Transcript Chapter 15 Acids and Bases
Weak
Bases
in weak bases, only a small fraction of molecules accept H ’ s weak electrolyte most of the weak base molecules do not take H + from water much less than 1% ionization in water [HO – ] << [weak base] finding the pH of a weak base solution is similar to finding the pH of a weak acid
NH 3 + H 2 O NH 4 + + OH -
1
2
Find the pH of 0.100 M NH 3 (
aq
) solution given
K
b NH 3 for = 1.76 x 10 -5 Write the reaction for the base with water Construct an ICE table for the reaction Enter the initial concentrations – assuming the [OH ] from water is ≈ 0
NH
3
+ H
2
O
[NH 3 ] ] 0.100
NH
4 +
+ OH
[NH 4 + ] ] 0 [OH ] ] ≈ 0 since no products initially,
Q
c = 0, and the reaction is proceeding forward 3
Find the pH of 0.100 M NH 3 (
aq
) solution given
K
b NH 3 for = 1.76 x 10 -5 represent the change in the concentrations in terms of
x
sum the columns to find the equilibrium concentrations in terms of
x
initial change equilibrium substitute into the equilibrium constant expression [NH 3 ] 0.100
-
x
0.100 -
x
[NH 4 + ] 0 +
x x
[OH ] 0 +
x x K
b = [NH 4 + ][OH ] [ NH 3 ] = ( 1.00
´ 10 1 -
x
) 4
Find the pH of 0.100 M NH 3 (
aq
) solution given
K
b NH 3 for = 1.76 x 10 -5 determine the value of
K
b from Table 15.8
since
K
b is very small, approximate the [NH 3 ] eq = [NH 3 ] init and solve for
x
initial change equilibrium [NH 3 ] 0.100
-
x
0.100
x
[NH 4 + ] 0 +
x x
[OH ] ≈ 0 +
x x K K
b b = = éé [ ] [ éé [ ] [ ] 3 ] éé = 1 .
76 ´ 10 5 = ( (
x
2 ( ) ( ) ´ é 10 1 1 ) 1 .
00 ´ 10 1
x
)
x
= ( 1.76
´ 10 5 ) ( 1.00
´ 10 1 )
x
= 1.33
´ 10 3 5
Find the pH of 0.100 M NH 3 (
aq
) solution given
K
b NH 3 for = 1.76 x 10 -5 check if the approximation is valid by seeing if
x
< 5% of [NH 3 ] init initial change equilibrium [NH 3 ] 0.100
-
x
0.100
x
= 1.33 x 1 .
33 1 .
00 ´ 10 3 ´ 10 1 ´ 100 10 -3 % = 1 .
33 % < 5 % the approximation is valid [NH 4 + ] 0 +
x x
[OH ] ≈ 0 +
x x
6
Find the pH of 0.100 M NH 3 (
aq
) solution given
K
b NH 3 for = 1.76 x 10 -5 substitute
x
into the equilibrium concentration definitions and solve [NH 3 ] -
x
[NH 4 + ] +
x
[OH ] +
x
0.100
x
1.33E-3
x
1.33E-3
x
[
NH
3 ] =
0.100
-
x x
= 1.33 x =
0.100
10 -3 (
1.33
´
10
3 ) =
0.099 M
[ NH 4 + ] = [OH ] =
x
= 1 .
33 ´ 10 3 M 7
Find the pH of 0.100 M NH 3 (
aq
) solution given
K
b NH 3 for = 1.76 x 10 -5 use the [OH ] to find the [H 3 O + ] using
K
w substitute [H 3 O + ] into the formula for pH and solve initial change [NH 3 ] 0.100
-
x
[NH 4 + ] 0 +
x
[OH ] ≈ 0 +
x
equilibrium 0.099
1.33E-3 1.33E-3
K
w = [H 3 O + ][OH ] [H 3 O + ] = [H 3 O + ] = 1.00
1.33
´ ´ 10 14 10 3 7.52
´ 10 12 pH = = ( ( 3 O + ) ´ 10 12 ) = 11.124
8
Find the pH of 0.100 M NH 3 (
aq
) solution given
K
b NH 3 for = 1.76 x 10 -5 check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated
K
b to the given
K
b initial change equilibrium though not exact, the answer is reasonably close [NH 3 ] 0.100
-
x
0.099
[NH 4 + ] 0 +
x
1.33E-3
K
b = = [ ] [ ] [ NH 3 ] ( 1.33
´ 10 3 ( 0.099
) ) 2 = 1.8
´ 10 5 [OH ] ≈ 0 +
x
1.33E-3 9
Practice – Find the pH of a 0.0015 M morphine solution,
K
b
= 1.6 x 10
-6 10
Practice – Find the pH of a 0.0015 M morphine solution,
K
b = 1.6 x 10 -6 Write the reaction for the base with water Construct an ICE table for the reaction
B + H
2
O
[B]
BH
+
+ OH
[BH + ] [OH ] initial 0.0015
0 ≈ 0 Enter the initial concentrations – assuming the [OH ] from water is ≈ 0 change equilibrium since no products initially,
Q
c = 0, and the reaction is proceeding forward 11
Find the pH of a 0.0015 M morphine solution,
K
b = 1.6 x 10 -6 represent the change in the concentrations in terms of
x
sum the columns to find the equilibrium concentrations in terms of
x
initial change equilibrium substitute into the equilibrium constant expression [B] 0.0015
-
x
0.0015 -
x
[BH + ] 0 +
x x
[OH ] 0 +
x x K
b = [BH + ][OH ] [ ] = ( 1.5
´ 10 3 -
x
) 12
Find the pH of a 0.0015 M morphine solution,
K
b = 1.6 x 10 -6 determine the value of
K
b since
K
b is very small, approximate the [B] eq [B] init and solve for
x
= initial change equilibrium [B] 0.0015
-
x
0.0015
x
[BH + ] 0 +
x x
[OH ] ≈ 0 +
x x K K
b b = = éé 1 .
6 [ ] ´ 10 éé éé [ ] 6 [ ] = éé = = 1 .
5 ´ (
x
2 ( 10 ( ) ( ) ´ é 10 3 3 1 )
x
)
x
= ( 1.6
´ 10 6 ) ( 1.5
´ 10 3 )
x
= 4.9
´ 10 5 13
Find the pH of a 0.0015 M morphine solution,
K
b = 1.6 x 10 -6 check if the approximation is valid by seeing if
x
< 5% of [B] init initial change equilibrium [B] 0.0015
-
x
0.0015
[BH + ] 0 +
x x x
= 4.9 x 10 -5 4 .
9 1 .
5 ´ 10 5 ´ 10 3 ´ 100 % = 3 .
3 % the approximation is valid < 5 % [OH ] ≈ 0 +
x x
14
Find the pH of a 0.0015 M morphine solution,
K
b = 1.6 x 10 -6 substitute
x
into the equilibrium concentration definitions and solve [BH + ] ] [OH ] ] equilibrium equilibrium -
x
-
x
0.0015
x
+
x
+
x
+
x
+
x x
= 4.9 x 10 -5 [
Morphine
] =
0.0015
[ BH + ] = -
x
[OH = ]
0.0015
=
x
= ( 4 .
9
4.9
´
10
5 ´ 10 5 M ) =
0.0015 M
15
Find the pH of a 0.0015 M morphine solution,
K
b = 1.6 x 10 -6 use the [OH ] to find the [H 3 O + ] using
K
w substitute [H 3 O + ] into the formula for pH and solve initial change equilibrium
K
w = [H 3 O + ][OH ] [H 3 O + ] [H 3 O + ] = = 1.00
4 .
9 ´ 10 14 ´ 10 5 2 .
0 ´ 10 10 [B] 0.0015
-
x
0.0015
[BH + ] 0 +
x
4.9E-5 [OH ] ≈ 0 +
x
4.9E-5
pH
= = ( ( 3
O
+ ) ´
10
10 ) =
9.69
16
Find the pH of a 0.0015 M morphine solution,
K
b = 1.6 x 10 -6 check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated
K
b to the given
K
b initial [B] 0.0015
[BH + ] 0 change -
x
+
x
equilibrium 0.0015
K
b = [ ] [ ] 4.9E-5 the answer matches the given
K
b = (
4.9
´
10
5 (
0.0015
) ) 2 =
1.6
´
10
6 [OH ] ≈ 0 +
x
4.9E-5 17
Acid-Base Properties of
Salts
salts are water soluble ionic compounds salts that contain the cation of a strong base and an anion that is the conjugate base of a weak acid are basic
Example
: NaHCO 3 Na + solutions are basic is the cation of the strong base NaOH HCO 3 − is the conjugate base of the weak acid H 2 CO 3 Conversely salts that contain cations that are the conjugate acid of a weak base and an anion of a strong acid are acidic
Example
: NH 4 Cl solutions are acidic NH Cl − 4 + is the conjugate acid of the weak base NH is the anion of the strong acid HCl 3 18
Anions
as Weak Bases
every anion can be thought of as the conjugate base of an acid therefore, every anion can potentially be a base A − (
aq
) + H 2 O(
l
) HA(
aq
) + OH − (
aq
)
the stronger the acid HA is, the weaker the conjugate base A is
an anion that is the conjugate base of a strong acid is pH neutral Cl − (
aq
) + H 2 O(
l
) HCl(
aq
) + OH − (
aq
) an anion that is the conjugate base of a weak acid is basic F − (
aq
) + H 2 O(
l
) HF(
aq
) + OH − (
aq
) since HF is a weak acid, the position of this equilibrium favors the right 19
Use the Table to Determine if the Given Anion Is Basic or Neutral a) b) c) NO 3 − the conjugate base of a strong acid, therefore neutral HCO 3 − the conjugate base of a weak acid, therefore basic PO 4 3 − the conjugate base of a weak acid, therefore basic 20
Relationship between
K
a
of an Acid and
K
b
of its Conjugate Base many reference books only give tables of
K
a can be found from them values because
K
b values when you add equations, you multiply the
K
’ s HA(
aq
) + H 2 O(
l
) Û A (
aq
) + H 2 O(
l
) Û A (
aq
) + H 3 O + (
aq
)
K
a = [ A ][H 3 O + ] [HA] HA(
aq
) + OH (
aq
)
K
b = [ HA][H [A ] 3 O + ]
K
a ´
K
b = [A ][H 3 O + ] [HA] ´ [HA][OH [A ] ]
K
a ´
K
b = [H 3 O + ][OH ] =
K
w 21
Find the pH of 0.100 M NaCHO 2 (
aq
) solution assuming
K
a =1.8x10
-4 for HCOOH Na + is the cation of a strong base – pH neutral. The CHO 2 − is the anion of a weak acid – pH basic HCO 2 − + H 2 O HCHO 2 + OH Write the reaction for the anion with water initial change [CHO 2 − ] 0.100
[HCHO 2 ] 0 [OH ] ≈ 0 Construct an ICE table for the reaction equilibrium Enter the initial concentrations – assuming the [OH ] from water is ≈ 0 22
Find the pH of 0.100 M NaCHO 2 (
aq
) solution assuming
K
a =1.8x10
-4 for HCOOH represent the change in the concentrations in terms of
x
sum the columns to find the equilibrium concentrations in terms of
x
Calculate the value of
K
b from the value of
K
a =1.8x10
-4 substitute into the equilibrium constant expression initial change equilibrium [CHO 2 − ] 0.100
-
x
0.100 -
x
[HCHO 2 ]
K K
b a ´
K
b =
K
w = 1 .
0 1 .
8 ´ 10 14 ´ 10 4 = 5 .
6 ´ 10 11 0 +
x x
[OH ]
K
b = [HCHO 2 ][OH ] [ CHO 2 ] = ( ( ) ( ) 1.00
´ 10 1 -
x
) ≈ 0 +
x x
23
Find the pH of 0.100 M NaCHO 2 (
aq
) solution assuming
K
a =1.8x10
-4 for HCOOH since
K
b is very small, approximate the [CHO 2 − ] eq [CHO 2 − ] init and solve for
x
= initial change
K
b for CHO 2 − = 5.6 x 10 -11 equilibrium [CHO 2 − ] 0.100
-
x
0.100 -
x
[HCHO 2 ] 0 +
x x
[OH ] ≈ 0 +
x x K
b b = = [HCHO 2 2 ][OH ] ] [ éé CHO 2 2 éé ] = = ( ( ( ) ( ) 1.00
´ 10 1 ) -
x
)
x
= 5 .
6 ´ 10 11 = 1 .
00
x
2 ´ 10 1 ( 5.6
´ 10 11 ) ( 1.00
´ 10 1 )
x
= 2.4
´ 10 6 24
Find the pH of 0.100 M NaCHO 2 (
aq
) solution check if the approximation is valid by seeing if
x
< 5% of [CHO 2 − ] init
K
b for CHO 2 − = 5.6 x 10 -11 initial change equilibrium [CHO 2 − ] 0.100
-
x
0.100
[HCHO 2 ] 0 +
x x
[OH ] ≈ 0 +
x x x
= 2.4 x 10 -6 2 .
4 1 .
00 ´ 10 ´ 6 10 1 ´ 100 % = 0 .
0024 % < 5 % the approximation is valid 25
Find the pH of 0.100 M NaCHO 2 (
aq
) solution substitute
x
into the equilibrium concentration definitions and solve
K
b for CHO 2 − = 5.6 x 10 -11 [CHO 2 − ] [HCHO 2 ] [OH ] -
x
0.100 −
x
+
x
2.4E-6
x
+
x
2.4E-6
x x
= 2.4 x 10 -6 [ CHO 2 ] = 0.100
-
x
= 0.100
( 2.4
´ 10 6 ) = 0.100 M [HCHO 2 ] = [OH ] =
x
= 2.4
´ 10 6 M 26
Find the pH of 0.100 M NaCHO 2 (
aq
) solution use the [OH ] to find the [H 3 O + ] using
K
w substitute [H 3 O + ] into the formula for pH and solve
K
b for CHO 2 − = 5.6 x 10 -11 initial change equilibrium
K
[H [H w = [H 3 O + ][OH ] 3 O 3 O + + ] ] = = 1.00
´ 10 14 4 .
2 .
4 2 ´ ´ 10 10 9 6 [CHO 2 − ] 0.100
-
x
0.100
[HCHO 2 ] 0 +
x
2.4E-6 [OH ] ≈ 0 +
x
2.4E-6 pH = ( 3 O + = ( ´ 10 9 ) ) = 8.38
27
Find the pH of 0.100 M NaCHO 2 (
aq
) solution check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated
K
b to the given
K
b initial change equilibrium [CHO 2 − ] 0.100
-
x
[HCHO 2 ] 0 +
x
[OH ] ≈ 0 +
x
though not exact, the answer is reasonably close 0.100
2.4E-6 2.4E-6
K
b = = [ HCHO 2 [ ] [ ] CHO 2 ] ( 2.4
´ 10 6 ( 0.100
) ) 2 = 5.8
´ 10 11
K
b for CHO 2 − = 5.6 x 10 -11 28
Polyatomic Cations
as Weak Acids some polyatomic cations can be thought of as the conjugate acid of a base therefore, some cations can potentially be an acid MH + (
aq
) + H 2 O(
l
) MOH(
aq
) + H 3 O + (
aq
)
the stronger the base MOH is, the weaker the conjugate acid MH + is
a cation that is the
counterion of a strong base
pH neutral (Na + , K + etc) is a cation that is the
conjugate acid of a weak base
is acidic NH 4 + (
aq
) + H 2 O(
l
) NH 3 (
aq
) + H 3 O + (
aq
) since NH 3 the right is a weak base, the position of this equilibrium favors 29
Metal Cations
as Weak Acids cations of small, highly charged metals are weakly acidic alkali metal cations and alkali earth metal cations pH neutral cations are hydrated Al(H 2 O) 6 3+ (
aq
) + H 2 O(
l
) Al(H 2 O) 5 (OH) 2+ (
aq
) + H 3 O + (
aq
) 30
Determine if the Given Cation is Acidic or Neutral a) C 2 H 5 NH 3 + (K b of C 2 H 5 NH 2 = 5.6 x 10 -4 ) the conjugate acid of a weak base, therefore acidic b) Ca 2+ (K b of Ca(OH) 2 = 3.74 x 10 -3 ) the counterion of a (not so) strong base, therefore neutral c) Cr 3+ a highly charged metal ion, therefore acidic 31
Classifying Salt Solutions as Acidic, Basic, or Neutral if the
salt cation
is the
counterion of a strong base
and the anion is the conjugate base of a strong acid, it will form a
neutral
solution NaCl Ca(NO 3 ) 2 KBr if the
salt cation
is the
counterion of a strong base
and the anion is the conjugate base of a weak acid, it will form a
basic
solution NaF Ca(C 2 H 3 O 2 ) 2 KNO 2 if the
salt cation
is the
conjugate acid of a weak base
and the anion is the conjugate base of a strong acid, it will form an
acidic
solution NH 4 Cl if the
salt cation
is a
highly charged metal ion
and the anion is the conjugate base of a strong acid, it will form an
acidic
solution Al(NO 3 ) 3 32
Classifying Salt Solutions as Acidic, Basic, or Neutral if the
salt cation
is the
conjugate acid of a weak base
and the
anion
is the
conjugate base of a weak acid
, the pH of the solution depends on the relative strengths of the acid and base NH 4 F HF is a stronger acid than NH 4 +
K
a of NH be acidic 4 + is larger than
K
b of the F − ; therefore the solution will Tro, Chemistry: A Molecular Approach 33
a) b) c) Determine whether a solution of the following salts is acidic, basic, or neutral SrCl 2 Sr 2+ is the counterion of a strong base, pH neutral Cl − is the conjugate base of a strong acid, pH neutral solution will be
pH neutral
AlBr 3 Al 3+ is a small, highly charged metal ion, weak acid Br − is the conjugate base of a strong acid, pH neutral solution will be
acidic
CH 3 NH 3 NO 3 CH 3 NH 3 + NO 3 − is the conjugate acid of a weak base, acidic is the conjugate base of a strong acid, pH neutral solution will be
acidic
34
Determine whether a solution of the following salts is acidic, basic, or neutral d) e) NaCHO 2 Na + is the counterion of a strong base, pH neutral CHO 2 − is the conjugate base of a weak acid, basic solution will be
basic
NH 4 F NH 4 + F − is the conjugate acid of a weak base, acidic is the conjugate base of a weak acid, basic
K
a (NH 4 + ) >
K
b (F − ); solution will be
acidic
35
Polyprotic
Acids
since polyprotic acids ionize in steps, each H has a separate
K
a
K
a1 >
K
a2 >
K
a3 generally, the difference in the pH
K
a values is great enough so that the second ionization does not happen to a large enough extent to affect most pH problems just do first ionization except H 2 SO 4 use [H 2 SO 4 ] o = [H 3 O + ] for the second ionization [A 2 ] =
K
a2 as long as the second ionization is negligible 36
37
Find the pH of 0.0100 M H 2 SO 4 (
aq
) solution @ 25 ° C Write the reactions for the acid with water Construct an ICE table for the reaction Enter the initial concentrations – assuming the [HSO 4 − ] and [H 3 O + ] is ≈ [H 2 SO 4 ] H 2 SO 4 + H 2 O HSO 4 + H 3 O + K a1 = big!
HSO 4 + H 2 O SO 4 2 + H 3 O + K a2 = 0.012
initial change equilibrium [HSO 4 ] 0.0100
[SO 4 2 ] 0 [H 3 O + ] 0.0100
38
Find the pH of 0.0100 M H 2 SO 4 (
aq
) solution @ 25 ° C represent the change in the concentrations in terms of
x
initial [HSO 4 ] 0.0100
change sum the columns to find the equilibrium concentrations in terms of
x
substitute into the equilibrium constant expression equilibrium −
x
0.0100 −
x K
a = [SO 4 2 ][H 3 O + ] [HSO 4 ] = [SO 4 2 ] 0 ( +
x x
( 0.0100
0.0100
[H 3 O + ] 0.0100
0.0100 −
x
+
x
)
x
) +
x
39
Find the pH of 0.0100 M H 2 SO 4 (
aq
) solution @ 25 ° C expand and solve for
x
using the quadratic formula
K
a for HSO 4 − = 0.012
K
a = [SO 4 2 ][H 3 O + ] [HSO 4 ] = ( ) ( ( 0.0100
+ 0.0100
-
x
)
x
) 0.012
= ( ( 0.0100
+
x
)
x
0.0100
-
x
) 1.2
´ 10 4 1.2
´ 10 2
x
= 1.00
´ 10 2
x
+
x
2 0 =
x
2 + 0.022
x
0.00012
x
= 0 .
022 ± ( 0 .
022 ) 2 4 ( 1 )(
x
= 2 ( 1 ) 0 .
027 or 0.0045
0 .
00012 ) 40
Find the pH of 0.0100 M H 2 SO 4 (
aq
) solution @ 25 ° C substitute
x
into the equilibrium concentration definitions and solve [HSO 4 ] [SO 4 2 ] [H 3 O + ] −
x
0.0100 −
x
+
x
+
x
0.0100 −
x x
= 0.0045
[ HSO 4 ] = [ H 3 O + ] = 0 .
0100 ( 0 .
0100 +
x
=
x
) = 0 .
0100 0 .
0145 M ( 0 .
0045 ) = 0 .
0055 M [ SO 2 4 ] =
x
= 0 .
0045 M 41
Find the pH of 0.0100 M H 2 SO 4 (
aq
) solution @ 25 ° C substitute [H 3 O + ] into the formula for pH and solve initial [HSO 4 ] 0.0100
change −
x
equilibrium pH = = ( 0.0055
( 3 O + ) ) = 1.839
[SO 4 2 ] 0 +
x
0.0045
[H 3 O + ] 0.0100
+
x
0.0145
42
Find the pH of 0.0100 M H 2 SO 4 (
aq
) solution @ 25 ° C check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated
K
a to the given
K
a initial [HSO 4 ] 0.0100
[SO 4 2 ] 0 [H 3 O + ] 0.0100
change −
x
+
x
+
x
equilibrium the answer matches 0.0055
0.0045
0.0145
K
a = = [ ] [ [ H 3 O + HSO 4 ] ( 0.0045
) ( 0.0145
( 0.0055
) ) ] = = 1.2
´ 10 2
K
a for HSO 4 − = 0.012
43
Strengths of
Binary Acids
the more + H X δ- polarized the bond, the more acidic the bond the stronger the H-X bond, the weaker the acid binary acid strength increases to the right across a period H-C < H-N < H-O < H-F binary acid strength increases down the column H-F < H-Cl < H-Br < H-I 44
Strengths of
Oxyacids
, H-O-Y
the more electronegative the Y atom, the stronger the acid helps weakens the H-O bond the more oxygens attached to Y, the stronger the acid further weakens and polarizes the H-O bond 45
Lewis Acid - Base
Theory
electron sharing electron donor = Lewis Base = nucleophile must have a lone pair of electrons electron acceptor = Lewis Acid = electrophile electron deficient when Lewis Base gives electrons from lone pair to Lewis Acid, a covalent bond forms between the molecules Nucleophile
: +
Electrophile Nucleophile
:
Electrophile product called an
adduct
other acid-base reactions also Lewis Tro, Chemistry: A Molecular Approach 46
Example - Complete the Following Lewis Acid-Base Reactions Label the Nucleophile and Electrophile
H OH C H + OH -1 OH H C Electrophile H •• + OH -1 Nucleophile H OH C H OH 47 Tro, Chemistry: A Molecular Approach
Practice - Complete the Following Lewis Acid-Base Reactions Label the Nucleophile and Electrophile
CaO + SO 3 KI + I 2 Tro, Chemistry: A Molecular Approach 48
Practice - Complete the Following Lewis Acid-Base Reactions Label the Nucleophile and Electrophile BF 3 + HF H +1 BF 4 -1 Elec Nuc CaO + SO 3 Nuc Nuc Elec Elec •• H F •• + F B F F Ca +2 SO 4 -2 Ca +2 KI + I 2 KI K +1 O •• O -2 •• + S O O •• I -1 •• + I I H +1 F F -1 B F F Ca +2 O O -2 S O O K +1 I I I -1 49
What
is
Acid Rain
?
natural rain water has a pH of 5.6
naturally slightly acidic due mainly to CO 2 rain water with a pH lower than 5.6 is called acid rain acid rain is linked to damage in ecosystems and structures 50
What
Causes
Acid Rain
?
many natural and pollutant gases dissolved in the air are nonmetal oxides CO 2 , SO 2 , NO 2 nonmetal oxides are acidic 2 SO 2 CO 2 + O 2 + H 2 O + 2 H 2 O H 2 CO 3 2 H 2 SO 4 processes that produce nonmetal oxide gases as waste increase the acidity of the rain natural – volcanoes and some bacterial action man-made – combustion of fuel weather patterns may cause rain to be acidic in regions other than where the nonmetal oxide is produced 51
Damage
from
Acid Rain
52
pH of Rain in Different Regions
Sources of SO 2 from Utilities
Damage
from
Acid Rain
acids react with metals, and materials that contain carbonates acid rain damages bridges, cars, and other metallic structures acid rain damages buildings and other structures made of limestone or cement acidifying lakes affecting aquatic life dissolving and leaching more minerals from soil making it difficult for trees 55
Acid Rain
Legislation
1990 Clean Air Act attacks acid rain force utilities to reduce SO 2 result is acid rain in northeast stabilized and beginning to be reduced 56