Question

Q: A 50 mL solution containing 0.10 M 1-Naphtoic acid (pK_a = 3.70) and 0.15 M arsenic acid (pK_a1 = 2.24; pK_a2 = 6.96) is titrated with 0.20 M KOH.

a. How many end points will be observed?

b. What is the pH of the solution before adding any KOH?

c. Calculate the volume of KOH required to reach the first equivalence point.

d. Calculate the pH after the addition of 100 mL of KOH (assume that the OH^-contribution from water autoprotolysis is negligible).

Answer 1

(a) There are 2 end points.

b) An acid having lower pKa dissociates at a lower pH, so low pH = low pK_a=more acidic.

So the arsenic acid (pKa1) will be the first deprotonated.

It means that the pH at 0mL of titrant added can be calculated using Ka1 of arsenic acid. Then Ka1 = ([A^-][H⁺])/[HA] = x²/(F-x)
(5.7544 x 10^-3) = x²/(0.15 - x)
Use quadratic to find x = 0.02664M = [H⁺],

therefore pH=1.574

c) I think both 1-Naphtoic acid (NA) and arsenic acid (AA) are deprotonated, NA being first. At V_e , the moles of the neutralized acids = the moles of KOH used.

So since both are neutralized up to the first end point:

total moles = 0.005 + 0.0075 = 0.0125 moles = moles KOH

0.0125 moles / 0.2(moles/L) = 0.0625L = 62.5mL