Question

A 10.5 mL sample of HNO3 was diluted to a volume of 103.37 mL. Then, 23 mL of that diluted solution were needed to neutralize 53.9 mL of 0.74 M KOH. What was the concentration of the original nitric acid?

Answer 1

The balanced equation for the reaction between HNO3 and KOH is HNO3+ KOHKN >KNO3 H20 (l) Number of moles of KOH = M*V = 0.74 M * 53.9 mL = 39.886 mmol Volume of HNO3 taken for titration- 23 mL According to balanced equation, 1 mole of KOH reacts with 1 mole of HNO3 So, 39.886 mmol of KOH would react with 39.886 mmol of HNO3 Number of moles of HNO3 39.886 mmol Volume of HNO3 taken for the reaction = 23 mL Molarity of HNO3 moles/volume 39.886 mmol/23 mL 1.734174 M Molarity of the diluted HNO3 solution M 1.734174 M Volume of the diluted HNO3 solution V1-103.37 mL Volume of concentrated solution-V2-10.5 mL Molarity of concentrated HNO3 solution M2? We know that M2 = M,V1/V2= 1.734174 M * 103.37 mL/10.5 mL = 17.07253 M Concentration of original solution 17.07 M