Question

Consider the titration of 40.00 mL of 0.500 M NaIO (sodium hypoiodite) with 0.750 M HNO3. What volume of HNO3 will give pH = pKa?

Answer 1

Solution.

Firstly, we assume that the ionization constant of nitric acid is equal to one. Therefore, the number is equal to one. And this means that the acid will be in excess.

pH = -lg[H+]

lg[H+] = -1

[H+] = 0.1

n(NaIO) = 0.04 liters * 0.500 M = 0.02 moles

x = liters of HNO3

n((H+) in excess) = x*0.750-0.02

V(total) = 0.04 + x

[H+] = (n((H+) in excess))/V(total)

0.1 = (x*0.750-0.02)/(0.04+x)

x = 0.036 liters

x = 36.00 mL

Answer:

36.00 mL