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### Topic: titrations  (Read 22420 times)

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#### chemical-idiot

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##### titrations
« on: May 14, 2005, 07:36:24 PM »
I need help on the following:

10 mL of vinegar diluted with water to reach 100 mL.  Then take 10 mL of that, and place into erlenmeyer flask.

Titrate.

Conc.  NaOH:  0.0715 mol/L
Conc. Vinegar(CH3COOH):  ? (to calculate)

Volume of NaOH used: 14.3 mL

I did:  CH3COOH   +   NaOH  --------->   CH3COONa  +  H20

0.07125 mol/L  x  0.0413 L of NaOH  =  0.00102 mol   [1:1 ratio]

conc.  CH3COOH:  0.00102 mol   /  0.01 L of vinegar =  0.102 mol/L

I was wondering if that is correct if i used ---- 10 mL of vinegar diluted with water to reach 100 mL.  Then take 10 mL of that, and place into erlenmeyer flask.

and if the question asks for concentration of ACETIC ACID is that the same thing, meaning 0.102 mol/L or is it asking for the 5% of the acetic acid found in vinegar (if so how do u figure this out)..

and following up on the acetic acid, is the moles of acid in 10 mL = 0.00102 mol or something else?

#### Grumples

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##### Re:titrations
« Reply #1 on: May 14, 2005, 10:30:22 PM »
Well, right off the bat you had a math error.  remember, 14.3 mL becomes 0.0143 liters, not .0413 liters.

Now, firstly you have to convert moles of NaOH to moles of CH3COOH, which you did.
That will give you the number of moles of acetic acid in your 10 mL sample.  since you took 1/10 of the original, take that number and multiply by 10, then divide by the original volume (10 mL).  Here are the calcualtions:

V NaOH x M NaOH = n NaOH

n NaOH = n CH3COOH

n CH3COOH x 10 = original amount of CH3COOH

original # of moles/original volume = conc. CH3COOH in original sample

Remember, since you are doing an acid/base titration, the stuff that is being titrated must be acid, since it is reacting with a base.  normal vinegar is a roughly 5% sol'n of acetic acid, along with various other stuff (sugar, water, salt, etc.).  You cannot find the conc. of vinegar, since it is not a pure substance.

#### Borek

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##### Re:titrations
« Reply #2 on: May 15, 2005, 04:16:59 AM »
conc.  CH3COOH:  0.00102 mol   /  0.01 L of vinegar =  0.102 mol/L

I was wondering if that is correct if i used ---- 10 mL of vinegar diluted with water to reach 100 mL.  Then take 10 mL of that, and place into erlenmeyer flask.

0.102M is a concentration in diluted sample, in the original sample acetic acid cocnetration was 10 times larger.

Quote
and if the question asks for concentration of ACETIC ACID is that the same thing, meaning 0.102 mol/L or is it asking for the 5% of the acetic acid found in vinegar (if so how do u figure this out)..

I suppose the question is about the original concentration in vinegar.

Quote
and following up on the acetic acid, is the moles of acid in 10 mL = 0.00102 mol or something else?

You calculated it. Nothing else
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info

#### Borek

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##### Re:titrations
« Reply #3 on: May 15, 2005, 04:24:34 AM »
Well, right off the bat you had a math error.  remember, 14.3 mL becomes 0.0143 liters, not .0413 liters.

There are two typos, but the result is right.

Quote
You cannot find the conc. of vinegar, since it is not a pure substance.

He is not trying to determine 'vinegar concentration' but acetic acid concentration and that is perfectly doable as long as there are no other acids in the sample.
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info

#### chemical-idiot

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##### Re:titrations
« Reply #4 on: May 15, 2005, 10:01:14 AM »
yes, that 0.0413 was a typo.  Thanks a lot, i appericiate all the help.

#### chemical-idiot

• Guest
##### Re:titrations
« Reply #5 on: May 15, 2005, 11:38:26 AM »
so if i were to get the concentration of the acetic acid... which is 5% of the vinegar.  Do i have to do anything with that 5% or is my concentration (re-calculated) of 1.02 mol/L the correct concentration of acetic acid.

#### Borek

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##### Re:titrations
« Reply #6 on: May 15, 2005, 01:35:05 PM »
1.02M is the answer - it is 6.08% w/w (check my CASC for easy conversions between different concentration units/measures). This conversion is not very precise, as the density of the solution is in vinegar modified by other dissolved substances (see Grumples post).
« Last Edit: May 15, 2005, 01:47:52 PM by Borek »
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#### GCT

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##### Re:titrations
« Reply #7 on: May 15, 2005, 02:04:35 PM »
I need help on the following:

10 mL of vinegar diluted with water to reach 100 mL.  Then take 10 mL of that, and place into erlenmeyer flask.

Titrate.

Conc.  NaOH:  0.0715 mol/L
Conc. Vinegar(CH3COOH):  ? (to calculate)

Volume of NaOH used: 14.3 mL

I did:  CH3COOH   +   NaOH  --------->   CH3COONa  +  H20

0.07125 mol/L  x  0.0413 L of NaOH  =  0.00102 mol   [1:1 ratio]

conc.  CH3COOH:  0.00102 mol   /  0.01 L of vinegar =  0.102 mol/L

I was wondering if that is correct if i used ---- 10 mL of vinegar diluted with water to reach 100 mL.  Then take 10 mL of that, and place into erlenmeyer flask.

and if the question asks for concentration of ACETIC ACID is that the same thing, meaning 0.102 mol/L or is it asking for the 5% of the acetic acid found in vinegar (if so how do u figure this out)..

and following up on the acetic acid, is the moles of acid in 10 mL = 0.00102 mol or something else?