Question

A sample of an ionic compound NaA, where A- is the anion of a
weak acid, was dissolved in enough water to make 100.0 mL of
solution and was then titrated with 0.100 M HCl. After 500.0 mL
of HCl was added, the pH was measured and found to be 5.00.
The experimenter found that 1.00 L of 0.100 M HCl was required
to reach the stoichiometric point of the titration.
a. What is the Kb value for A?
b. Calculate the pH of the solution at the stoichiometric point
of the titration.

Answer 1

Answer:

a) Kb = 10^-9

b) pH = 3.02

Explanation:

a) pH 5.0 titration with a 100 mL sample containing 500 mL of 0.10 M HCl, or 0.05 moles of HCl. Therefore we have the following:

[NaA] and [A-] = 0.05/0.6 = 0.083 M

Kb = Kw/Ka = 10^-14/[H+] = 10^-14/10^-5 = 10^-9

b) For the stoichiometric point in the titration, 0.100 moles of NaA have to be found in a 1.1L solution, and this is equal to:

[A-] = [H+] = (0.1 L)*(1 M)/1.1 L = 0.091 M

pKb = 10^-9

Ka = 10^-5

HA = H+ + A-

Ka = 10^-5 = ([H+]*[A-])/[HA] = [H+]^2/(0.091 - [H+])

[H+]^2 + 10^5 * [H+] - 10^-5 * 0.091 = 0

Clearing [H+]:

[H+] = 0.00095 M

pH = -log([H+]) = -log(0.00095) = 3.02

Answer 2

a) The Kb value of A = 10^-9

b) The pH of the solution should be = 3.02

Calculation of the Kb value and pH of the solution:

a.

Since pH 5.0 titration with a 100 mL sample comprise of 500 mL of 0.10 M HCl, or 0.05 moles of HCl.

So,

[NaA] and [A-] = 0.05/0.6

= 0.083 M

Now

Kb = Kw/Ka

= 10^-14/[H+]

= 10^-14/10^-5

= 10^-9

b)Now the pH should be

[A-] = [H+]

= (0.1 L)*(1 M)/1.1 L

= 0.091 M

Now

pKb = 10^-9

Ka = 10^-5

HA = H+ + A-

Now

Ka = 10^-5 = ([H+]*[A-])/[HA] = [H+]^2/(0.091 - [H+])

[H+]^2 + 10^5 * [H+] - 10^-5 * 0.091 = 0

Now

Clearing [H+]:

[H+] = 0.00095 M

pH = -log([H+])

= -log(0.00095)

= 3.02

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