I'm doing an Equivalent Weight and Normality experiment and I got very confused on the discussion questions.[ the unknown is titrated with the KMnO4]
1) In this experiment MnO4- is reduced to Mn2+ by the unknown. What is the equivalent weight of MnO4- for this experiment.
Given: Molarity of KMnO4 : 0.079812M
amount of unknown: 1mL
Change in volume of titration: 26.01mL
The equivalent weight is the molar mass (molecular weight) divided by the change in the number of electrons in the titration. The molar mass of KMnO4 is approximately 158 (You need to check it out to get an exact number) and the change in the number of electrons is 5. Mn in MnO4^- is +7 and in Mn^+2 it is +2. From 7 to 2 is 5. One thing confusing about equivalent weights is that a substance can have more than one; for example, KMnO4 can have an equivalent weight = 158/5 in this reaction but if it was carried out in a neutral or basic solution, KMnO4 goes to MnO2 which is a change from +7 to +4 and the equivalent weight is 158/3.
If 2 moles of electrons are required to oxidize 1 mole of unknown, what is the molarity of the unknown solution??
is the molarity reduce to half??
To determine the equivalent weight of MnO4- in this experiment, you'll need to consider the stoichiometry of the reaction between MnO4- and the unknown substance. The balanced chemical equation for the reduction of MnO4- to Mn2+ is:
MnO4- + 8H+ + 5e- -> Mn2+ + 4H2O
From the equation, you can see that it takes 5 moles of electrons (5e-) to reduce 1 mole of MnO4-.
To determine the number of moles of MnO4- used in the titration, you can use the following equation:
moles of MnO4- = Molarity of KMnO4 × volume of KMnO4 titrated (in liters)
Given:
Molarity of KMnO4: 0.079812 M
Volume of KMnO4 titrated: 26.01 mL (0.02601 L)
Plugging in the values:
moles of MnO4- = 0.079812 M × 0.02601 L = 0.00207973212 moles
Now, we know that it takes 5 moles of electrons to reduce 1 mole of MnO4-. Therefore, the moles of electrons involved in the titration can be calculated as:
moles of electrons = 5 × moles of MnO4- = 5 × 0.00207973212 moles = 0.0103986606 moles
Since the unknown is titrated with 1 mL, we can assume that it's the limiting reagent in this reaction. Therefore, the moles of electrons involved in the reaction are equal to the moles of the unknown substance.
Now, to calculate the equivalent weight of the unknown substance, divide the mass of the unknown substance used by the moles of electrons:
equivalent weight = mass of unknown substance (in grams) / moles of electrons
Since we don't have the mass of the unknown substance, we can't calculate the exact equivalent weight without additional information.