1. Chapter 19
Ionic Equilibria in Aqueous Systems

2. Ionic Equilibria in Aqueous Systems
19.1 Equilibria of Acid-Base Buffer Systems
19.2 Acid-Base Titration Curves
19.3 Equilibria of Slightly Soluble Ionic Compounds
19.4 Equilibria Involving Complex Ions
19.5 Application of Ionic Equilibria to Chemical Analysis

3. Buffered solutions
A solution that resists a change in its pH when either OH- or H+ ions are added.
Ex: Blood, absorbs acid and bases without change in pH.
The components of a buffer are the conjugate acid-base pair of a weak acid.
WA and its salt-HF and NaF
WB and its salt- NH3 and NH4Cl.

4. The effect of addition of acid or base to …
Figure 19.1
The effect of addition of acid or base to …
an unbuffered solution
acid added
base added
Figure 19.2
or a buffered solution
acid added
base added

5. Common ion effect
The shift in equilibrium position that occurs because of the addition of an ion already involved in equilibrium reaction is called common ion effect.
Consider a weak acid HF and its salt NaF.
NaF (s) _______ Na+ (aq)+ F- (aq) , Major species: Na+ F- HF H2O
HF (aq)↔ H+ (aq) + F- (aq)

6. Common ion effect F- is the common ion.
F- from the added NaF moves the equ position to the left according to LC principle.
If we dissolve HF in a NaF solution, the F-ion and H+ ion enter the solution. The F- ion already present combines with the H+, which lowers the H+.

7. Buffered Solution Characteristics
Buffers contain relatively large amounts of weak acid and corresponding base.
Added H+ is consumed by A-
Added OH is consumed by HA
The pH is determined by the ratio of the concentrations of the weak acid and its conjugate base.

8. The Effect of Added Acetate Ion on the Dissociation of Acetic Acid
Table 19.1
The Effect of Added Acetate Ion on the Dissociation of Acetic Acid
[CH3COOH]initial
[CH3COO-]added
% Dissociation* pH
0.10
0.00
0.050
0.15
1.3
2.89
0.036
4.44
0.018
4.74
0.012
4.92
* % Dissociation =
[CH3COOH]dissoc
[CH3COOH]initial
x 100

9. How a buffer works. Figure 19.3 Buffer after addition of H3O+
H2O + CH3COOH H3O+ + CH3COO-
Buffer with equal concentrations of conjugate base and acid
Buffer after addition of OH-
CH3COOH + OH H2O + CH3COO-
H3O+
OH-

10. Problem
P.1. A buffered solution contains 0.50 M acetic acid (Ka= 1.8 x 10-5) and 0.50 M sodium acetate.
a) Calculate pH of this solution.
b) Calculate the change in pH that occurs when mol solid NaOH is added to 1.0 L of buffered solution described in the above problem.Compare this pH change with the one that occurs when mol solid NaOH is added to 1.0 L of water.
c) Calculate the change in pH that occurs when mol solid HCl is added to 1.0 L of buffered solution.

11. The Henderson-Hasselbalch Equation
HA + H2O H3O+ + A-
Ka =
[H3O+] [A-]
[HA]
Ka [HA]
[A-]
[H3O+] =
[A-]
[HA]
- log[H3O+] = - log Ka + log
pH = pKa + log
[base]
[acid]

12. Problems
P. 2. Calculate the pH of a solution 0.75 M lactic acid (Ka= 1.4 x10-4 ) and 0.25 M sodium lactate.
P.3. A buffer solution contains 0.25 M NH3 (Kb= 1.8 x 10-5) and 0.40 M NH4Cl.
Calculate pH of this solution.

13. Buffering Capacity
It represents the amount of H+ or OH the buffer can absorb without a significant change in pH.
More concentrated the components of a buffer , the greater the buffer capacity.
pH of a buffered solution is determined by the ratio [A-]/[HA] .
If an acid or a base is added , the concentration ratio changes less when the buffer component concentrations are similar than they are different.

14. Buffering Capacity:
Buffer has highest capacity when component concentrations are equal.
pH=pKa highest buffer capacity.
The Buffer Range is the pH range over which the buffer works effectively.
The further the buffering component concentration ratio is from 1 less effective is the buffering action.

15. Buffer Capacity and Buffer Range
Buffer capacity is the ability to resist pH change.
The more concentrated the components of a buffer, the greater
the buffer capacity.
The pH of a buffer is distinct from its buffer capacity.
A buffer has the highest capacity when the component
concentrations are equal.
Buffer range is the pH range over which the buffer acts effectively.
Buffers have a usable range within ± 1 pH unit of the pKa of
its acid component.

16. The relation between buffer capacity and pH change.
Figure 19.4
The relation between buffer capacity and pH change.

17. Preparing a Buffer
Choose the conjugate acid-base pair. Ratio of concentrations should be close to1. pH=pKa
Calculate ratio of buffer component concentrations. Use the Henderson Hasselbalch equation.
Determine the buffer concentration. Higher the buffer concentrations greater the buffer capacity.
Mix the solutions and adjust the pH by adding strong acid or strong base with the help of a pH probe.

18. Preparing a Buffer 1. Choose the conjugate acid-base pair.
2. Calculate the ratio of buffer component concentrations.
3. Determine the buffer concentration.
4. Mix the solution and adjust the pH.

19. Problem
P.4. A chemist needs a solution buffered at pH= 4.30 and can choose from the following acids and their sodium salts:
chloroacetic acid Ka= 1.35 x 10-3
2. propanoic acid Ka= 1.3 x 10-5
3. benzoic acid Ka= 6.4 x 10-5
4. hypochlorous acid. Ka= 3.5 x 10-8
Calculate the ratio [HA]/[A-] required for each system to yield a pH of 4.30, which system will work the best?

20. Acid base Titration Curves
A plot of pH of the solution being analyzed as a function of the amount of titrant added.

21. Acid-Base Indicators:
It marks the end point of a titration by changing color.
Ex: phenolphthalein is colorless in its HIn form and pink in In- form, or basic form.
Indicator changes color over a range of about 2 pH units.

22. Colors and approximate pH range of some common acid-base indicators.
Figure 19.5 pH

23. The color change of the indicator bromthymol blue.
Figure 19.6
The color change of the indicator bromthymol blue.
basic
change occurs over ~2pH units
acidic

24. Strong Acid- Strong Base Titration curves:
pH is low initially, as base is added the pH increases slowly.
The pH rises steeply when the moles of OH- nearly equals the moles of H3O+
The additional drop of base neutralizes the tiny excess acid and introduces a tiny excess of base.
Then pH increases smoothly as more base is added.

25. Equivalence (stoichiometric) point
Enough titrant has been added to react exactly with the solution being analyzed.
Equivalence point is nearly the vertical portion of the curve.
The point at which the number of moles of added OH- =the number of moles of H3O+
For a SA vs SB titration pH =7(neutral ions)

26. End point
It occurs when indicator changes color. The indicator chosen should be close to the equivalence point.

27. Curve for a strong acid-strong base titration
Figure 19.7
Curve for a strong acid-strong base titration

28. Calculating pH: Initial pH is the pH of the acid.
Before the equ. Point find initial moles present, moles reacted, change in volume and then the pH.
At the equ point, pH =7
After the equ point find the excess of OH- present and then calculate pH.

29. Problem
P mL of m HNO3 is titrated with 0.100M NaOH. Calculate the pH of the solution at selected points during the course of titration, where 0 mL, 20.0 mL, mL, mL of M NaOH has been added.

30. Weak acid-Strong Base Titrations:
Initial pH is high as weak acid dissociates slightly.
Buffer Region: A gradual rising portion of the curve appearing before the equ.point.
At midpoint of the buffer region pH=pKa.
pH at equ point is > 7.
Beyond equ point pH increases slowly as excess OH- is added.

31. Curve for a weak acid-strong base titration
Titration of 40.00mL of M HPr with M NaOH
Figure 19.8
Curve for a weak acid-strong base titration
pH = 8.80 at equivalence point
pKa of HPr = 4.89
methyl red
[HPr] = [Pr-]

32. Calculating pH
1. When only HA present, Use ICE table calculate H+ and pH like a weak acid type calculation.
2. As we add base, find change in moles and solve for H+
3. At equ. Point pH >7, Find Kb from Ka and then calculate OH- and pH.
4. Beyond the equ. point , calculate excess of OH- present and calculate pH.

33. Problem P.6. 50.0 mL of 0.1 M acetic acid
(Ka=1.8 x 10-5) with 0.10 M NaOH. Calculate the pH at various points representing volumes of 0mL, 10.0 mL , mL, 50.0 mL , 60.0 mL of added NaOH.

34. Weak base –strong acid Titration curves:
Same shape curve as WA vs SB , but inverted.
Initially pH above 7 as it is a weak base.
pH decreases in buffer region.
At midpoint pH=pKa
After buffer region curve drops vertically to equ.point .
pH at equ.point is below 7
Beyond equ point , pH decreases slowly as more acid is added.

35. Curve for a weak base-strong acid titration
Titration of 40.00mL of M NH3 with M HCl
Figure 19.9
pKa of NH4+ = 9.25
Curve for a weak base-strong acid titration
pH = 5.27 at equivalence point

36. Problems
P mL of M NH3 is titrated with 0.10 M HCl. Calculate the pH at various points: 1) before adding HCl 2) Before equ.point 3) At equ point (setup)
P.8 If 2.00 mmol of solid acid in mL water is titrated with M NaOH. After 20.0 mL NaOH has been added the pH is What is the Ka value for acid?

37. Titration curves of polyprotic acids:
Number of curves =Number of H+ ions.

38. Titration of 40.00mL of 0.1000M H2SO3 with 0.1000M NaOH
Figure 19.10
Curve for the titration of a weak polyprotic acid.
pKa = 7.19
pKa = 1.85
Titration of 40.00mL of M H2SO3 with M NaOH

39. Sample Problem 19.3
Calculating the pH During a Weak Acid-
Strong Base Titration
PROBLEM:
Calculate the pH during the titration of mL of M propanoic acid (HPr; Ka = 1.3x10-5) after adding the following volumes of M NaOH:
(a) 0.00mL
(b) mL
(c) mL
(d) mL
PLAN:
The amounts of HPr and Pr- will be changing during the titration. Remember to adjust the total volume of solution after each addition.
SOLUTION:
(a) Find the starting pH using the methods of Chapter 18.
Ka = [Pr-][H3O+]/[HPr]
[Pr-] = x = [H3O+]
[Pr-] = x = [H3O+]
x = 1.1x10-3 ; pH = 2.96
(b)
HPr(aq) + OH-(aq) Pr-(aq) + H2O (l)
Amount (mol)
Before addition - -
Addition - - -
After addition -

40. Sample Problem 19.3
Calculating the pH During a Weak Acid-
Strong Base Titration
continued
[H3O+] = 1.3x10-5
mol
mol
= 4.3x10-6M
pH = 5.37
(c) When 40.00mL of NaOH are added, all of the HPr will be reacted and the [Pr -] will be
( mol)
( L) + ( L)
= M
Ka x Kb = Kw
Kb = Kw/Ka = 1.0x10-14/1.3x10-5 = 7.7x10-10
[H3O+] = Kw / = 1.6x10-9M
pH = 8.80
(d) 50.00mL of NaOH will produce an excess of OH-.
M = ( )
(0.0900L)
mol XS base = (0.1000M)( L L) = mol
[H3O+] = 1.0x10-14/ = 9.0x10-11M
M =
pH = 12.05

41. Sickle shape of red blood cells in sickle cell anemia.
Figure 19.11
Sickle shape of red blood cells in sickle cell anemia.

42. Equilibria of slightly soluble ionic compounds:
An ionic substance dissociates completely in water to form hydrate cations and anions.
Both the forward and reverse reactions take place.
A stage is reached when no more solid dissolves. This is called as dynamic equilibrium.
CaF2 (s) ↔Ca2+ (aq) + 2F- (aq)
Ksp= [Ca2+ ] [F-]2
Ksp is called as the solubility product constant or solubility product for the equilibrium expression.

43. Equilibria of slightly soluble ionic compounds:
For the hypothetical compound, MpXq
At equilibrium Qsp = [Mn+]p [Xz-]q= Ksp
Higher the Ksp greater the solubility for formulas containing same total number of ions.
If salts being compared produce different number of ions.
Compare their solubilities, which is opposite of Ksp

44. At equilibrium Qsp = [Mn+]p [Xz-]q = Ksp
Ion-Product Expression (Qsp)
and Solubility Product Constant (Ksp)
For the hypothetical compound, MpXq
At equilibrium Qsp = [Mn+]p [Xz-]q = Ksp

45. Sample Problem 19.4
Writing Ion-Product Expressions for Slightly
Soluble Ionic Compounds
PROBLEM:
Write the ion-product expression for each of the following:
(a) Magnesium carbonate
(b) Iron (II) hydroxide
(c) Calcium phosphate
(d) Silver sulfide
PLAN:
Write an equation which describes a saturated solution. Take note of the sulfide ion produced in part (d).
SOLUTION:
(a) MgCO3(s) Mg2+(aq) + CO32-(aq)
Ksp = [Mg2+][CO32-]
(b) Fe(OH)2(s) Fe2+(aq) + 2OH- (aq)
Ksp = [Fe2+][OH-] 2
(c) Ca3(PO4)2(s) Ca2+(aq) + 2PO43-(aq)
Ksp = [Ca2+]3[PO43-]2
(d) Ag2S(s) Ag+(aq) + S2-(aq)
S2-(aq) + H2O(l) HS-(aq) + OH-(aq)
Ksp = [Ag+]2[HS-][OH-]
Ag2S(s) + H2O(l) Ag+(aq) + HS-(aq) + OH-(aq)

46. Table 19.2 Solubility-Product Constants (Ksp) of Selected Ionic
Compounds at 250C
Name, Formula
Ksp
Aluminum hydroxide, Al(OH)3
3 x 10-34
Cobalt (II) carbonate, CoCO3
1.0 x 10-10
Iron (II) hydroxide, Fe(OH)2
4.1 x 10-15
Lead (II) fluoride, PbF2
3.6 x 10-8
Lead (II) sulfate, PbSO4
1.6 x 10-8
Mercury (I) iodide, Hg2I2
4.7 x 10-29
Silver sulfide, Ag2S
8 x 10-48
Zinc iodate, Zn(IO3)2
3.9 x 10-6

47. Problems
P.10. CuBr has a measured solubility of 2.0 x 10-4 mol/L at 25 C. Calculate Ksp.
P.11. Calculate Ksp value for Bi2S3 which has a solubility of 1.0 x mol/L at 25 C.
P.12. Calculate the solubility of copper (II) iodate at 25 C, if Ksp = 1.4 x 10-7

48. Table 19.3 Relationship Between Ksp and Solubility at 250C
No. of Ions
Formula
Cation:Anion
Ksp
Solubility (M) 2
MgCO3
1:1
3.5 x 10-8
1.9 x 10-4 2
PbSO4
1:1
1.6 x 10-8
1.3 x 10-4 2
BaCrO4
1:1
2.1 x 10-10
1.4 x 10-5 3
Ca(OH)2
1:2
5.5 x 10-6
1.2 x 10-2 3
BaF2
1:2
1.5 x 10-6
7.2 x 10-3 3
CaF2
1:2
3.2 x 10-11
2.0 x 10-4 3
Ag2CrO4
2:1
2.6 x 10-12
8.7 x 10-5

49. Common ion Effect:
Presence of a common ion decreases solubility of a slightly soluble ionic compound.

50. The effect of a common ion on solubility
Figure 19.12
The effect of a common ion on solubility
PbCrO4(s) Pb2+(aq) + CrO42-(aq)
CrO42- added
PbCrO4(s) Pb2+(aq) + CrO42-(aq)

51. Problem
.13. Calculate the solubility of CaF2 (Ksp=4.0 x 10-11) in a M NaF solution.

52. Test for the presence of a carbonate.
Figure 19.13
Test for the presence of a carbonate.

53. pH and solubility:
If the compound contains the anion of a weak acid, addition of H3O+ increases its solubility.

54. Sample Problem 19.8
Predicting the Effect on Solubility of Adding
Strong Acid
PROBLEM:
Write balanced equations to explain whether addition of H3O+ from a strong acid affects the solubility of these ionic compounds:
(a) Lead (II) bromide
(b) Copper (II) hydroxide
(c) Iron (II) sulfide
PLAN:
Write dissolution equations and consider how strong acid would affect the anion component.
SOLUTION:
(a) PbBr2(s) Pb2+(aq) + 2Br-(aq)
Br- is the anion of a strong acid.
No effect.
(b) Cu(OH)2(s) Cu2+(aq) + 2OH-(aq)
OH- is the anion of water, which is a weak acid. Therefore it will shift the solubility equation to the right and increase solubility.
(c) FeS(s) Fe2+(aq) + S2-(aq)
S2- is the anion of a weak acid and will react with water to produce OH-.
FeS(s) + H2O(l) Fe2+(aq) + HS-(aq) + OH-(aq)
Both weak acids serve to increase the solubility of FeS.

55. Predicting formation of precipitate:
If Qsp > Ksp ppt occurs
If Qsp < Ksp solution is unsaturated, no ppt occurs.
If Qsp = Ksp solution is saturated and no change occurs.

56. Sample Problem 19.9
Predicting Whether a Precipitate Will Form
PROBLEM:
A common laboratory method for preparing a precipitate is to mix solutions of the component ions. Does a precipitate form when 0.100L of 0.30M Ca(NO3)2 is mixed with 0.200L of 0.060M NaF?
PLAN:
Write out a reaction equation to see which salt would be formed. Look up the Ksp valus in a table. Treat this as a reaction quotient, Q, problem and calculate whether the concentrations of ions are > or < Ksp. Remember to consider the final diluted solution when calculating concentrations.
SOLUTION:
CaF2(s) Ca2+(aq) + 2F-(aq) Ksp = 3.2x10-11
mol Ca2+ = L(0.30mol/L) = 0.030mol
[Ca2+] = 0.030mol/0.300L = 0.10M
mol F- = L(0.060mol/L) = 0.012mol
[F-] = 0.012mol/0.300L = 0.040M
Q = [Ca2+][F-]2 =
(0.10)(0.040)2 = 1.6x10-4
Q is >> Ksp and the CaF2 WILL precipitate.

57. Equilibria Involving Complex Ions:
Complex ion: A central metal ion covalently bonded to two or more anions, or molecules called ligands.
Ligand: Lewis base that contains a lone pair of electron that can be donated to an empty orbital on the metal ion to form a covalent bond.
Ex: H2O, NH3, CN-,Cl-

58. Equilibria Involving Complex Ions:
Coordination number: Number of ligands attached to a metal ion.ex; 6,4,2
Formation constants/Stability Constants : Metal ions add ligands one at a time in steps.
A ligand increases the solubility of a slightly soluble ionic compound if it forms a complex ion with the cation.

59. Cr(NH3)63+, a typical complex ion.
Figure 19.14
Cr(NH3)63+, a typical complex ion.

60. The stepwise exchange of NH3 for H2O in M(H2O)42+.
Figure 19.15
M(H2O)42+
The stepwise exchange of NH3 for H2O in M(H2O)42+.
NH3
M(H2O)3(NH3)2+
3NH3
M(NH3)42+

62. Sample Problem 19.10
Calculating the Concentration of a Complex Ion
PROBLEM:
An industrial chemist converts Zn(H2O)42+ to the more stable Zn(NH3)42+ by mixing 50.0L of M Zn (H2O)42+ and 25.0L of 0.15M NH3. What is the final [Zn (H2O)42+]?
Kf of Zn(NH3)42+ is 7.8x108.
PLAN:
Write the reaction equation and Kf expression. Use a reaction table to list various concentrations. Remember that components will be diluted when mixed as you calculate final concentrations. It is obvious that there is a huge excess of NH3 and therefore it will drive the reaction to completion.
SOLUTION:
Zn(H2O)42+(aq) + 4NH3(aq) Zn(NH3)42+(aq) + 4H2O(l)
Kf =
[Zn(NH3)42+]
[Zn(H2O)42+][NH3]4
[Zn(H2O)42+]initial = (50.0L)(0.0020M)
75.0L
= 1.3x10-3 M
[NH3]initial =
(25.0L)(0.15M)
75.0L
= 5.0x10-2 M

63. Sample Problem 19.10
Calculating the Concentration of a Complex Ion
continued
Since we assume that all of the Zn(H2O)42+ has reacted, it would use 4 times its amount in NH3.
[NH3]used = 4(1.3x10-3M) = 5.2x10-3M
[Zn(H2O)42+]remaining = x(a very small amount)
Zn(H2O)42+(aq) + 4NH3(aq) Zn(NH3)42+(aq) + 4H2O(l)
Concentration(M)
Initial
1.3x10-3
5.0x10-2 -
Change
~(-1.3x10-3)
~(-5.2x10-3)
~(+1.3x10-3) -
Equilibrium x
4.5x10-2
1.3x10-3 -
Kf =
[Zn(NH3)42+]
[Zn(H2O)42+][NH3]4
= 7.8x108 =
(1.3x10-3)
x(4.5x10-2)
x = 4.1x10-7M

64. Sample Problem 19.11
Calculating the Effect of Complex-Ion Formation
on Solubility
PROBLEM:
In black-and-white film developing, excess AgBr is removed from the film negative by “hypo”, an aqueous solution of sodium thiosulfate (Na2S2O3), through formation of the complex ion Ag(S2O3)23-. Calculate the solubility of AgBr in (a) H2O; (b) 1.0M hypo. Kf of Ag(S2O3)23- is 4.7x1013 and Ksp AgBr is 5.0x10-13.
PLAN:
Write equations for the reactions involved. Use Ksp to find S, the molar solubility. Consider the shifts in equilibria upon the addition of the complexing agent.
SOLUTION:
AgBr(s) Ag+(aq) + Br-(aq) Ksp = [Ag+][Br-] = 5.0x10-13
(a)
S = [AgBr]dissolved = [Ag+] = [Br-]
Ksp = S2 = 5.0x10-13 ; S = 7.1x10-7M
(b)
AgBr(s) Ag+(aq) + Br-(aq)
Ag+(aq) + 2S2O32-(aq) Ag(S2O3)23-(aq)
AgBr(s) + 2S2O32-(aq) Br -(aq) + Ag(S2O3)23-(aq)

65. Sample Problem 19.11
Calculating the Effect of Complex-Ion Formation
on Solubility
continued
[Br-][Ag(S2O3]23-
[AgBr][S2O32-]2
Koverall = Ksp x Kf =
= (5.0x10-13)(4.7x1013)
= 24
AgBr(s) + 2S2O32-(aq) Br-(aq) + Ag(S2O3)23-(aq)
Concentration(M)
Initial -
1.0
Change -
-2S +S +S
Equilibrium -
1.0-2S S S
Koverall = S2
(1.0-2S)2
= 24 S
1.0-2S
= (24)1/2
S = [Ag(S2O3)23-] = 0.45M

66. The amphoteric behavior of aluminum hydroxide.
Figure 19.16
The amphoteric behavior of aluminum hydroxide.
3H2O(l) + Al(H2O)3(OH)3(s)
Al(H2O)3(OH)3(s)
Al(H2O)3(OH)4-(s) + H2O(l)

67. Sample Problem 19.12
Separating Ions by Selective Precipitation
PROBLEM:
A solution consists of 0.20M MgCl2 and 0.10M CuCl2. Calculate the [OH-] that would separate the metal ions as their hydroxides. Ksp of Mg(OH)2= is 6.3x10-10; Ksp of Cu(OH)2 is 2.2x10-20.
PLAN:
Both precipitates are of the same ion ratio, 1:2, so we can compare their Ksp values to determine which has the greater solubility.
It is obvious that Cu(OH)2 will precipitate first so we calculate the [OH-] needed for a saturated solution of Mg(OH)2. This should ensure that we do not precipitate Mg(OH)2. Then we can check how much Cu2+ remains in solution.
SOLUTION:
Mg(OH)2(s) Mg2+(aq) + 2OH-(aq) Ksp = 6.3x10-10
Cu(OH)2(s) Cu2+(aq) + 2OH-(aq) Ksp = 2.2x10-20
[OH-] needed for a saturated Mg(OH)2 solution =
= 5.6x10-5M

68. Sample Problem 19.12
Separating Ions by Selective Precipitation
continued
Use the Ksp for Cu(OH)2 to find the amount of Cu remaining.
[Cu2+] = Ksp/[OH-]2 = 2.2x10-20/(5.6x10-5)2 =
7.0x10-12M
Since the solution was 0.10M CuCl2, virtually none of the Cu2+ remains in solution.

69. The general procedure for separating ions in qualitative analysis.
Figure 19.17
The general procedure for separating ions in qualitative analysis.
Add precipitating ion
Add precipitating ion
Centrifuge
Centrifuge

70. A qualitative analysis scheme for separating cations into five ion groups.
Figure 19.18
Acidify to pH 0.5; add H2S
Centrifuge
Add NH3/NH4+ buffer(pH 8)
Centrifuge
Add (NH4)2HPO4
Centrifuge
Add
6M HCl
Centrifuge

71. Step 5 Dissolve in HCl and add KSCN
Extra:
A qualitative analysis scheme for Ag+,Al3+,Cu2+, and Fe3+
Step 1 Add NH3(aq)
Centrifuge
Centrifuge
Step 2 Add HCl
Step 3 Add NaOH
Centrifuge
Step 4 Add HCl, Na2HPO4
Step 5 Dissolve in HCl and add KSCN

72. A view inside Carlsbad Caverns, New Mexico
Figure B19.1
A view inside Carlsbad Caverns, New Mexico

73. Formation of acidic precipitation.
Figure B19.3
Formation of acidic precipitation.

74. A forest damaged by acid rain
Figure B19.4
A forest damaged by acid rain

75. The effect of acid rain on marble statuary.
Figure B19.5
The effect of acid rain on marble statuary.
1944
1994
Location: New York City