Transcript Slide 1

pH calculations
strong acids complete dissociation
HA  H+ + AWhat is the pH of a 0.040 M solution of HClO4?
HClO4  H+ + ClO4[H+] = [ClO4-] = 0.040 M
pH = - log [H+]
pH = 1.40
[H+] from H2O = 1.0 x 10-7 M
ignore autoionization if [HA] > 10-7 M
pH calculations
strong bases complete dissociation
What is the pH of a 0.011 M solution of Ca(OH)2?
Ca(OH)2+ H2O  Ca2+ (aq) + 2(OH-) (aq)
[OH-] = 2 [Ca2+] = 0.022 M
pH + pOH = 14.00
pOH = - log [OH-]
pOH = 1.67
pH = 12.33
Na2O + H2O 2 Na+ (aq) + 2 OH- (aq)
pH calculations
weak acids incomplete dissociation
HA  H+ + A-
[HA]eq
Ka = [H+]eq [A-]eq = x2
[HA]
x
[HA]
i
eq
= [HA]i - x
Ka  x2
[HA]i
Ka very small
if
x
[HA]i
assume x << [HA]i
< 5%
approximation O.K.
assume H+ from HA >> H+ from H2O (1 x 10-7 M)
pH calculations
weak acids incomplete dissociation
What is the Ka of a 0.10 M HCOOH, pH = 2.38
HCOOH  H+ + HCOO- [H+]eq = 10-2.38 = 4.2 x 10-3
+][HCOO-]
[H
= x2 = (4.2 x 10-3)2 =1.8 x 10-4
Ka =
[HCOOH]
0.10 - x 0.10
[HA] (M) [H+] (M) [A-] (M)
4.2x10-3
0.00
0.00
Initial
0.10
0.10
-x
+x
+x
Change
4.2 %
Equil.
0.10 - x
x
x
% ionization
Calculate the pH of a 0.100 M HF solution
HF  H+ + FKa = 6.8 x 10-4 = [H+][F-]
[HF]
[HF] [H+]
0.100 0.00
I
C
-x
+x
E 0.100 - x x
[F-]
0.00
+x
x
6.8 x 10-4 = x2
0.100 - x
x = [H+] = 8.24 x 10-3
pH = 2.08
0.00824 x 100 = 8.25 % 5% assumption not good
0.100
solve quadratic or successive iterations
.00790
.00791
-3
pH = 2.10
.00791 x = 7.91 x 10
Calculate the pH of a 0.100 M HF solution pH = 2.10
% ionization = [H+] x 100 = 7.91 x 10-3 x 100 = 7.91 %
[HF]
0.10
Calculate the pH of a 0.010 M HF solution pH = 2.64
[H+] = 2.3 x 10-3
% ionization =
2.3 x 10-3 x 100 = 23%
0.010
HF (aq)  H+ (aq) + F- (aq)
1 mol solute 2 mol solute
increase concentration
Find the pH of a 0.0037 M solution of H2CO3
H2CO3  H+ + HCO3-
Ka1 = 4.3 x 10-7
HCO3-  H+ + CO32-
Ka2 = 5.6 x 10-11
x = 4.0 x 10-5 x 100 = 1.10 %
4.3 x 10-7 = x2
-3
-3
3.7
x
10
3.7x10
pH = 4.40
+
[H2CO3]
[H ]
[HCO3 ]
3.7 x 10-3 0.00
0.00
I
[CO32-] =
C
-x
+x
+x
x
E 3.7x10-3-x
x
3.7 x 10-3 4.0 x 10-5 4.0 x 10-5
Find the pH of a 0.0037 M solution of H2CO3
H2CO3  H+ + HCO3-
Ka1 = 4.3 x 10-7
HCO3-  H+ + CO32-
Ka2 = 5.6 x 10-11
5.6 x 10-11=(4.0 x 10-5 + y)(y) y = 5.6 x 10-11 = [CO32-]
4.0 x 10-5 - y
[HCO3-]
[H+]
[CO32-]
4.0 x 10-5 4.0 x 10-5 0.00
I
C
-y
+y
+y
E 4.0 x 10-5-y 4.0 x 10-5 +y y
% ionization
1.4 x 10-4 %
pH calculations
weak bases incomplete dissociation
What is the [OH-] of a 0.15 M solution of NH3
-5
+
K
=
1.8
x
10
NH3 + H2O  NH4 + OH
b
+][OH-]
[NH
= x2 = 1.8 x 10-5
Kb =
4
[NH3]
0.15 - x
x = 1.64 x 10-3 = [OH-] pOH = 2.79
pH = 11.21
[NH3] (M) [NH4+] (M) [OH-] (M)
0.00
0.00 1.64x10-3
Initial
0.15
-x
+x
+x
Change
0.15
Equil.
0.15 - x
x
x
1.1 %