Acetic acid, CH_2COOH, is a weak acid whose Ka is 1.8 times 10^-6 at 25degreeC.
ID: 1001612 • Letter: A
Question
Acetic acid, CH_2COOH, is a weak acid whose Ka is 1.8 times 10^-6 at 25degreeC. a) Write the equation that represents the reaction of acetic acid with water, lie sure to include all charges on ions and the designations for the phases of all species (i.e., solid liquid, gas, aqueous) b) What is the pH of a 2.0 M solution of acetic acid in water? c) What is the pH of a 2.0 M solution acetate, NaCH_1COO, in water? Now consider a solution with 2.0 M of acetic acid and 2.0 M of sodium acetate. a) What is the pH of the solution described above?Explanation / Answer
Solution:-(4)
(a) Acetic acid is a weak acid and donates H+ to water. The reaction could be written as...
CH3COOH(aq) + H2O(l) <-------> CH3COO-(aq) + H3O+(aq)
(b) To calculate the pH of a weak acid or base we make the ICE table as..
CH3COOH(aq) + H2O(l) <-------> CH3COO-(aq) + H3O+(aq)
I 2.0 0 0
C -X +X +X
E (2.0 - X) X X
Ka = (X)2/(2.0 - X)
1.8 x 10-5 = (X)2/(2.0 - X)
on cross multiply....
1.8 x 10-5(2.0 - X) = (X)2
3.6 x 10-5 - 1.8 x 10-5X = X2
on rearrangement...
X2 + 1.8 x 10-5X - 3.6 x 10-5 = 0
on solving this quadratic equation...
X = 0.00599
So, H+ = 0.00599M
pH = - log[H+] = - log(0.00599)
pH = 2.22
(c) Sodium acetate, CH3COONa is a salt of weak acid and strong base so it would be baic in nature. Na+ would be the spectator ion so it could be ignored....
CH3COO-(aq) + H2O(l) ----------> CH3COOH(aq) + OH-
I 2.0 0 0
C -X +X +X
E (2.0 -X) X X
Here we have a base so we need Kb value. Kb = 1.0 x 10-14/Ka
Kb = 1.0 x 10-14/1.8 x 10-5 = 5.56 x 10-10
5.56 x 10-10 = (X)2/(2.0-X)
Kb value is very low so (2.0 - X) could be taken as 2.0. So,...
5.56 x 10-10 = (X)2/2.0
on cross multiply...
(X)2 = 1.11 x 10-9
on doing square root on both sides...
X = 3.33 x 10-5
So, OH- = 3.33 x 10-5
From this OH- we would calculate pOH and then pH.
pOH = - log[OH-] = -log(3.33 x 10-5) = 4.48
pH = 14 - 4.48 = 9.52
(5)
(a) Now we have a solution of acetic acid and sodium acecate which acts as a buffer solution and the pH of buffer could be calculated by using Handerson equation,
pH = Pka + log(base/acid)
Pka = - log(Ka) = - log(1.8 x 10-5) = 4.74
so, pH = 4.74 + log(2.0/2.0)
pH = 4.74