Question

The molarity of an aqueous solution of potassium hydroxide ( KOH ) is determined by titration against a 0.171 M nitric acid ( HNO₃ ) solution.

If 25.0 mL of the base are required to neutralize 10.4 mL of nitric acid , what is the molarity of the potassium hydroxide solution?

Answer 1

1)

the reaction is

KOH + HN03 --> KN03 + H20

we can see that

moles of KOH present = moles of HN03 added

now

moles = molarity x volume (ml) / 1000

so

moles of HN03 added = 0.171 x 10.4 / 1000

moles of HN03 added = 1.7784 x 10-3

now

we know that

moles of KOH present = moles of HN03 added

so

moles of KOH present = 1.7784 x 10-3

now

we know that

molarity = moles x 1000 / volume (ml)

given

volume of KOH = 25 ml

so

molarity of KOH = 1.7784 x 10-3 x 1000 / 25

molarity of KOH = 0.071136

so

the molarity of KOH solution is 0.071136 M