Question

Calculate [H+] and Ka or Kb for each solution given the molarity and pH of the following solutions:

1. 0.10 M NaCH₃CO₂ with 6.51 pH

2. 0.10 M Na₂CO₃ with pH 11.38

3. 0.10 M HCH₃CO₂ with pH 2.93

Answer 1

1. CH3COONa , sodium acetate is basic nature

Give pH = 6.51

As pH = 14 -pOH

pOH = 14 - 6.51 = 7.49

WE knw that pOH = -log[OH-] therefore 7.49 = -log [OH-]

10^-7.49 = OH- = 3.23 x 10^-8

CH3COONa -----------> CH3COO- + Na+

CH3COO- + H2O ---------------> CH3COOH + OH-

Construct an ICE table as follows:

	CH3COO- +	H2O-------------------->	CH3COOH +	OH-
Initial	0.10M		x	x
Change	-x		+x	+x
Equilibrium	0.1 -x		x	x

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

As x = conc. of OH- which is equal to 3.23 x10^-8

Kb = 3.23 x10^-8 X 3.23x10^-8/0.1 = 1.06 X 10^-14

pH = -log[H+]

H+ = 10^-6.51 = 3.09 x 10^-7

2. Na2CO3 is a basic salt. IT dissociate as follows:

Na2CO3 ----------> 2Na+ + CO₃^2-

CO₃^2- + H2O --------> HCO₃^- + OH-

As pH = 11.38

pH = -log[H+]

H+ = 10^-11.38 =4.16 X 10^-12

pH = 14 -pOH

pOH = 14 - 11.38 = 2.68

pOH = -log[OH-]

OH- = 10^-2.68 = 0.002

Kb = [HCO₃^-][OH-]/[CO₃^2-] .......you can cnstruct ICE table for this like above

Kb = [0.002][0.002]/[0.1] = 4 X 10^-5

3. CH3COOH is an acid

pH = 2.93 therfore H+ = 10^-2.93 = 0.001174 or 1.17X 10^-3

CH3COOH + H2O ----> CH3COO- + H3O+

COnstruct ICE table

Ka = [CH3COO-][H3O+]/CH3COOH = 1.17 X 10^-3 x 1.17 X 10^-3 / 0.1 = 1.37 X 10^-5