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.

4. 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