Question

500.0mL of 0.1350 M KOH is titrated with HNO₃.

a) how many moles of KOH are present in the sample?

b) What is the mole ratio of KOH: HNO₃ ?

c) How many moles of HNO₃ are used for this reaction? How many grams of HNO₃ is this?

d) If the HNO₃ is available as a solution whose concentration is 0.04655M, how many mL of the solution will be neded for the reaction?

Answer 1

a) moles of KOH = 0.1350mol/L * (500/1000)L = 0.0675 mol
b) HNO₃ + KOH ----> KNO₃ + H₂O
   After writing balanced stoichiometric equation we can understand that
   the mole ratio of KOH:HNO₃ = 1:1
c) Since the mole ratio of KOH to HNO₃ is 1 we can say that 0.0675 mol
   of HNO₃ are used for this reaction
Molar mass of HNO₃ = 63.01 g/mol
grams of HNO₃ = 0.0675 mol * 63.01 gm/mol = 4.25317 grams
d) 0.04655 mol/L * (volume in L) = 0.0675 mol
   volume in L = 1.45005 L
   mL of the solution needed for the reaction = 1.45005*1000 = 1450.0537 mL