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?

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

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the answer is option a.