Cliffs AP Chemistry, 3rd Edition

Total volume of solution = 75.0 mL + 75.0 mL = 150.0 mL

M 1 V 1 = M 2 V 2

(4.00 × 10 –4mole/liter)(0.0750 liter) = (x)(0.1500 liter)

x = [Mg2+] = 2.00 × 10 –4M

The same would be true for [OH–].

Q= [Mg2+][OH–]^2 = (2.00 × 10 –4)^3 = 8.00 × 10 –12

Ksp= 1.58 × 10 –11

A precipitate would notform because Q < Ksp.

3. Acetic acid, HC 2 H 3 O 2 , which is represented as HA, has an acid ionization constant Kaof
   1.74 × 10 –5.

(a) Calculate the hydrogen ion concentration, [H+], in a 0.50-molar solution of acetic

acid.

(b) Calculate the pH and pOH of the 0.50-molar solution.

(c) What percent of the acetic acid molecules do not ionize?

(d) A buffer solution is designed to have a pH of 6.50. What is the [HA]:[A–] ratio in this

system?

(e) 0.500 liter of a new buffer is made using sodium acetate. The concentration of

sodium acetate in this new buffer is 0.35 M. The acetic acid concentration is 0.50 M.

Finally, 1.5 grams of LiOH is added to the solution. Calculate the pH of this new

buffer.

Answer

3. Given: Kafor HA is 1.74 × 10 –5.
   (a) Restatement: [H+] in 0.50 M HA.

Step 1: Write the balanced equation for the ionization of acetic acid, HA.

HA(aq) ↔H+(aq) + A–(aq)

Step 2: Write the equilibrium expression for Ka.

HA

HA

a=

+ -

K

6

77

@

AA

Part II: Specific Topics