Question

1) A solution of sodium acetate (NaCH3COO) has a pH of 9.70. What is the molarity of the solution?

2) Predict which member of each pair produces the more acidic aqueous solution: (a) K+ or Cu2+, (b) Fe2+ or Fe3+, (c) Al3+ or Ga3+

Answer 1

1)

Let the initial concentration of NaCH3COO be C

now

we know that

NaCH3COO is a weak base

so

CH3COO- + H20 ---> CH3COOOH + OH-

Kb = [CH3COOH] [OH-] / [CH3COO-]

using ICE table

Kb = [y] [y] / [C-y]

given

pH = 9.7

so

pOH = 14 - pH

so

pOH = 14 - 9.7

pOH = 4.3

now

pOH = -log [OH-]

so

4.3 = -log [OH-]

[OH-] = 5.012 x 10-5

so

y = [OH-] = 5.012 x 10-5

now

Kb for NaCH3COO is 5.7544 x 10-10

so

5.7544 x 10-10 = [5.012 x 10-5 ] [5.012 x 10-5] / [C - 5.012 x 10-5]

so

C = 4.365

so

the molarity of NaCH3COO is 4.365 M

2)

a) Cu+2 ; higher cation charge more acidic

b) Fe+3 ; higher cation charge more acidic

c) Al+3 ; smaller cations are more acidic than larger cations