1. Applications of Aqueous Equilibria

2. “Common Ions”
When we dissolve acetic acid in water, the following equilibrium is established:
CH3COOH  CH3COO- + H+
If sodium acetate were dissolved in solution, which way would this equilibrium shift?

3. “Common Ions”
When we dissolve acetic acid in water, the following equilibrium is established:
CH3COOH  CH3COO- + H+
If sodium acetate were dissolved in solution, which way would this equilibrium shift?
Adding acetate ions from a strong electrolyte would shift this equilibrium left (Le Chatelier’s principle).

4. Common Ion Effect
Whenever a weak electrolyte (acetic acid) and a strong electrolyte (sodium acetate) share a common ion, the weak electrolyte ionizes less than it would if it were alone. (Le Chatelier’s)
This is called the common-ion effect.

5. Steps for Common-Ion Problems
1. Consider which solutes are strong electrolyte and weak electrolytes.
2. Identify the important equilibrium (weak) that is the source of H+ and therefore determines pH.
3. Create an ICE chart using the equilibrium and strong electrolyte concentrations.
4. Use the equilibrium constant expression to calculuate [H+] and pH.

6. What is the concentration of silver and chromate in a solution with silver chromate in 0.1 M silver nitrate
Ag2CrO4(s)  2Ag CrO4-2 I
C x x
E x x
Ksp = 9.0 x = [Ag+1]2 • [CrO4-2]
= 9.0 x = [.1 +2x]2 • [x]
9.0 x = 0.12 • x
[CrO4-2] = x = 9.0 x M
[Ag+1] = x = .1M
[Ag2CrO4] = 9.0 x 10-10
What are the sources of Ag+1?
Why is this a plus sign?

7. Sample Problem
What 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 1.0 L of solution?
Ka for acetic acid = 1.8 x 10-5
pH = 4.74

8. Sample Problem
Calculate the fluoride concentration and pH of a solution that is 0.20 M in HF and 0.10 M in HCl
Ka for HF = 6.8 x 10-4.
[F-] = 1.2x10-3M
pH = 1.00

9. HX(aq) + H2O(l) H+(aq) + X-(aq)
Buffers
Solutions that resist pH change when small amounts of acid or base are added.
Two types
weak acid and its salt
weak base and its salt
HX(aq) + H2O(l) H+(aq) + X-(aq)
Add OH Add H+
shift to right shift to left
Based on the common ion effect.

10. Buffers
buffered
unbuffered pH
ml HCl added

11. Buffers and blood Control of blood pH CO2 (aq) + H2O H2CO3 (aq)
Oxygen is transported primarily by hemoglobin in the red blood cells.
CO2 is transported both in plasma and the red blood cells.
CO2 (aq) + H2O
H2CO3 (aq)
H+(aq) + HCO3-(aq)

12. Buffers Composition and Action of Buffered Solutions
The Ka expression is
A buffer resists a change in pH when a small amount of OH- or H+ is added.

13. Buffered Solutions Equation not necessary, since you know
Addition of Strong Acids or Bases to Buffers
With the concentrations of HX and X- (note the change in volume of solution) we can calculate the pH from the Henderson-Hasselbalch equation
Equation not necessary, since you know
Ka = [X-]
[H+] [HX]

14. Buffers
Action of Buffered Solutions

15. Buffers Buffer Capacity and pH
Buffer capacity is the amount of acid or base neutralized by the buffer before there is a significant change in pH.
The greater the amounts(molarity) of the conjugate acid-base pair, the greater the buffer capacity.
The pH of the buffer depends on Ka

16. Buffered Solutions
Addition of Strong Acids or Bases to Buffers

17. Buffer Addition of Strong Acids or Bases to Buffers
Break the calculation into two parts: stoichiometric and equilibrium.
The amount of strong acid or base added results in a neutralization reaction:
X- + H+  HX + H2O
HX + OH-  X- + H2O.
By knowing how much H+ or OH- was added (stoichiometry) we know how much HX or X- is formed.

18. Buffers
The final [HX] and [X-] after the neutralization reaction are used as the initial concentrations for the equilibrium reaction.
HX H+ X-
Initial conc., M
Change, DM
Eq. Conc., M
Then the equilibrium constant expression is used to find [H+] and pH = - log [H+]
Ka = [H+] [X-]
[HX]

19. HBz(aq) + H2O(l) H+(aq) + Bz-(aq)
Buffer Example
Determine the initial pH of a buffer solution that has 0.10 M benzoic acid and 0.20 M sodium benzoate at 25 oC. Ka = 6.5 x 10-5
HBz(aq) + H2O(l) H+(aq) + Bz-(aq)
HBz H+ Bz-
Initial conc., M
Change, DM -x x x
Eq. Conc., M x x x

20. Buffer Example Solve the equilibrium equation in terms of x
Ka = 6.5 x =
x = (6.5 x )(0.10) / (0. 20)
(assuming x<<0.10)
= 3.2 x 10-5 M H+
pH = - log (3.2 x 10-5 M) = 4.5
initial pH
x ( x) x

21. Buffer Example
Determine the pH of a buffer solution that has 0.10 M benzoic acid and 0.20 M sodium benzoate at 25 oC after 0.05 moles of HCl is added.
First, find the concentrations of the HBz and Bz- after HCl is added.
HBz H+ Bz-
Initial conc., M
Change, DM -x x x
Eq. Conc., M x x x
The 0.05 mol HCl reacts completely with 0.05 mol Bz-(aq) to form 0.05 mol HBz(aq)
Then equilibrium will be re-established based on the new initial concentrations of 0.15 M HBz(aq) and 0.15 M Bz-(aq).

22. Buffer Example Solve the equilibrium equation in terms of x
Ka = 6.5 x =
x = (6.5 x )(0.15) / (0. 15)
(assuming x<<0.15)
= 6.5 x 10-5 M H+
pH = - log (6.5 x 10-5 M) = 4.2
after 0.05 mole HCl added
x ( x) x

23. Buffer Example
Determine the pH of a buffer solution that has 0.10 M benzoic acid and 0.20 M sodium benzoate at 25 oC after 0.05 moles of NaOH is added.
First, find the concentrations of HBz and Bz- after NaOH is added.
HBz H+ Bz-
Initial conc., M
Change, DM -x x x
Eq. Conc., M x x x
The 0.05 mol NaOH reacts completely with 0.05 mol HBz(aq) to form 0.05 mol Bz-(aq)
Then equilibrium will be re-established based on the new initial concentrations of 0.05 M HBz(aq) and 0.25 M Bz-(aq).

24. Buffer Example Solve the equilibrium equation in terms of x
Ka = 6.5 x =
x = (6.5 x )(0.05) / (0. 25)
(assuming x<<0.05)
= 1.3 x 10-5 M H+
pH = - log (1.3 x 10-5 M) = 4.9
after 0.05 mole NaOH added
x ( x) x

25. Acid-Base Titrations Strong Acid-Base Titrations
The plot of pH versus volume during a titration is a titration curve.

26. Remember the Acid-Base Titration Curves?
Buffer
Zone
HC2H3O2
HCl

27. Acetic Acid/Acetate Ion Buffer Lab
For this experiment, you will prepare a buffer that contains acetic acid and its conjugate base, the acetate ions. The equilibrium equation for the reaction is shown below:
HC2H3O2(aq) + H2O(l) <=> H+ (aq)+ C2H3O2- (aq)
The equilibrium expression for this reaction, Ka, has a value of 1.8 x 10-5 at 25ºC.

28. Acetic Acid/Acetate Ion Buffer Lab
The ratio between the molarity of the acetate ions to the molarity of the acetic acid in your buffer must equal the ratio between the Ka value and 10- assigned pH.
This ratio should be reduced , so that either the [HC2H3O2] or [C2H3O2- ] has a concentration of 0.10 M, and the concentration of the other component must fall within a range from 0.10 M to 1.00 M.
Complete the calculations only that are needed to prepare mL of your assigned buffer solution that has these specific concentrations. Can you predict the final pH when a strong acid or base is added to the buffer solution?

29. Acetic Acid/Acetate Ion Buffer Lab

30. Acetic Acid/Acetate Ion Buffer Lab

31. Making a Buffer Calculations
You want to prepare mL of a buffer with a pH = 10.00, with both the acid and conjugate base having molarities between 0.10 M to 1.00 M. You may choose from any of the acids listed below:
You must select an acid with a Ka value close to 10- assigned pH.
The only two options are ammonium or the hydrogen carbonate ions.

32. Making a Buffer Calculations

33. Making a Buffer Calculations
Place ~250 mL of distilled water in a 500 mL volumetric flask. Add 2.68 g NH4Cl and dissolve. Then mix in 18.9 mL of 14.8 M NH3. Fill with distilled water to the 500 mL mark on the flask.
If there is no concentrated NH3 available, the NH3 can be produced by neutralizing additional NH4Cl with 1.00 M NaOH.
Dissolve g NH4Cl (2.68 g g) in 280. mL of 1.00 M NaOH. Then dilute with distilled water and fill to the 500 mL mark on the flask.

34. Making a Buffer Calculations
Place ~250 mL of distilled water in a 500 mL volumetric flask. Add 2.68 g NH4Cl and dissolve. Then mix in 18.9 mL of 14.8 M NH3. Fill with distilled water to the 500 mL mark on the flask.
If there is no NH4Cl available, the NH4+ can be produced by neutralizing additional NH3 with 1.00 M HCl.
Place ~250 mL distilled water in volumetric flask. Add 50.0 mL of 1.00 M HCl and mix. Then add 22.3 mL of concentrated NH3 (18.9 mL mL). Mix and fill with distilled water to the 500 mL mark on the flask.

35. Making a Buffer Calculations
Prepare 500. mL of the buffer that has [CO32-] = M and [HCO31-] = M.

36. Making a Buffer Calculations
Prepare 500. mL of the buffer that has [CO32-] = M and [HCO31-] = M.
Place ~250 mL distilled water in a 500 mL volumetric flask. Add 5.30 g Na2CO3 and 7.52 g NaHCO3 and dissolve. Fill with distilled water to the 500 mL mark on the flask.

37. Acid-Base Titrations Strong Acid-Base Titrations
The plot of pH versus volume during a titration is a titration curve.

38. Indicator examples
Acid-base indicators are weak acids that undergo a color change at a known pH. pH
phenolphthalein

39. Indicator examples
Select the indicator that undergoes a color change closest to the pH at the equivalence point, where all of the acid has been neutralized by the base.
bromthymol blue
methyl red

40. Acid-Base Titrations Strong Acid-Base Titrations
Consider adding a strong base (e.g. NaOH) to a solution of a strong acid (e.g. HCl).
Before any base is added, the pH is given by the strong acid solution. Therefore, pH < 7.
When base is added, before the equivalence point, the pH is given by the amount of strong acid in excess. Therefore, pH < 7.
At equivalence point, the amount of base added is stoichiometrically equivalent to the amount of acid originally present. Therefore, the pH is determined by the salt solution. Therefore, pH = 7.

41. Acid-Base Titrations Strong Acid-Base Titrations
Consider adding a strong base (e.g. NaOH) to a solution of a strong acid (e.g. HCl).

42. Acid-Base Titrations Strong Acid-Base Titrations
We know the pH at equivalent point is 7.00.
To detect the equivalent point, we use an indicator that changes color somewhere near 7.00.
Usually, we use phenolphthalein that changes color between pH 8.3 to 10.0.
In acid, phenolphthalein is colorless.
As NaOH is added, there is a slight pink color at the addition point.
When the flask is swirled and the reagents mixed, the pink color disappears.
At the end point, the solution is light pink.
If more base is added, the solution turns darker pink.

43. Acid-Base Titrations Strong Acid-Base Titrations
The equivalence point in a titration is the point at which the acid and base are present in stoichiometric quantities.
The end point in a titration is the observed point.
The difference between equivalence point and end point is called the titration error.
The shape of a strong base-strong acid titration curve is very similar to a strong acid-strong base titration curve.

44. Acid-Base Titrations
Strong Acid-Base Titrations

45. Acid-Base Titrations Strong Acid-Base Titrations
Initially, the strong base is in excess, so the pH > 7.
As acid is added, the pH decreases but is still greater than 7.
At equivalence point, the pH is given by the salt solution (i.e. pH = 7).
After equivalence point, the pH is given by the strong acid in excess, so pH < 7.

46. HC2H3O2(aq) + NaOH(aq)  C2H3O2-(aq) + H2O(l)
Acid-Base Titrations
Weak Acid-Strong Base Titrations
Consider the titration of acetic acid, HC2H3O2 and NaOH.
Before any base is added, the solution contains only weak acid. Therefore, pH is given by the equilibrium calculation.
As strong base is added, the strong base consumes a stoichiometric quantity of weak acid:
HC2H3O2(aq) + NaOH(aq)  C2H3O2-(aq) + H2O(l)

47. Acid-Base Titrations
Weak Acid-Strong Base Titrations

48. Acid-Base Titrations Weak Acid-Strong Base Titrations
There is an excess of acetic acid before the equivalence point.
Therefore, we have a mixture of weak acid and its conjugate base.
The pH is given by the buffer calculation.
First the amount of C2H3O2- generated is calculated, as well as the amount of HC2H3O2 consumed. (Stoichiometry.)
Then the pH is calculated using equilibrium conditions. (Henderson-Hasselbalch.)

49. Acid-Base Titrations Weak Acid-Strong Base Titrations
At the equivalence point, all the acetic acid has been consumed and all the NaOH has been consumed. However, C2H3O2- has been generated.
Therefore, the pH is given by the C2H3O2- solution.
This means pH > 7.
More importantly, pH  7 for a weak acid-strong base titration.
After the equivalence point, the pH is given by the strong base in excess.

50. Acid-Base Titrations Weak Acid-Strong Base Titrations
For a strong acid-strong base titration, the pH begins at less than 7 and gradually increases as base is added.
Near the equivalence point, the pH increases dramatically.
For a weak acid-strong base titration, the initial pH rise is more steep than the strong acid-strong base case.
However, then there is a leveling off due to buffer effects.

51. Acid-Base Titrations Weak Acid-Strong Base Titrations
The inflection point is not as steep for a weak acid-strong base titration.
The shape of the two curves after equivalence point is the same because pH is determined by the strong base in excess.
Two features of titration curves are affected by the strength of the acid:
the amount of the initial rise in pH, and
the length of the inflection point at equivalence.

52. Acid-Base Titrations Weak Acid-Strong Base Titrations
The weaker the acid, the smaller the equivalence point inflection.
For very weak acids, it is impossible to detect the equivalence point.

53. Acid-Base Titrations Weak Acid-Strong Base Titrations
Titration of weak bases with strong acids have similar features to weak acid-strong base titrations.

54. Acid-Base Titrations Titrations of Polyprotic Acids
In polyprotic acids, each ionizable proton dissociates in steps.
Therefore, in a titration there are n equivalence points corresponding to each ionizable proton.
In the titration of Na2CO3 with HCl there are two equivalence points:
one for the formation of HCO3-
one for the formation of H2CO3.

55. Acid-Base Titrations
Titrations of Polyprotic Acids

56. Acid-Base Titrations Titrations of Polyprotic Acids Ka = [H+] [X-]
At the equivalence point, [H+] = [OH-]
At ½ the equivalence point, [X-] = [HX]
SOOOO…. At ½ equivalence point, Ka = [H+]
So pKa = pH

58. Solubility What happens when an ionic compound is dissolved in water?
What forces does water have to overcome?
Are all ionic compounds soluble?
Remember the solubility rules?
It’s time to review them!

60. Properties of aqueous solutions
There are two general classes of solutes.
Electrolytic
ionic compounds in polar solvents
dissociate in solution to make ions
conduct electricity
may be strong (100% dissociation) or weak (less than 10%)
Nonelectrolytic
do not conduct electricity
solute is dispersed but does not dissociate

61. Heterogeneous Equilibria
A(s) + H2O  B(aq) + C (aq)
Ksp = [B] • [C]
Why is this not divided by [A]?
Why is [H2O] not included?
Known as the Solubility Product = Ksp

62. Solubility Products, KSP
KSP expressions are used for ionic materials that are only slightly soluble in water.
Their only means of dissolving is by dissociation.
AgCl(s) Ag+ (aq) + Cl- (aq)
KSP =
[ Ag+] [ Cl-]

63. Solubility Products, KSP
At equilibrium, the system is a saturated solution of silver and chloride ions.
The only way to know that it is saturated it to observe some AgCl at the bottom of the solution.
As such, [AgCl] is a constant and KSP expressions do not include the solid form in the equilibrium expression. The [H2O] for solvation process is also excluded from the KSP expression.

64. Solubility Products, KSP
Determine the solubility of AgCl in water at 20 oC in terms of grams / 100 mL
KSP = [Ag+] [Cl-] = 1.7 x 10-10
At equilibrium, [Ag+] = [Cl-] so
1.7 x = [x] 2
[Ag+] = 1.3 x 10-5 M
g AgCl = 1.3 x 10-5 mol/L * 0.10 L * g/mol
Solubility = 1.9 x 10-4 g / 100 mL

65. Calculating Ksp
Calculate the Ksp for Bismuth Sulfide which has a solubility of 1.0 x
Bi2S3 → 2Bi S-2
Ksp =
For every Bi2S3 dissolved
2 Bi+3 and 3S-2 are formed
Ksp = [2 x 1.0 x ]2 • [3 x 1.0 x ]3
= 1.1 x 10-73
[Bi+3]2
• [S-2]3

66. Calculating Solubility
What is the molarity of a saturated solution of Cu(IO3)2?
Calculate the concentration of ions
for copper (II) iodate.
ksp = 1.4 x 10-7 at 25 °C
Cu(IO3)2(s)  Cu IO3-1
1.4 x 10-7 = [x] • [ 2x]2 = 4x3
x = 3.3 x 10-3 mol/L = [Cu+2]
2x = 6.6 x 10-3 mol/L = [IO3-1]
3.3 x 10-3 mol/L = [Cu(IO3)2]

67. Solubility Products, KSP
Another example
Calculate the silver ion concentration when excess silver chromate is added to a M sodium chromate solution.
KSP Ag2CrO4 = 1.1 x 10-12
Ag2CrO4 (s) Ag+ + CrO42-

68. Solubility Products, KSP
KSP = 1.1 x = [ Ag+ ]2 [ CrO42- ]
[CrO42-] = [CrO42-]Ag2CrO4 + [CrO42-]Na2CrO4
With such a small value for KSP, we can assume that the [CrO42-] from Ag2CrO4 is negligible.
If we’re wrong, our silver concentration will be significant (>1% of the chromate concentration.)
Then you’d use the quadratic approach.

69. Solubility Products, KSP
KSP = 1.1 x = [ Ag+ ]2 [ CrO42- ]
[CrO42-] = M
[ Ag+ ] = ( KSP / [ CrO42- ] ) 1/2
= (1.1 x / M ) 1/2
= 1.1 x M
[ Ag+ ] << [ CrO42- ]
so our assumption was valid.

70. Relative Solubility
Be careful when comparing solubility products (Ksp)
Compare
AgI Ksp = 1.5 x 10-16
CuI Ksp = 5.0 x 10-12
Which is more soluble?
Copper (I) iodide is more soluble. Each compound produces the same number ions

71. Relative Solubility CuS Ksp = 8.5 x 10-45 Ag2S Ksp = 1.6 x 10-49
Bi2S3 Ksp = 1.1 x 10-73
Which is more soluble?
Each has a different number of ions,
so calculate the solubility of each.
Estimate (guess)
three groups and calculate the solubility

72. Factors influencing solubility
Common ion and salt effects.
As with other equilibria we’ve discussed, adding a ‘common’ ion will result in a shift of a solubility equilibrium.
AgCl (s) Ag+ (aq) + Cl- (aq)
KSP = [Ag+] [Cl-]
Adding either Ag+ or Cl- to our equilibrium system will result in driving it to the left.

73. Factors That Affect Solubility
Common-Ion Effect

74. Factors influencing solubility
Complex ion formation.
The solubility of slightly soluble salts can be increased by complex ion formation.
Example. Addition of excess Cl-
AgCl(s) Ag+(aq) + Cl-(aq) +
2Cl-(aq)
AgCl2-(aq)
A large excess of chloride results in the formation of the complex.
More AgCl will dissolve as a result.

75. Factors That Affect Solubility
Formation of Complex Ions

76. Factors That Affect Solubility
Formation of Complex Ions
Consider the addition of ammonia to AgCl (white precipitate): +
The overall reaction is +
Effectively, the Ag+(aq) has been removed from solution.
By Le Châtelier’s principle, the forward reaction (the dissolving of AgCl) is favored.
Knet = Ksp•Kf = (1.8 x 10-10)(1.7 x 107) = 3.1 x 10-3 at 25ºC

77. Factors That Affect Solubility
Amphoterism
Amphoteric oxides will dissolve in either a strong acid or a strong base.
Examples: hydroxides and oxides of Al3+, Cr3+, Zn2+, and Sn2+.
The hydroxides generally form complex ions with four hydroxide ligands attached to the metal:
Hydrated metal ions act as weak acids. Thus, the amphoterism is interrupted:

78. Factors That Affect Solubility
Amphoterism
Hydrated metal ions act as weak acids. Thus, the amphoterism is interrupted:

79. Factors influencing solubility
Hydrolysis.
If the anion of a weak acid, or cation of a weak base is part of a KSP, solubilities are greater than expected.
AgCN(s) Ag+(aq) + CN-(aq) +
H2O(l)
HCN(aq) + OH-(aq)
This competing
equilibrium causes the CN- to be lower than expected.
More AgCN will dissolve as a result.

80. Factors That Affect Solubility
Solubility and pH
Again we apply Le Châtelier’s principle:
If the F- is removed, then the equilibrium shifts towards the decrease and CaF2 dissolves.
F- can be removed by adding a strong acid:
As pH decreases, [H+] increases and solubility increases.
The effect of pH on solubility is dramatic.

81. Factors That Affect Solubility
Solubility and pH

82. Solubility Equilibria
Solubility-Product Constant, Ksp
Consider
for which
Ksp is the solubility product. (BaSO4 is ignored because it is a pure solid so its concentration is constant.)

83. Solubility Equilibria
Solubility-Product Constant, Ksp
In general: the solubility product is the molar concentration of ions raised to their stoichiometric powers.
Solubility is the amount (grams) of substance that dissolves to form a saturated solution.
Molar solubility is the number of moles of solute dissolving to form a liter of saturated solution.

84. Solubility Equilibria
Solubility and Ksp
To convert solubility to Ksp
solubility needs to be converted into molar solubility (via molar mass);
molar solubility is converted into the molar concentration of ions at equilibrium (equilibrium calculation),
Ksp is the product of equilibrium concentration of ions.

85. Solubility Equilibria
Solubility and Ksp

86. Precipitation Reactions
We will use “Q” and Ksp
If Q is bigger than Ksp, what will happen?
Smaller
The Q in a solubilty problem is called the
Ion Product

87. A solution prepared by adding 750 mL of 4. 00 x 10-3 M Ce(NO3)3 to 300
A solution prepared by adding 750 mL of 4.00 x 10-3 M Ce(NO3)3 to mL of 2.00 x 10-2 M KIO3. Will Ce(IO3)3 precipitate? (Ksp = 1.9 x 10-10)
Step one: Calculate the concentration of Ce+3 and IO3-1 before a reaction occurs
[Ce+3]0 = (750.0 mL)(4.00x10-3 M) = x 10-3 M ( mL)
[IO3-1]0 = (300.0 mL)(2.00x10-3 M) = x 10-3 M ( mL)
Q = (2.86 x 10-3 M) x (5.71 x 10-3 M)3 = 5.32 x 10-10
Q > Ksp therefore the precipitation reaction will occur

88. Precipitation and Separation of Ions
At any instant in time, Q = [Ba2+][SO42-].
If Q < Ksp, precipitation occurs until Q = Ksp.
If Q = Ksp, equilibrium exists.
If Q > Ksp, solid dissolves until Q = Ksp.
Based on solubilities, ions can be selectively removed from solutions.
Consider a mixture of Zn2+(aq) and Cu2+(aq). CuS (Ksp= 610-37) is less soluble than ZnS(Ksp=210-25), CuS will be removed from solution before ZnS.

89. Precipitation and Separation of Ions
As H2S is added to the green solution, black CuS forms in a colorless solution of Zn2+(aq).
When more H2S is added, a second precipitate of white ZnS forms.
Selective Precipitation of Ions
Ions can be separated from each other based on their salt solubilities.
Example: if HCl is added to a solution containing Ag+ and Cu2+, the silver precipitates(Ksp for AgCl is 1.810-10) while the Cu2+ remains in solution, since CuCl2.
Removal of one metal ion from a solution is called selective precipitation.

90. Qualitative Analysis for Metallic Elements
Qualitative analysis is designed to detect the presence of metal ions.
Quantitative analysis is designed to determine how much metal ion is present.

91. Qualitative Analysis for Metallic Elements
We can separate a complicated mixture of ions into five groups:
Add 6 M HCl to precipitate insoluble chlorides (AgCl, Hg2Cl2, and PbCl2).
To the remaining mix of cations, add H2S in 0.2 M HCl to remove acid insoluble sulfides (e.g. CuS, Bi2S3, CdS, PbS, HgS, etc.).
To the remaining mix, add (NH4)2S at pH 8 to remove base insoluble sulfides and hydroxides (e.g. Al(OH)3, Fe(OH)3, ZnS, NiS, CoS, etc.).
To the remaining mixture add (NH4)2HPO4 to remove insoluble phosphates (Ba3(PO4)2, Ca3(PO4)2, MgNH4PO4).
The final mixture contains alkali metal ions and NH4+.