Question

In the following reaction the iodide ion (I-) in HI is oxidized by the permanganate ion (MnO4-). A 5.0 mL solution of 0.20M HI requires 4.0 mL of KMnO4 to completely oxidize all the iodide. What is the molarity of the KMnO4 solution?
10 HI + 2 KMnO4 + 3 H2SO4 → 5 I2 + 2 MnSO4 + 8 H2O
1)0.05 M
2)0.16 M
3)0.25 M
4)0.8 M

Answer 1

Answer:

The molarity of the KMnO4 solution is 0.05M (option 1)

Explanation:

10 HI + 2 KMnO4 + 3 H2SO4 → 5 I2 + 2 MnSO4 + 8 H2O

This is the base.

10 moles of HI need 2 moles of KMnO4, to make 5 moles of I2 (gas)

Our solution of HI is 0,20 M which means 0,2 moles in 1 L of solution but we used 5 mL, so how many moles did we use?

Molarity . volume = Moles

0.2 moles/L  . 0.005L = 0.001 moles

Notice, we had to convert 5 mL into 0.005 L

So, let's go back to begining: 10 moles of HI need 2 moles of KMnO4.

How many moles of salt, do we need for 0.001 moles of HI.

The rule of three is:

10 moles of HI ___ need ___ 2 moles of KMnO4

0.001 moles of HI __ need ___ (0.001 . 2 ) / 10 = 0.0002 moles

This is our quantity of moles, that we need from KMnO4 but this moles are in 4 mL.

Molarity = moles / L but we can also take account molarity as mM / mL (Molarity/1000)

0.0002 moles . 1000 = 0.2 mM

0.2 mM / 4mL = 0.05 M