Given The Concentration And The Ka, Calculate The Percent Dissociation

Given the concentration and the Ka, calculate the percent dissociation

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Problem #1: Calculate the percent dissociation of a weak acid in a 0.050 M HA solution. (Ka = 1.60 x 10-5)

Solution:

1) Calculate the [H+]:

1.60 x 10-5 = [(x) (x)] / 0.050

x = 8.9443 x 10-4 M

2) Divide the [H+] by the concentration, then multiply by 100:

(8.9443 x 10-4 M / 0.050 M) x 100 = 1.79% dissociated

Problem #2: A certain weak acid, HA , has a Ka value of 9.2 x 10¯7

1) Calculate the percent dissociation of HA in a 0.10 M solution. 2) Calculate the percent dissociation of HA in a 0.010 M solution. 3) Calculate the percent dissociation of HA in a 0.0010 M solution.

Solution to part one:

Step #1: Calculate the [H+]:

9.2 x 10¯7 = [(x) (x)] / (0.10 - x)

neglect the minus x

x = 3.03315 x 10¯4 M (note that I kept some guard digits, I'll round off the final answer.)

Step #2: Divide the [H+] by the concentration, then multiply by 100:

(3.03315 x 10¯4 M / 0.10 M) x 100 = 0.303% dissociated

Solution to part two:

Step #1: Calculate the [H+]:

9.2 x 10¯7 = [(x) (x)] / (0.010 - x)

neglect the minus x

x = 9.591663 x 10¯5 M (note that I kept some guard digits, I'll round off the final answer.)

Step #2: Divide the [H+] by the concentration, then multiply by 100:

(9.591663 x 10¯5 M / 0.010 M) x 100 = 0.959% dissociated

Solution to part three:

Step #1: Calculate the [H+]:

9.2 x 10¯7 = [(x) (x)] / (0.0010 - x)

neglect the minus x

x = 3.03315 x 10¯4 M (note that I kept some guard digits, I'll round off the final answer.)

Step #2: Divide the [H+] by the concentration, then multiply by 100:

(3.03315 x 10¯5 M / 0.0010 M) x 100 = 3.03% dissociated

Yes, as the solution becomes more dilute, the percent dissociation increases. However, as the solution becomes more and more dilute, the above technique breaks down. (At 0.00000010 M, the percent dissociation is 303%.) This is because the [H+] is becoming larger and larger compared to the concentration of the acid. This means that the neglecting 'x' approximation introduces more and more error.

This means that, as the dilution increases, we must shift to more sophisticated calculational techniques which are beyond the scope of this web site.

Problem #3: Calculate the degree of ionization of acetic acid in the following solutions:

solution 1 : 0.10 M HC2H3O2 solution 2 : 5 mL 0.10 M HC2H3O2 + 5 mL H2O solution 3 : 1 mL 0.10 M HC2H3O2 + 99 mL H2O

Solution to part one:

1) Calculate the [H+]:

[H+] = √(Ka times concentration)

[H+] = √(1.77 x 10¯5 times 0.10) = 0.00133 M

2) Divide the [H+] by the concentration, then multiply by 100:

(0.00133 /0.1) x 100 = 1.33%

Solution to part two:

1) Calculate the [H+]:

[H+] = √(Ka times concentration)

[H+] = √(1.77 x 10¯5 times 0.050) = 0.000941 M

2) Divide the [H+] by the concentration, then multiply by 100:

(0.000941 /0.05) x 100 = 1.88%

Comment: note how the solution was diluted from 5 mL to 10 mL, thus cutting the concentration in half.

Solution to part three:

1) Calculate the new molarity using M1V1 = M2V2:

(0.10) (1) = (x) (100)

x = 0.0010 M

2) Calculate the [H+]:

[H+] = √(Ka times concentration)

[H+] = √(1.77 x 10¯5 times 0.0010) = 0.000133 M

3) Divide the [H+] by the concentration, then multiply by 100:

(0.000133 /0.0010) x 100 = 13.3%

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