Step 4: Rewrite the equilibrium expression in terms of the single unknown.
K.. P
P
p^05035
CO
CO^2
==-
^h
Step 5: Rewrite this relationship in terms of the quadratic equation so that you can solve for the
unknown x, the pressure of the CO.
x^2 = 1.75 −0.50x
Putting this equation into the standard form, ax^2 + bx + c= 0, you get
x^2 + 0.50x−1.75 = 0
Step 6: Use the quadratic equation to solve for x.
...
x
21
0 50! 0 50^2 4 1 1 75
=
---
^
^^^
h
8 hhhB
For Examples 4 and 5, use the following information:
A student prepared a 1.00 M acetic acid solution (HC 2 H 3 O 2 ). The student found the pH of the
solution to be 2.00.
- What is the Kavalue for the solution?
A 3.00 × 10 –7
B 2.00 × 10 –6
C 2.00 × 10 –5
D 1.00 × 10 –4
E 1.00 × 10 –3
Answer: D
Step 1: Write the balanced equation in a state of equilibrium.
HC 2 H 3 O2(aq)?H+(aq)+ C 2 H 3 O 2 – (aq)
Step 2: Write the equilibrium expression.
K
HC H O
HCHO
a 232
(^232)
- 6
77
@
AA
Step 3: Use the pH of the solution to determine [H+].
pH = −log[H+] = 2.00
H+= 10–2.00 = 0.0100 M
Part II: Specific Topics
- 6