Question

How many milliliters of 0.731 M HIO4 are needed to titrate each of the following solutions to the equivalence point?

(a) 40.0 mL of 0.731 M RbOH

mL=

(b) 28.6 mL of 0.950 M KOH

mL =

(c) 798.0 mL of a solution that contains 8.76 g of NaOH per liter

mL=

Answer 1

As, HIO4 is mono basic acid [H+] = 0.731M

(a) RbOH is monoacidic base then [OH] = 0.731M

For neutralization of acids and base:

Macid X Vacid = Mbase X Vbase

0.731 X Vacid = 40.0 X 0.731

Vacid = 29.24 / 0.731

Vacid = 40.0 mL

(b) Similarly, KOH is a bono acidic base, [OH-] = 0.950 M

Macid X Vacid = Mbase X Vbase

0.731 X Vacid = 28.6 X 0.950

Vacid = 27.17 / 0.731 = 37.17 mL

(c) We need Molarity of this Solution when 8.76 g of NaOH present in 798.0 mL

Moles of NaOHpresent = 8.76 g /40 (MW of NaOH)

Moles of NaOH present = 0.219

Now, Molarity of this 798 mL solution = 0.219 X 1000 /798 = 0.274M

So, Mbase = 0.274M

Now, from equation:

Macid X Vacid = Mbase X Vbase

0.731 X Vacid = 0.274 X 798

Vacid = 218.652 / 0.731

Vacid = 299.11 mL