Question

a) Set up an ICE table and calculate the pH of a solution that is 0.150 M propanoic acid, HC₃H₅O₂. Ka = 1.30 x 10^-5 for this weak acid. Calculate the % dissociation and justify whether or not "x" can be neglected.

b) Another solution is 0.150 M of the conjugate base, sodium propanoate (NaC₃H₅O₂). Calculate pOH and then pH for this basic solution. Use Ka from part (a).

c) What would the pH of the acidic solution in part (a) become if 0.0180 mol HCl was added to one liter of the solution? (Assume A^- reacts with H⁺ to make HA.)

Answer 1

Answer: A)

CH₃CH₂COOH --> H⁺ + CH₃CH₂COO^-

Initial 0.150 0 0

Change -a a a   

Equi. 0.15 - a a a

Ka = [H⁺][CH₃CH₂COO^-]/[CH₃CH₂COOH] or 1.3 x 10^-5 = a²/(0.15-a) ..... (1)

Case I: 0.15 >> a, we can neglect the "a" and equation 1 becomes

1.3 x 10^-5 = a²/0.15 a² = 0.195 x 10^-5 a = 1.396 x 10^-3 M

pH = -log(a) ~~ -log[H⁺] = 2.85

Case II: polynomial equation will be a² +1.3 x 10^-5a - 0.195 x 10^-5 = 0

which on solving gives a = [H⁺] ~~  0.0013899 M or pH = 2.85

Hence "X or a" can be neglected for the calculation of pH.

Percent dissociation = [H⁺]/[HA] = (1.396 x 10^-3 / 0.15) x 100 = 0.93%

B) From the relation between Ka.Kb = Kw

We can calculate the Kb for the conjugate base pair of the following acid;

Kb = 10^-14/(1.3x10^-5) = 7.692 x 10^-10

Now the reaction can be written as;

CH₃CH₂COO^- + H₂O ---> CH₃COOH + OH^-

Initial 0.15 0 0

Change -b excess b b

Equi. 0.15-b b b

Kb = b²/(0.15-b) or   b² + 7.692 x 10^-10b - 1.154 x 10^-10 = 0

b = [OH-] ~~ 1.0741 x 10^-5 M or pOH = 4.96 and pH = 14 -4.96 = 9.04

C) From part (a) we know that at equilibrium we have the following equation; Considering the total volume to be 1 L, the molarities will be

CH₃CH₂COOH --> H⁺ + CH₃CH₂COO^-

Initial 0.150 0.018 0

Change -a a a

Equilibrium 0.150 -a 0.018+a a

Ka = [a][0.018+a ]/[0.15-a]

or

1.3 x 10^-5 (0.15 -a) = 0.018a + a² or a² + 1.80 x 10-2a - 0.195 x 10^-5 = 0

a =  0.00010769 and [H⁺] = 0.0180 + 0.000107 = 0.018107M     pH = 1.74  

The maximum contribution comes from HCl only even if H⁺ from HCl reacts with the CH₃COO^-.

As due to the strong nature of HCl, the maximum contribution of H+ comes from HCl only.

Please let me know, if you have any doubts by commenting below the answer.

Thanks