REAL.docx

SOLUTIONS:
1. Find the equilibrium pH of 1000 mg/L of NH4OH (ammonium hydroxide). The Kb for the base is equal to 10-5.
NH4OH NH4+ + OH-
Given:
Equilibrium pH = ?
1000 mg/L of NH4OH
Kb for the base is equal to 10-5
a. Convert 1000 mg/L to M.
Molecular Weight of NH4OH:
N = 1 x 14 = 14 g/mol
H = 5 x 1 = 5 g/mol
O = 1 x 16 = 16 g/mol
35 g/mol NH4OH
1000 mg of NH4OH x 1 g of NH4OH x 1 mol NH4OH = 0.029 mol NH4OH or 0.029 M NH4OH
L of NH4OH 1000 mg NH4OH 35 g NH4OH L
b. Get the value of x.

NH4OH

NH4+
+
OH-
I
0.029

0

0
C
- x

+ x

+ x
E
0.029 - x

x

x

10-5 = [x] [x]
[0.029-x]
x = 5.335 x 10-4
c. Substitute the value of x.
OH = x
OH = 5.335 x 10-4

d. Find equilibrium pH.
pOH = -log[5.335 x 10-4]
pOH = 3.273
pH = 14-3.273

pH = 10.727

2. What would be the equilibrium pH if 200 milligrams of Hydrofluoric acid (HF) were dissolved in 1 liter of solution? The pKa for the acid is equal to 3.2. (Hint: Convert pKa to Ka)
HF H+ + F-
Given:
pH = ?
200 mg of HF
1 L of HF
3.2 pKa
a. Find the concentration of HF (M).
Molecular Weight of HF:
H = 1 x 1 g/mol = 1 g/mol
F = 1 x 19 g/mol = 19 g/mol
20 g/mol HF
M = moles
L of sol'n
M = 200 mg HF x 1 g HF x 1 mol HF = 0.01 mol HF or 0.01 M HF
1 L HF 1000 mg HF 20 g HF L
b. Convert pKa to Ka.
3.2 = -log(Ka)
Ka = 6.310 x 10-4
c. Get the value of x.

HF

H+
+
F-
I
0.01

0

0
C
- x

+ x

+ x
E
0.01 - x

x

x

6.310 x 10-4 = [x] [x]
[0.01-x]
x = 2.217x10-3
c. Substitute the value of x.
H = x
H = 2.217x10-3
d. Find pH.
pH = -log[2.217x10-3]
pH = 2.654

3. One method to remove metals from water is to raise the pH and cause them to precipitate as their metal hydroxides.
Cd(OH)2 Cd2+ + 2OH- Ksp = 2.5x10-14
The pH of water initially was 6.8 and then was raised to 8.0. Is the dissolved cadmium concentration reduced to below 100 mg/L at the final pH? Assume the temperature of the water is 25°C.
Given:
pH of water initially was 6.8 and then was raised to 8.
100 mg/L Cd at the final pH
25°C
Ksp = 2.5x10-14
Find: Is the dissolved cadmium concentration reduced to below 100 mg/L at the final pH?

Initial
Final
pH = 6.8
pH = 14-6.8
pOH= 7.2
pH = 8
pH = 14-8
pOH = 6
Way 1:
7.2 = -log(x)
x = 6.310x10-8
[OH-] = 6.310x10-8

Way 2:
(Using Ion Product Constant)

[H+] = 10-pH (Formula)
Kw= 1x10-14

Kw= [H+] [OH-]
[OH-] = 1x10-14
10-6.8
[OH-] = 6.310x10-8
Way 1:
6=-log(x)
x = 1x10-6
[OH-] = 1x10-6

Way 2:
(Using Ion Product Constant)

[H+] = 10-pH (Formula)
Kw= 1x10-14

Kw= [H+] [OH-]
[OH-] = 1x10-14
10-8
[OH-] = 1x10-6
Ksp = [Cd2+] [OH]2
Cd2+ = x
Ksp = [x] [OH]2

x = 2.5x10-14
[6.310x10-8]2
x = 6.279 M
Cd2+ = 6.279 M



x = 2.5x10-14
[1x10-6]2
x = 0.025 M
Cd2+ = 0.025 M

6.279 mol Cd x 112.4 g Cd x 1000 mg = 705759.6 mg/L
L Cd 1 mol Cd 1 g

0.025 mol Cd x 112.4 g Cd x 1000 mg = 2810 mg/L Cd
L Cd 1 mol Cd 1 g


Cd = 2810 mg/L is the final concentration.
So, NO.
Cd = 2810 mg/L is the final concentration.
So, NO.