Question

32.33 mL of 1.031 M potassium hydroxide were required to reach the endpoint of a titration of 50.00 mL of nitric acid. What was the original concentration of nitric acid in that 50.00 mL sample?

Answer 1

The molar concentration of the nitric acid solution was 0.6666 mol/L.

Balanced equation: KOH + HNO_3 → KNO_3 + H_2O

Moles of KOH: 32.33 mL KOH × (1.031 mmol KOH /1 mL KOH)

= 33.33 mmol KOH

Moles of HNO_3: 33.33 mmol KOH× (1 mmol HNO_3/1 mmol KOH)

= 33.33 mmol HNO_3

Concentration of KOH: c = "moles"/"litres" = 33.33 mmol/50.00 mL

= 0.6666 mol/L