Question

How many milliliters of 0.705 M HIO₄ are needed to titrate each of the following solutions to the equivalence point?




(a) 51.5 mL of 0.564 M RbOH

_________________mL



(b) 71.8 mL of 0.987 M NaOH

_________________ mL



(c) 172.0 mL of a solution that contains 31.6 g of KOH per liter

______________________ mL

Answer 1

a)

Balanced neutralization reaction is

RbOH + HIO₄   $\rightarrow$ RbIO₄ + H₂O

no. of moles = molarity X volume of solution in lliter

moalrity of RbOH = 0.564 M

volume of RbOH = 0.0515 liter (1 ml = 0.001 L then 51.5 ml = 51.5 X 0.001 = 0.0515 liter)

moles of RbOH = 0.564 X 0.0515 = 0.029046 moles

According to balanced neutralization reaction HIO₄ and RbOH react in equimolar proportion therefore to react with 0.029046 moles of RbOH required HIO₄ are 0.029046 moles

moles of HIO₄ = 0.029046 moles

molarity of HIO₄ = 0.705 M

volume of solution in liter = no. of moles / molarity

volume of HIO₄ solution = 0.029046 / 0.705 = 0.0412 liter

1 liter = 1000 ml then 0.0412 liter = 0.0412 X 1000 = 41.2 ml

to reach equivalence point 41.2 ml of 0.705 M HIO₄ required

Ans = 41.2 ml

b)

Balanced neutralization reaction is

NaOH + HIO₄   $\rightarrow$ NaIO₄ + H₂O

no. of moles = molarity X volume of solution in lliter

moalrity of NaOH = 0.987 M

volume of NaOH = 0.0718 liter (1 ml = 0.001 L then 71.8 ml = 71.8 X 0.001 = 0.0718 liter)

moles of NaOH = 0.987 X 0.0718 = 0.070867 moles

According to balanced neutralization reaction mHIO₄ and NaOH react in equimolar proportion therefore to react with 0.070867 moles of NaOH required HIO₄ are 0.070867 moles

moles of HIO₄ = 0.070867 moles

molarity of HIO₄ = 0.705 M

volume of solution in liter = no. of moles / molarity

volume of HIO₄ solution = 0.070867 / 0.705 = 0.10052 liter

1 liter = 1000 ml then 0.10052 liter = 0.10052 X 1000 = 100.52 ml

to reach equivalence point 100.52 ml of 0.705 M HIO₄ required

Ans = 100.52 ml

c)

Balanced neutralization reaction is

KOH + HIO₄   $\rightarrow$ KIO₄ + H₂O

molar mass of KOH = 56.1056 g/mol

gm of KOH = 31.6 gm

no. of moles = gm of compound / molar mass

moles of KOH in 1 liter solution = 31.6/56.1056 = 0.563224 moles

molarity of solution = no.of mole/ volume of solution in liter

molarity of KOH = 0.563224/1 = 0.563224 M

no. of moles = molarity X volume of solution in lliter

moalrity of KOH = 0.563224 M

volume of KOH = 0.172 liter (1 ml = 0.001 L then 172 ml = 172 X 0.001 = 0.172 liter)

moles of KOH = 0.563224 X 0.172 = 0.09687 moles

According to balanced neutralization reaction HIO₄ and KOH react in equimolar proportion therefore to react with 0.09687 moles of KOH required HIO₄ are 0.09687 moles

moles of HIO₄ = 0.09687 moles

molarity of HIO₄ = 0.705 M

volume of solution in liter = no. of moles / molarity

volume of HIO₄ solution = 0.09687 / 0.705 = 0.13741 liter

1 liter = 1000 ml then 0.13741 liter = 0.13741 X 1000 = 137.41 ml

to reach equivalence point 137.41 ml of 0.705 M HIO₄ required

Ans = 137.41 ml