Question

50.00 mL of 0.100M of a weak acid (Ka=1.3x10^-5) is titrated with 0.100M NaOH.

            a. Compute the volume of NaOH required to reach the equivalence point.

            b. Calculate the pH of the original solution before any NaOH has been added.

            c. After 30.00 mL of NaOH has been added, what is the pH of the solution?

            d. What is the pH at the equivalence point?

            e. Write a brief explanation as to why it is not exactly neutral.

            f. What is the pH when 60.00 mL of NaOH has been delivered?

Answer 1

LEt's assume the weak acid is a monoprotic acid, therefore in the equivalence point: moles of A = moles of B

a) In the equivalence point, the above expression is equal: MaVa = MbVb so, solving for Vb:

Vb = 0.1 * 50 / 0.1 = 50 mL

b) In the original solution we have:

HA --------> A^- + H⁺

i. 0.1 0 0

e. 0.1-x x x

1.3x10^-5 = x² / 0.1-x But x is small because ka is low, so:

(1.3x10^-5 * 0.1)^1/2 = x

x = 1.14x10^-3 M = [H⁺]

pH = -log(1.14x10^-3) = 2.94

c) After 30 mL:

moles of NaOH = 0.1 * 0.030 = 0.003 moles

moles of HA = 0.1 * 0.05 = 0.005 moles

HA + NaOH ----------> NaA + H₂O

i. 0.005 0.003 0 0

e. 0.002 0 0.003

In this part, we use the HH equation which is: pH = pKa + log(S/A)

pH = -log(1.3x10^-5) + log(0.003/0.002)

pH = 4.89 + 0.18

pH = 5.07

d) pH at equivalence point (V = 50 mL, total volume of 100 mL)

HA + OH^- ----------> A^- + H₂O

0.005 0.005 0 0

0 0.005 0.005

Now, in the equivalence all the moles of acid are already react with the moles of the base, so now, the hydrolisis is occuring here:

A^- + H₂O ------------> HA + OH^-   Kb = 1x10^-14 / 1.3x10^-5 = 7.69x10^-10

i. 0.005/0.1 0 0

e. 0.05-x x x

7.69x10^-10 = x² / 0.05-x

7.69x10^-10 * 0.05 = x²

x = [OH^-] = 6.2x10^-8 M

pOH = -log(6.2x10^-8) = 5.21

pH = 14 - 5.21 = 8.79

e) This solution it's not exactly neutral basically because it's a reaction between a weak acid and a strong base, and the weak acid does not (as seen in previous part) dissociates completely with NaOH. If both solutions were strong (strong acid and base) then, the H from the acid will dissociates completely and the concentration would be complete. But in this part, you are only dissociating as you are adding base, just a part of the acid and not all of it. That's why the pH it's not neutral.

f) after 60 mL of NaOH added, we only have excess of NaOH so,

moles of NaOH = 0.1 * 0.06 = 0.006

moles of NaOH remanent after reacting with acid = 0.006 - 0.005 = 0.001

[OH] = 0.001 / 0.110 = 0.0091 M

pOH = -log(0.0091) 2.04

pH = 14 - 2.04 = 11.96

Hope this helps.