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### Topic: Tritration Problem  (Read 5880 times)

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#### Yusuf

• Regular Member
•   • Posts: 14
• Mole Snacks: +0/-0 ##### Tritration Problem
« on: September 16, 2012, 10:24:47 AM »
I have a .0941 M base of NAOH and I tritrate a .4401g sample of unknown acid, and the amount of tritration required is 77.1 mL. The Initial amount of base solution before the tritration was 100mL. I need to find the molar mass of the unknown acid.

So, C1V1= C2V2    C2= C1V1/V2;   C2 = .0941*100mL/ 177.1 mL; C2 = .0531M

M=n/v n= M*v  .0531M * 177.1 * 10^-3L = 4.09401*10^-3

g/mol .4401/4.09401*10^-3
=
107.5 g/mol of unknown acid.

Does this process look, right? I think it's flawed..

or would it be, .4401/(77.1*10^-3L * .0941)? Which gives me 60.6 g/mol
« Last Edit: September 16, 2012, 10:52:39 AM by Yusuf »

#### Borek ##### Re: Tritration Problem
« Reply #1 on: September 16, 2012, 10:56:51 AM »
177.1 mL

Any logic behind using this number?
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#### Yusuf

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• Mole Snacks: +0/-0 ##### Re: Tritration Problem
« Reply #2 on: September 16, 2012, 11:10:18 AM »
177.1 mL

Any logic behind using this number?

V2 > V1

100mL = initial amount of solution before tritration

171.1 mL = initial amount of solution before tritration +  amount of solution after tritration

thus V2 = 177.1

#### Borek ##### Re: Tritration Problem
« Reply #3 on: September 16, 2012, 11:21:33 AM »
It has nothing to do with the stoichiometry of the titration, please compare

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

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•   • Posts: 14
• Mole Snacks: +0/-0 ##### Re: Tritration Problem
« Reply #4 on: September 16, 2012, 11:33:11 AM »
It has nothing to do with the stoichiometry of the titration, please compare

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

Alright, so if I'm getting this right the concentration of NAOH should equal the concentration of unknown acid, but if that's true then what's the point of C1V1=C2V2? Is it just for general purpose dilution?

according to this the molar mass should be

n = .0941M *77.1*10^-3Ml

n = .00725511 moles

g/mol    .4401/ .00725511

= 60.6 g/mol of unknown acid?

I think I got it :?

#### Borek ##### Re: Tritration Problem
« Reply #5 on: September 16, 2012, 11:46:18 AM »
Alright, so if I'm getting this right the concentration of NAOH should equal the concentration of unknown acid

Titration is not about finding equal concentrations, but about finding equivalent amount of substance. In some particular cases calculations look identical.

Quote
but if that's true then what's the point of C1V1=C2V2? Is it just for general purpose dilution?

Apparently this equation served its purpose - it is shown to students to confuse them, and it worked in your case.

You add NaOH to the acid solution till amount added is exactly that needed to neutralize the acid.

Number of moles of monoprotic acid can be calculated using nA=CAVA. Same can be told about number of moles of monoprotic base nB=CBVB. As they react 1:1, at the equivalence point we have used exactly the same number of base as there was acid present, which can be written as CAVA=CBVB. Note this works ONLY for substances that react 1:1 (so it won't work for the titration of sulfuric acid, as explained on the page I linked to earlier). Unfortunately, instead of using indexes like A and B, teachers use 1 and 2, which are the same as used in dilution calculation. This equation LOOKS like the one used in dilution, but it is completely unrelated.
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#### Yusuf

• Regular Member
•   • Posts: 14
• Mole Snacks: +0/-0 ##### Re: Tritration Problem
« Reply #6 on: September 16, 2012, 12:38:07 PM »
Alright, so if I'm getting this right the concentration of NAOH should equal the concentration of unknown acid

Titration is not about finding equal concentrations, but about finding equivalent amount of substance. In some particular cases calculations look identical.

Quote
but if that's true then what's the point of C1V1=C2V2? Is it just for general purpose dilution?

Apparently this equation served its purpose - it is shown to students to confuse them, and it worked in your case.

You add NaOH to the acid solution till amount added is exactly that needed to neutralize the acid.

Number of moles of monoprotic acid can be calculated using nA=CAVA. Same can be told about number of moles of monoprotic base nB=CBVB. As they react 1:1, at the equivalence point we have used exactly the same number of base as there was acid present, which can be written as CAVA=CBVB. Note this works ONLY for substances that react 1:1 (so it won't work for the titration of sulfuric acid, as explained on the page I linked to earlier). Unfortunately, instead of using indexes like A and B, teachers use 1 and 2, which are the same as used in dilution calculation. This equation LOOKS like the one used in dilution, but it is completely unrelated.

I did this

CA = x
CB = .0941

VA = 77.1 mL
VB = 100 mL

x= .112 M

C= N/V

N = C*V

(.112M * 77.1 * 10^-3L)
=
9.41*10^-3 mols

G/Mol = molar mass

.4401/ 9.41*10^-3 mols = 46.77 g/ mol of unknown acid

although, this one

http://i.imgur.com/nD39p.png

Gives me 60.66 g/mol.
« Last Edit: September 16, 2012, 12:55:14 PM by Yusuf »

#### Borek ##### Re: Tritration Problem
« Reply #7 on: September 16, 2012, 01:36:46 PM »
VA = 77.1 mL
VB = 100 mL

You got them reversed.
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#### Yusuf

• Regular Member
•   • Posts: 14
• Mole Snacks: +0/-0 ##### Re: Tritration Problem
« Reply #8 on: September 16, 2012, 02:24:31 PM »
VA = 77.1 mL
VB = 100 mL

You got them reversed.

Thank you very much! I got this down now!