Percent Dissociation of a base

 When we discussed about acids, I explained that the extent to which an acid is dissociated in solution is shown by the acid dissociation constant. Then I went ahead to explain that the percent dissociation of the acid can be obtained by the formula;

α = √Ka/C * 100/1

Where 

α = percent dissociation

Ka = acid dissociation constant

C = concentration.

Refer to the post on June 22, 2023 found on the page timeline for more details on this.


Now, for a base we have the percent dissociation as;

α = √Kb/C * 100/1

The only new term here is Kb which we have described as the base dissociation constant.


Example 3;

Calculate the percent dissociation of a 0.6M aniline solution if the Kb for aniline is 3.8 * 10^-10.


Solution;

From the parameters in the question;

Kb = 3.8 * 10^-10

C = 0.6 M

Substituting the parameters into the formula given above;

α = √3.8 * 10^-10/0.6 * 100/1

= 0.0025%

Thus the percent dissociation of 0.6 M aniline is 0.0025%.

I believe that the solution to this problem is clear. I would now leave you with another problem to try on your own;


Self test question II

Calculate the Kb for a 0.0222 M compound HB which is 0.15% dissociated in solution.

(This question is a little tricky yet it is straight forward. Can you easily pick up the cues?)


Answer to self test I (July 10, 2023);

Kindly refer to our post yesterday to see the question. We would just show the step by step solution to the problem here.


Step 1

Set up the ICE table as follows;

            HCOOH(aq) + H2O(l) --- HCOOH2^+(aq) + OH^-

I           0.05                                         0                            0

C         -x                                              +x                         +x

E      0.05 – x                                      x                             x

Recall that Methanoic acid is the base here because it accepted a proton.


Step 2

We also know that;

pKa + pKb = 14

pKb = 14 – pKa

pKb = 14 – 3.75 (pKa was given in the question)

pKb = 10.25

Kb = Antilog (-pKb)

Kb = Antilog (-10.25)

Kb = 5.6 * 10^-11


Step 3

Then

Kb = [HCOOH2^+] [OH^-]/[HCOOH]

Note that  [HCOOH2^+] = [OH^-] = x

5.6 * 10^-11 = x^2/(0.05 – x)

5.6 * 10^-11(0.05 – x) = x^2

2.8 * 10^-12 - 5.6 * 10^-11x = x^2

x^2 + 5.6 * 10^-11x - 2.8 * 10^-12 = 0

Solving by the use of a quadratic calculator(https://www.mathpapa.com/quadratic-formula/).

x = 1.67 * 10^-6 M (We discarded the negative result because concentration can not be negative)

[HCOOH2^+] = [OH^-] = x = 1.67 * 10^-6 M

Thus the hydroxide ion concentration in the solution is 1.67 * 10^-6 M


Kindly reach out to us if any confusion still persists.


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