Calculations on pKa and pH

 Hello! I heartily welcome you all to today’s lesson. We are going to look at the last two calculations that pertain to pH, acids and bases today and tomorrow we would commence our discussion on salts to pave way for a discussion on buffers. I hope all the topics so far are making sense? Your feedback would be highly appreciated.


Example 8;

What is the pH of 0.2 M solution of sodium acetate. The pKa of acetic acid is 1.8 * 10^-5.


Solution;

We have that the hydrolysis of sodium acetate occurs as follows;

CH3CH2COONa(aq) <-> CH3CH2COO^-(aq) + Na^+(aq)

   CH3CH2COO^-(aq) + H2O(l) <-> CH3CH2COOH + OH^-(aq)

We can see that that the acetate ion acted as a base hence it accepted a proton. We would have to obtain the Kb

Kb = 1 * 10^-14/Ka

Kb = 1 * 10^-14/1.8 * 10^-5

Kb = 5.6 * 10^-10


Setting up the ICE TABLE, we have that;

         CH3CH2COO^-(aq) + H2O(l) <-> CH3CH2COOH + OH^-(aq)

I            0.2                                                      0                         0

C           -x                                                        +x                       +x

E        0.2 – x                                                    x                        x


Kb = [CH3CH2COOH] [OH^-]/[CH3CH2COO^-]

Given that; [CH3CH2COOH] = [OH^-] = x

Substituting values;

5.6 * 10^-10 = x^2/0.2 – x     

5.6 * 10^-10(0.2 – x) = x^2

1.1 * 10^-10 - 5.6 * 10^-10x =x^2

x^2 +  5.6 * 10^-10x - 1.1 * 10^-10= 0

x = 1.05 * 10^-5 M

pOH = - log (1.05 * 10^-5 M)

= 4.98

pH = 14 – 4.98

pH = 9.02

The pH of the solution is 9.02.


Example 9;

What will be the pH at which the ammonium ion (pKa = 9.2) will be 95% deprotonated?


Solution;

We must first obtain the initial concentration of the of the ammonium ion.

Recall that

 percent dissociation =√Ka/C * 100

Ka = acid dissociation constant

C = Initial concentration of acid

Ka = Antilog(-pKa)

Ka = Antilog (-9.2)

Ka = 6.3 * 10^-10

Thus;

95 = √6.3 * 10^-10/C * 100

95/100 = √6.3 * 10^-10/C

(0.95)^2 = 6.3 * 10^-10/C

 C = 6.3 * 10^-10/(0.95)^2

C = 6.98 * 10^-10 M


  Setting up the ICE TABLE;

               NH4^+(aq) + H2O(l)     <->  H3O^+(aq) + NH3(aq)

I              6.98 * 10^-10                                 0                     0

C                -x                                               +x                   +x

E             6.98 * 10^-10 – x                           x                     x

Ka = [H3O^+] [NH3]/[ NH4^+]

[H3O^+] = [NH3] = x

6.3 * 10^-10  = x^2/(6.98 * 10^-10 – x)

6.3 * 10^-10(6.98 * 10^-10 – x) = x^2

4.4 * 10^-19 - 6.3 * 10^-10x =x^2

x^2 + 6.3 * 10^-10x - 4.4 * 10^-19= 0

x = 4.19 * 10^-10 M

pH = -log(4.19 * 10^-10 M)

pH = 9.4

The pH of the required solution is 9.4.


Now try this own your own. You can solve it on a piece of paper, snap the solution and upload in the comments section under this post.

Question;

Determine the acid dissociation constant for 0.02 M solution of a monoprotic weak acid that has a pH of 2.74.


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