Question

A. A 25.00 mL sample of nitric acid requires 27.75 mL of 0.088 M NaOH to reach the end point of the titration. What is the molarity of the nitric acid solution?

B. 0.5106 grams of KHP were added to 100.0mL of water. What is the molarity of the KHP solution? (Do not type units with your answer.)

Answer 1

A) balanced equation is

HNO3 + NaOH ---- > H2O + NaNO3

molarity of NaOH = number of moles of NaOH / volume of solution in L

number of moles of NaOH = 0.088M * 0.02775 L = 0.0024 mole of NaOH

from the balanced equation we can say that

1 mole of NaOH requires 1 mole of HNO3 so

0.0024 mole of NaOH will require 0.0024 mole of HNO3

molarity of HNO3 = number of moles of HNO3 / volume of solution in L

molarity of HNO3 = 0.0024 mole / 0.025 L = 0.096 M

Therefore, the molarity of nitric acid is 0.096 M

B)

molarity of KHP = number of moles of KHP / volume of solution in L

number of moles of KHP = 0.5106g / 204.22 g/mol = 0.0025 mole

Therefore, molarity of KHP = 0.0025 mole / 0.1 L = 0.025