Question

(8) When 100.0 mL of weak acid HA were titrated with 0.09381 M RbOH, 27.36 mL were required to reach the equivalence point.

(a) Find the molarity of HA.

(b) What is the formal concentration of A- at the equivalence point?

(c) The pH at the equivalence point was 10.99. Find pK_a for HA.

(d) What was the pH when only 19.47 mL of RbOH has been added?

Answer 1

a)

We know that HA + RbOH- -> A^- Rb⁺ + H2O

   27.36 ml RbOH X (0.09831 mole RbOH / 1000 ml NaOH Soln.) X (1 mole HA/ 1 mole RaOH)

    = 2.689 x 10-3 moles

  Molarity    = 2.689 x 10-3 moles / 0.12736 L = 0.021 M

b) Since the acid will completely reacts at end point, hence the formal concentartion of A- is equal to HA = 0.021 M

c) and d)

You are titrating a weak acid with a strong base and want to determine the pH before the equivalence point. Thus you have a buffer before the equivalence point.

You can use the Henderson-Hasselbalch equation

pH = pKa + log([A-]/[HA]) (assume activities are ignored).

In order to use this equation, you need to know the pKa of the acid which is not given. However, you can determine the Ka of the acid from the given information. I will call the weak acid HA and its conjugate base will be A-. Note the conjugate base of a weak acid is a weak base. Assume it's a monoprotic acid.

The reaction is HA + OH- -> A- + H2O.

At equivalence point, the solution will contain only A- and no HA is left. You can't use the HA equation here since there is no HA. However, you can consider the reaction of the A- with H2O (remember weak base can react with H2O) which is:
...A- + H2O <-> HA + OH-
I:.F M...............0 M...0 M
C:x..................x.......x
E:F-x...............x.......x
((I: inital, C: change, E: equilibrium, F: formal concentration))

Write the equilibrium constant for the base: Kb = Kw/Ka = [HA]*[OH-]/[A-] = x*x/(F-x) = x^2/(F-x)

Write the pH = -log[H+] = -log(Kw/[OH-])
or Kw/[OH-] = 10^-10.99 (Kw = 10E-14) or [OH-] = 9.77E-4. From what I wrote above, [OH-] = x = 9.77E-4

So Kw/Ka = x^2/(F-x) = (9.77E-4)^2/(F-9.77E-4) (*)

Now this is a tricky one. How do you find F? First, F is not the initial concentration of the acid since the solution is diluted, so lets call it F' instead which is the concentration of the diluted acid and F is the concentration of the non diluted acid. Second, to find F' you need to know F.

Find the mole of NaOH at equivalence point: n = 0.09381 * 27.63/1000 = 0.00259 mole.

(Note this is also the mole of the initial HA since it's 1:1 ratio. Therefore the initial concentration of the HA is M = 0.00259*1000/100 = 0.0259 M.)

Therefore F' = F*dilution factor = F*(initial volume of HA/total volume of solution) = 0.0259*[100/(100+27.63)] = 0.0203 M

(*) 10E-14/Ka = (9.77E-4)^2/(0.0203-9.77E-4) or Ka = 2.02E-10 or

                pKa = -logKa = 9.69

Before equivalence point:
....HA + OH- -> A- + H2O
RI: 1.......B.......0
RF:C......0.......B
((RI: relative initial quantities, RF: relative final quantities, B = 19.47/27.63, C = 1- B))

So pH = pKa + log([A-]/[HA])

              = 9.69 + log(B/C) = 9.69 + log[(19.47/27.63)/(1-19.47/27.63)]

             = 10.06