pKa for HC2H3O2 = 4.745
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31. Basic Buffers B:(aq) + H2O(l)  H:B+(aq) + OH−(aq)
buffers can also be made by mixing a weak base, (B:), with a soluble salt of its conjugate acid, H:B+Cl−
H2O(l) + NH3 (aq)  NH4+(aq) + OH−(aq)
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32. Buffer Solutions Consider HOAc/OAc- to see how buffers work.
ACID USES UP ADDED OH-.
We know that OAc- + H2O HOAc + OH- has Kb = 5.6 x 10-10
Therefore, the reverse reaction of the WEAK ACID with added OH- has Kreverse = 1/ Kb = 1.8 x 109
Kreverse is VERY LARGE, so HOAc completely uses up the OH- !!!!

33. Buffer Solutions Consider HOAc/OAc- to see how buffers work.
CONJUGATE BASE USES UP ADDED H+ HOAc + H2O OAc- + H3O+ has Ka = 1.8 x 10-5.
Therefore, the reverse reaction of the WEAK BASE with added H+has Kreverse = 1/ Ka = 5.6 x 104
Kreverse is VERY LARGE, so OAc- completely uses up the H+ !

34. Buffer Solutions
Problem: What is the pH of a buffer that has [HOAc] = M and [OAc-] = M? in 1 L
HOAc + H2O OAc- + H3O+
Ka = 1.8 x 10-5
[HOAc] [OAc-] [H3O+]
initial
change
equilib
-x +x x
x x x

35. Buffer Solutions
Problem: What is the pH of a buffer that has [HOAc] = M and [OAc-] = M?
HOAc + H2O OAc- + H3O+
Ka = 1.8 x 10-5
[HOAc] [OAc-] [H3O+]
equilib x x x
Assuming that x << and 0.600, we have K a =
1.8 x 10 -5 [H 3 O +
](0.600)
0.700
[H3O+] = 2.1 x 10-5 and pH = 4.68

36. Adding an Acid to a Buffer
Problem: What is the new pH when mL of 1.00 M HCl is added to:
a) L of pure water (before HCl, pH = 7.00)
b) L of buffer that has [HOAc] = M and [OAc-] = M (pH = 4.68)
Solution to Part (a)
Calculate [HCl] after adding 1.00 mL of HCl to 1.00 L of water
M1 • V1 = M2 • V2
M2 = 1.00 x 10-3 M
pH = 3.00

37. Adding an Acid to a Buffer
What is the pH when 1.00 mL of 1.00 M HCl is added to:
a) 1.00 L of pure water (after HCl, pH = 3.00)
b) 1.00 L of buffer that has [HOAc] = M and [OAc-] = M (pH = 4.68)
Solution to Part (b)
Step 1 — do the stoichiometry
H3O+ (from HCl) + OAc- (from buffer) ---> HOAc (from buffer)
The reaction occurs completely because K is very large.

38. Adding an Acid to a Buffer
What is the pH when 1.00 mL of 1.00 M HCl is added to:
a) 1.00 L of pure water (after HCl, pH = 3.00)
b) 1.00 L of buffer that has [HOAc] = M and [OAc-] = M (pH = 4.68)
Solution to Part (b): Step 1—Stoichiometry
[H3O+] [OAc-] [HOAc]
Before rxn
Change
After rxn

39. Adding an Acid to a Buffer
What is the pH when 1.00 mL of 1.00 M HCl is added to;
a) 1.00 L of pure water (after HCl, pH = 3.00)
b) 1.00 L of buffer that has [HOAc] = M and [OAc-] = M (pH = 4.68)
Solution to Part (b): Step 2—Equilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [OAc-] [H3O+]
Before rxn
Change x +x +x
After rxn x x x

40. Adding an Acid to a Buffer
What is the pH when 1.00 mL of 1.00 M HCl is added to
a) 1.00 L of pure water (after HCl, pH = 3.00)
b) 1.00 L of buffer that has [HOAc] = M and [OAc-] = M (pH = 4.68)
Solution to Part (b): Step 2—Equilibrium
HOAc + H2O H3O+ + OAc-
[HOAc] [OAc-] [H3O+]
After rxn x x x
Because [H3O+] = 2.1 x 10-5 M BEFORE adding HCl, we again neglect x relative to and

41. Adding an Acid to a Buffer
What is the pH when 1.00 mL of 1.00 M HCl is added to
a) 1.00 L of pure water (after HCl, pH = 3.00)
b) 1.00 L of buffer that has [HOAc] = M and [OAc-] = M (pH = 4.68)
Solution to Part (b): Step 2—Equilibrium
HOAc + H2O OAc H3O+ [ H 3 O + ] =
[HOAc]
[OAc - • K a
0.701
0.599 ( 1 . 8
x 10 -5 )
[H3O+] = 2.1 x 10-5 M > pH = 4.68
The pH has not changed significantly upon adding HCl to the buffer!

42. REVIEW PROBLEMS
Calculate the pH of L of a buffer solution composed of 0.10 M sodium acetate and 0.15 M acetic acid, before and after adding 1.0 grams of sodium hydroxide solid (no volume change).

43. Sample Problem
Calculate the pH of L of a buffer solution composed of 0.10 M sodium acetate and 0.15 M acetic acid.
HC2H3O2 <---> H C2H3O2-
- x x x x
[C2H3O2-] [H+]
[HC2H3O2]
x(0.10)
(0.15)
Ka = =
= 1.8 x 10-5
X = 2.7 x 10-5 M = [H+] pH =

44. Sample Problem
Calculate the pH after adding 1.0 grams of sodium hydroxide solid.
HC2H3O (.500L)(.15M) = .075 mole
NaC2H3O (.500L)(.10M) = .050 mole
NaOH (1.0g)(40.0g/mole) = .025 mole
HC2H3O2 + OH- ---> HOH + C2H3O2-
.050
.500
.075
.500
[HC2H3O2] =
= 0.10
[C2H3O2-] =
= M

45. Sample Problem
Calculate the pH after adding 1.0 grams of sodium hydroxide solid.
HC2H3O2 <---> H C2H3O2-
- x x x x
[C2H3O2-][H+]
[HC2H3O2]
x(0.15)
(0.10)
Ka = =
= 1.8 x 10-5
X = 1.2 x 10-5 M = [H+] pH = 4.92

46. Chlorous Acid, HClO2 pKa = 1.95
Ex. 16.5a – Which of the following acids would be the best choice to combine with its sodium salt to make a buffer with pH 4.25?
Chlorous Acid, HClO2 pKa = 1.95
Nitrous Acid, HNO2 pKa = 3.34
Formic Acid, HCHO2 pKa = 3.74
Hypochlorous Acid, HClO pKa = 7.54
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47. Chlorous Acid, HClO2 pKa = 1.95
Ex. 16.5a – Which of the following acids would be the best choice to combine with its sodium salt to make a buffer with pH 4.25?
Chlorous Acid, HClO2 pKa = 1.95
Nitrous Acid, HNO2 pKa = 3.34
Formic Acid, HCHO2 pKa = 3.74
Hypochlorous Acid, HClO pKa = 7.54
The pKa of HCHO2 is closest to the desired pH of the buffer, so it would give the most effective buffering range.
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48. Buffering Capacity
buffering capacity is the amount of acid or base that can be added to a buffer without destroying its effectiveness
the buffering capacity increases with increasing absolute concentration of the buffer components
as the [base]:[acid] ratio approaches 1, the ability of the buffer to neutralize both added acid and base improves
buffers that need to work mainly with added acid generally have [base] > [acid]
buffers that need to work mainly with added base generally have [acid] > [base]
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49. Buffering Capacity
a concentrated buffer can neutralize more added acid or base than a dilute buffer
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50. Titration Curve a plot of pH vs. amount of added titrant
the inflection point of the curve is the equivalence point of the titration
prior to the equivalence point, the known solution in the flask is in excess, so the pH is closest to its pH
the pH of the equivalence point depends on the pH of the salt solution
equivalence point of neutral salt, pH = 7
equivalence point of acidic salt, pH < 7
equivalence point of basic salt, pH > 7
beyond the equivalence point, the unknown solution in the burette is in excess, so the pH approaches its pH
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51. Titration Curve: Unknown Strong Base Added to Strong Acid
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52. Tro, Chemistry: A Molecular Approach

53. Titration of 25 mL of 0.100 M HCHO2 with 0.100 M NaOH
HCHO2(aq) + NaOH(aq)  NaCHO2(aq) + H2O(aq)
Initial pH:
Ka = 1.8 x 10-4
[HCHO2]
[CHO2-]
[H3O+]
initial
0.100
0.000
≈ 0
change -x +x
equilibrium x x
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54. Titration of 25 mL of 0.100 M HCHO2 with 0.100 M NaOH
HCHO2(aq) + NaOH(aq)  NaCHO2 (aq) + H2O(aq)
initial mol of HCHO2 = L x mol/L = 2.50 x 10-3
before equivalence
added 5.0 mL NaOH HA A-
OH−
mols Before
2.50E-3
mols added -
5.0E-4
mols After
2.00E-3
≈ 0
Tro, Chemistry: A Molecular Approach

55. Titration of 25 mL of 0.100 M HCHO2 with 0.100 M NaOH
HCHO2(aq) + NaOH(aq)  NaCHO2 (aq) + H2O(aq)
initial mol of HCHO2 = L x mol/L = 2.50 x 10-3
at equivalence
CHO2−(aq) + H2O(l)  HCHO2(aq) + OH−(aq)
added 25.0 mL NaOH
Kb = 5.6 x 10-11
[HCHO2]
[CHO2-]
[OH−]
initial
0.0500
≈ 0
change +x -x
equilibrium x
5.00E-2-x HA A-
OH−
mols Before
2.50E-3
mols added -
mols After
≈ 0
[OH-] = 1.7 x 10-6 M

56. Adding NaOH to HCHO2 added 10.0 mL NaOH 0.00150 mol HCHO2 pH = 3.56
mol NaOH xs
pH = 12.22
added 30.0 mL NaOH
mol NaOH xs
pH = 11.96
added 25.0 mL NaOH
equivalence point
mol CHO2−
[CHO2−]init = M
[OH−]eq = 1.7 x 10-6
pH = 8.23
added 5.0 mL NaOH
mol HCHO2
pH = 3.14
initial HCHO2 solution
mol HCHO2
pH = 2.37
added 12.5 mL NaOH
mol HCHO2
pH = 3.74 = pKa
half-neutralization
added 40.0 mL NaOH
mol NaOH xs
pH = 12.36
added 15.0 mL NaOH
mol HCHO2
pH = 3.92
added 50.0 mL NaOH
mol NaOH xs
pH = 12.52
added 20.0 mL NaOH
mol HCHO2
pH = 4.34
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57. Tro, Chemistry: A Molecular Approach

58. Solubility Equilibria
all ionic compounds dissolve in water to some degree
however, many compounds have such low solubility in water that we classify them as insoluble
we can apply the concepts of equilibrium to salts dissolving, and use the equilibrium constant for the process to measure relative solubilities in water
Tro, Chemistry: A Molecular Approach

59. Solubility Product
the equilibrium constant for the dissociation of a solid salt into its aqueous ions is called the solubility product, Ksp
for an ionic solid MnXm, the dissociation reaction is:
MnXm(s)  nMm+(aq) + mXn−(aq)
the solubility product would be
Ksp = [Mm+]n[Xn−]m
for example, the dissociation reaction for PbCl2 is
PbCl2(s)  Pb2+(aq) + 2 Cl−(aq)
and its equilibrium constant is
Ksp = [Pb2+][Cl−]2
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60. Tro, Chemistry: A Molecular Approach

61. Molar Solubility
solubility is the amount of solute that will dissolve in a given amount of solution
at a particular temperature
the molar solubility is the number of moles of solute that will dissolve in a liter of solution
the molarity of the dissolved solute in a saturated solution
for the general reaction MnXm(s)  nMm+(aq) + mXn−(aq)
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62. PbCl2(s)  Pb2+(aq) + 2 Cl−(aq)
Ex 16.8 – Calculate the molar solubility of PbCl2 in pure water at 25C
Write the dissociation reaction and Ksp expression
Create an ICE table defining the change in terms of the solubility of the solid
PbCl2(s)  Pb2+(aq) + 2 Cl−(aq)
Ksp = [Pb2+][Cl−]2
[Pb2+]
[Cl−]
Initial
Change +S
+2S
Equilibrium S 2S
Tro, Chemistry: A Molecular Approach

63. Ex 16.8 – Calculate the molar solubility of PbCl2 in pure water at 25C
Substitute into the Ksp expression
Find the value of Ksp from Table 16.2, plug into the equation and solve for S
Ksp = [Pb2+][Cl−]2
Ksp = (S)(2S)2
[Pb2+]
[Cl−]
Initial
Change +S
+2S
Equilibrium S 2S
Tro, Chemistry: A Molecular Approach

64. CaF2(s)  Ca2+(aq) + 2 F−(aq)
Ex – Calculate the molar solubility of CaF2 in M NaF at 25C
Write the dissociation reaction and Ksp expression
Create an ICE table defining the change in terms of the solubility of the solid
CaF2(s)  Ca2+(aq) + 2 F−(aq)
Ksp = [Ca2+][F−]2
[Ca2+]
[F−]
Initial
0.100
Change +S
+2S
Equilibrium S S
Tro, Chemistry: A Molecular Approach

65. Ex 16. 10 – Calculate the molar solubility of CaF2 in 0
Ex – Calculate the molar solubility of CaF2 in M NaF at 25C
Substitute into the Ksp expression
assume S is small
Find the value of Ksp from Table 16.2, plug into the equation and solve for S
Ksp = [Ca2+][F−]2
Ksp = (S)( S)2
Ksp = (S)(0.100)2
[Ca2+]
[F−]
Initial
0.100
Change +S
+2S
Equilibrium S S
Tro, Chemistry: A Molecular Approach

66. Precipitation
precipitation will occur when the concentrations of the ions exceed the solubility of the ionic compound
if we compare the reaction quotient, Q, for the current solution concentrations to the value of Ksp, we can determine if precipitation will occur
Q = Ksp, the solution is saturated, no precipitation
Q < Ksp, the solution is unsaturated, no precipitation
Q > Ksp, the solution would be above saturation, the salt above saturation will precipitate
some solutions with Q > Ksp will not precipitate unless disturbed – these are called supersaturated solutions
Tro, Chemistry: A Molecular Approach

67. a supersaturated solution will precipitate if a seed crystal is added
precipitation occurs if Q > Ksp
Tro, Chemistry: A Molecular Approach

68. Selective Precipitation
a solution containing several different cations can often be separated by addition of a reagent that will form an insoluble salt with one of the ions, but not the others
a successful reagent can precipitate with more than one of the cations, as long as their Ksp values are significantly different
Tro, Chemistry: A Molecular Approach

69. precipitating may just occur when Q = Ksp
Ex What is the minimum [OH−] necessary to just begin to precipitate Mg2+ (with [0.059]) from seawater?
precipitating may just occur when Q = Ksp
Tro, Chemistry: A Molecular Approach

70. precipitating Mg2+ begins when [OH−] = 1.9 x 10-6 M
Ex What is the [Mg2+] when Ca2+ (with [0.011]) just begins to precipitate from seawater?
precipitating Mg2+ begins when [OH−] = 1.9 x 10-6 M
Tro, Chemistry: A Molecular Approach

71. precipitating Mg2+ begins when [OH−] = 1.9 x 10-6 M
Ex What is the [Mg2+] when Ca2+ (with [0.011]) just begins to precipitate from seawater?
precipitating Mg2+ begins when [OH−] = 1.9 x 10-6 M
precipitating Ca2+ begins when [OH−] = 2.06 x 10-2 M
when Ca2+ just begins to precipitate out, the [Mg2+] has dropped from M to 4.8 x M
Tro, Chemistry: A Molecular Approach

72. Qualitative Analysis
an analytical scheme that utilizes selective precipitation to identify the ions present in a solution is called a qualitative analysis scheme
wet chemistry
a sample containing several ions is subjected to the addition of several precipitating agents
addition of each reagent causes one of the ions present to precipitate out
Tro, Chemistry: A Molecular Approach

73. Qualitative Analysis
Tro, Chemistry: A Molecular Approach

75. Ag+(aq) + 2 H2O(l)  Ag(H2O)2+(aq)
Complex Ion Formation
transition metals tend to be good Lewis acids
they often bond to one or more H2O molecules to form a hydrated ion
H2O is the Lewis base, donating electron pairs to form coordinate covalent bonds
Ag+(aq) + 2 H2O(l)  Ag(H2O)2+(aq)
ions that form by combining a cation with several anions or neutral molecules are called complex ions
e.g., Ag(H2O)2+
the attached ions or molecules are called ligands
e.g., H2O
Tro, Chemistry: A Molecular Approach

76. Formation Constants
Tro, Chemistry: A Molecular Approach

77. SOLUBILITY AND COMPLEX IONS
If the metal cation can form a complex ion with the other species present, a new net equilibrium will exist. The process is similar to that in the previous slide.
If silver bromide is treated with ammonia solution, some of the solid dissolves and the complex ion is formed.
AgBr(s) + 2 NH3(aq) <=====> Ag(NH3)2+(aq) + Br-
Knet = Ksp . Kf =
( 3.3 x ) ( 1.6 x 107) = 5.3 x 10-6

78. Simultaneous Equilibria
1. If you add sufficient chromate ion to an aqueous suspension of PbCl2, can PbCl2 be converted to PbCrO4?
PbCl2 <--> Pb Cl-
Pb CrO42- <--> PbCrO4
1.7 x 10-5
1/1.8 x 10-14
9.4 x 108
PbCl2 + CrO42- <--> PbCrO Cl-
Yes!

79. Simultaneous Equilibria
2. Can AgCl be dissolved by adding a solution of NH3?
Write the overall equation and determine the K value.
AgCl <--> Ag Cl-
Ag NH3 <--> Ag(NH3)2+
1.8 x 10-10
1.6 x 107
2.9 x 10-3
AgCl + 2 NH3 <--> Ag(NH3) Cl-
No, unless very high [NH3]

80. Cu2+(aq) + 4 NH3(aq)  Cu(NH3)22+(aq)
Ex – mL of 1.5 x 10-3 M Cu(NO3)2 is mixed with mL of 0.20 M NH3. What is the [Cu2+] at equilibrium?
Write the formation reaction and Kf expression.
Look up Kf value
Determine the concentration of ions in the diluted solutions
Cu2+(aq) + 4 NH3(aq)  Cu(NH3)22+(aq)
Tro, Chemistry: A Molecular Approach

81. Cu2+(aq) + 4 NH3(aq)  Cu(NH3)22+(aq)
Ex – mL of 1.5 x 10-3 M Cu(NO3)2 is mixed with mL of 0.20 M NH3. What is the [Cu2+] at equilibrium?
Create an ICE table. Since Kf is large, assume all the Cu2+ is converted into complex ion, then the system returns to equilibrium
Cu2+(aq) + 4 NH3(aq)  Cu(NH3)22+(aq)
[Cu2+]
[NH3]
[Cu(NH3)22+]
Initial
6.7E-4
0.11
Change
-≈6.7E-4
-4(6.7E-4)
+ 6.7E-4
Equilibrium x

82. Cu2+(aq) + 4 NH3(aq)  Cu(NH3)22+(aq)
Ex – mL of 1.5 x 10-3 M Cu(NO3)2 is mixed with mL of 0.20 M NH3. What is the [Cu2+] at equilibrium?
Cu2+(aq) + 4 NH3(aq)  Cu(NH3)22+(aq)
Substitute in and solve for x
confirm the “x is small” approximation
[Cu2+]
[NH3]
[Cu(NH3)22+]
Initial
6.7E-4
0.11
Change
-≈6.7E-4
-4(6.7E-4)
+ 6.7E-4
Equilibrium x
since 2.7 x << 6.7 x 10-4, the approximation is valid
Tro, Chemistry: A Molecular Approach