need help with chem pH question please

Question

A)Determine the pH and percent dissociation of a 0.183M propionic (CH₃CH₂COOH) acid solution. Begin by writing the balanced chemical equation.

 

B) A portion of the above propionic acid solution was diluted 10 fold. Determine the pH and percent dissociation of the dilute solution. Comment on any differences as compared to the solution of higher concentration.

 

Answer 1

First, we cannot answer this question without knowing the Ka for propionic acid. I guess we could look it up, but it would probably be better for you to include it in the question so that we are sure we are both using the same Ka value.

Also, it would be expected that you at least make an attempt to answer to question, and then write in an explain where you are having difficulty. Otherwise, it appears as though you are simply asking tutors to do your homework. Just a suggestion.

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UPDATE:

Using Ka = 1.3x10^-5. Let propionic acid be HA. Then HA <==> H⁺ + A^-

Ka = 1.3x10^-5 = [H⁺][A^-]/[HA] = (x)(x) / 0.183-x (assume x is small relative to 0.183 and ignore it)

x² = 2.38x10^-6

x = [H⁺] = [A^-] = 1.54x10^-3 M (small relative to 0.183 M)

pH = -log [H⁺] = -log 1.54x10^-3

pH = 2.81

% ionization/dissociation = 1.54x10^-3 / 0.183 (x100%) = 0.84% (agrees with what you got. Good job).

Part B: Diluting 10 fold produces a [HA] = 0.1 x 0.183 M = 0.0183 M.

Do same thing as above. Ka = 1.3x10^-5 = [H⁺][A^-]/[HA] = (x)(x) / 0.0183-x (ignore x again)

x² = 2.38x10^-7

x = 4.88x10^-4 M = [H⁺] = [A^-]

pH = -log 4.88x10^-4

pH = 3.31

% ionization/dissociation = 4.88x10^-4 / 0.0183 (x100%) = 2.67% (greater than original, and this is what is to be expected for a more dilute weak acid)

Hope this all made sense.