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### Topic: CHEM II question - titration exercise  (Read 4529 times)

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#### Guilo

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##### CHEM II question - titration exercise
« on: February 12, 2013, 09:52:58 AM »
Hello,

I have no idea how to solve this...

The initial mass of an unknown sample containing iron (II) is 0.512 g.
12.6 mL of 0.01522 M KMnO4 is required to titrate the unknown to the endpoint.

a) How many moles of MnO4- were added?
b) How many moles of iron (II) must be present in the sample?
c) How many grams of iron (II) must be present in the sample?
d) What is the percent of iron present in the sample?

I would know b, c, d, if I would know how to solve a. b, c, d, questions are simple, but I totally don't understand how to solve a. I wrote the rest so people who don't know similar exercise could get something more out of that topic.
I know I shouldn't ask such a basics here, but I spent like an hour on YouTube and internet with no success. My professor is very unclear on every topic.

Thank you so much for helping.

Guilo

#### sjb

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##### Re: CHEM II question - titration exercise
« Reply #1 on: February 12, 2013, 10:11:53 AM »
Hello,

I have no idea how to solve this...

The initial mass of an unknown sample containing iron (II) is 0.512 g.
12.6 mL of 0.01522 M KMnO4 is required to titrate the unknown to the endpoint.

a) How many moles of MnO4- were added?
b) How many moles of iron (II) must be present in the sample?
c) How many grams of iron (II) must be present in the sample?
d) What is the percent of iron present in the sample?

I would know b, c, d, if I would know how to solve a. b, c, d, questions are simple, but I totally don't understand how to solve a. I wrote the rest so people who don't know similar exercise could get something more out of that topic.
I know I shouldn't ask such a basics here, but I spent like an hour on YouTube and internet with no success. My professor is very unclear on every topic.

Thank you so much for helping.

Guilo

What does 0.01522 M KMnO4 mean?

#### Guilo

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##### Re: CHEM II question - titration exercise
« Reply #2 on: February 12, 2013, 11:21:58 AM »
What does 0.01522 M KMnO4 mean?

The molarity of KMnO4 solution is 0.01522
Molarity = moles/1L

#### Borek

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##### Re: CHEM II question - titration exercise
« Reply #3 on: February 12, 2013, 11:42:30 AM »
Molarity = moles/1L

Yes and no. What if you have just 0.5 L, or 12.6 mL of the solution?
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#### Guilo

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##### Re: CHEM II question - titration exercise
« Reply #4 on: February 12, 2013, 11:54:15 AM »
Molarity = moles/1L

Yes and no. What if you have just 0.5 L, or 12.6 mL of the solution?

Yes, I understand what you mean. Molarity of KMnO4 solution solution in this exercise is 0.01522, so it is 0.01522 moles in 0.0126 L.

But, I don't have any idea how to start solving a)

0.01522 moles KMnO4/0.0126 L KMnO4=158 g KMnO4/1 mol KMnO4=2.40476/0.0126≈191g/L

And I have no idea what to do next...

#### Arkcon

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##### Re: CHEM II question - titration exercise
« Reply #5 on: February 12, 2013, 12:04:17 PM »
Quote
Yes, I understand what you mean. Molarity of KMnO4 solution solution in this exercise is 0.01522, so it is 0.01522 moles in 0.0126 L.

Incorrect.  The molarity of the solution was given as 0.01522 moles per 1 L.  The experiment uses less than one liter, you have to convert to determine how many moles were used.

You will also have to write a balanced chemical equation for the reaction between iron (II) and permanganate.  You will the relate the moles of permanganate to moles of iron.
Hey, I'm not judging.  I just like to shoot straight.  I'm a man of science.

#### Guilo

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##### Re: CHEM II question - titration exercise
« Reply #6 on: February 12, 2013, 12:18:07 PM »
the two half-reactions are

MnO4- + 8H+ + 5e- = Mn2+ + 4H2O
Fe2+ = Fe3+ + 1e-   // x5

the balanced equation is

MnO4- + 5Fe2+ + 8H+ = Mn2+ + 5Fe3+ + 4H2O

I don't understand why there is KMnO4 not just MnO4 (In the exercise question).

Is that right?
0.01522 moles/0.0126 L x 0.0126L ≈ 0.0002 = 2x10-4 moles

Still no idea what to do next...

#### Borek

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##### Re: CHEM II question - titration exercise
« Reply #7 on: February 12, 2013, 02:30:21 PM »
I don't understand why there is KMnO4 not just MnO4 (In the exercise question).

How are you going to prepare solution containing just MnO4-? It has to contain some counterion.

Quote
Is that right?
0.01522 moles/0.0126 L x 0.0126L ≈ 0.0002 = 2x10-4 moles

No. Basically you multiplied and divided by the same number, so it canceled out.

To speed things up, see here:

http://www.titrations.info/titration-calculation
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#### Guilo

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##### Re: CHEM II question - titration exercise
« Reply #8 on: February 14, 2013, 11:06:20 AM »
Hey Borek,

The initial mass of an unknown sample containing iron (II) is 0.512 g.
12.6 mL of 0.01522 M KMnO4 is required to titrate the unknown to the endpoint.

a) How many moles of MnO4- were added?
0.01522 moles/1 L x 0.0126L ≈ 0.0002 = 2x10-4 moles

b) How many moles of iron (II) must be present in the sample?
2x10-4 x 5 = 0.001

c) How many grams of iron (II) must be present in the sample?
0.001 moles of Fe2+ x 56g Fe/1 mole Fe = 0.056g Fe

d) What is the percent of iron present in the sample?
0.056g Fe/0.512 g unknown sample x 100 = 10.9%
« Last Edit: February 14, 2013, 11:19:09 AM by Guilo »

#### Borek

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##### Re: CHEM II question - titration exercise
« Reply #9 on: February 14, 2013, 02:06:50 PM »
Never round down intermediate results - that is, report them rounded, but use fulls precision for calculations.

You are off by almost 2% (or almost 20%, depending on how you look at it) because of roundings, but the approach is correct.
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