pH AVERAGE UNKNOW ACID: 0.201 M NaOH 0.1005M unknow acid pH AVERAGE 1 4.42 2 4.8

ID: 540913 • Letter: P

Question

pH AVERAGE

UNKNOW ACID: 0.201 M

NaOH 0.1005M

unknow acid

pH AVERAGE

1 4.42 2 4.84 3 5.48 4 10.1 5 12.11 urvalloT T Reeded Analysis (A1) Write the chemical equation (one that would lead to the Ka) and Ka expression (A2) Calculate Ka for each solution. At least for one solution, show your work clearly and completely so anyone can follow what you have done to calculate Ka (A3) Calculate average Ka. Postlab Questions: (Q1) What would happen to the final answer if yo as your total volume in your calculation? (Q2) The calculations are done assuming that OH- added is reacting completely. What would happen to the final result if not all OH- was used up? Hint: WHAT is the final result? What would happen to the VALUE? Why? u forgot to add water to test tube #1 but used 2000mL

Explanation / Answer

(A) Assume a monoprotic weak acid, HA. The acid dissociation reaction is

HA (aq) --------> H+ (aq) + A- (aq)

The acid dissociation constant, Ka is given as

Ka = [H+][A-]/[HA] …..(1)

Since we have a monoprotic acid, note that at equilibrium, [H+] = [A-]. Also, at equilibrium, [HA] >> [H+], [A-] and is approximately equal to [HA]i where i denotes the initial concentration.

(B) Take the acid 1 as an example. We have pH = 4.42.

We know that pH = -log [H+]; therefore, -log [H+] = 4.42

====> [H+] = [A-] = antilog (-4.42) = 3.8019*10-5 M.

[HA] = 0.201 M

Plug in values in the expression for Ka (do not include units) and obtain Ka.

Ka = (3.8019*10-5)*(3.8019*10-5)/(0.201) = 7.19126*10-9 7.19*10-9 (ans).

Fill out the table below.

Unknown Acid

[H+] = [A-] (M)

Coorected Ka (to two decimal places)

4.42

3.8019*10-5

7.19126*10-9

7.19*10-9

4.84

1.4454*10-5

1.03945*10-9

1.04*10-9

5.48

3.3113*10-6

5.45511*10-11

5.45*10-11

10.1

7.9433*10-11

3.13909*10-20

3.14*10-20

12.11

7.7625*10-12

2.99781*10-24

3.00*10-24

(C) You haven’t included any other information; I simply calculated the Ka values considering you have only HA and nothing else. I shall not calculate the average Ka since the individual Ka values are grossly different and the average will be seriously wrong.