Question

In a titration 34.20 mL HI, 21.78 mL of 0.250 M LiOH was needed to reach the equivalence point.

a.) What is the reaction equation for this reaction?

b.) How many moles of LiOH were added during the titration?

c.) How many moles of HI were present in the original sample?

d.) What was the HI concentration in the original sample?

Please show all steps!!! This is for my test's study guide

Answer 1

(a) The balanced reaction is : HI + LiOH ----> LiI + H₂O

(b) Number of moles of LiOH , n = Molarity x volume in L

                                              = 0.250 M x 21.78 mLx 10^-3 L/mL

                                              = 0.054 moles

(c) According to the balanced reaction ,

1 mole of LiOH reacts with 1 mole of HI

0.0054 moles of LiOH reacts with 0.0054 moles of HI

(d) The concentration of HI = number of moles of HI / volume of solution in L

                                      = 0.0054 mol / (34.20 mL x10^-3 L/mL)

                                      = 1.592 M