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#Post#: 1453--------------------------------------------------
Help me on titration problem
DIR By: jamesjpb
Date: December 7, 2012, 10:54 pm
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The legal minimum limit of acetic acid in vineger is 4.00 % by
mass . A 5.00 mL sample of vinegar was titrated with 38.08mL of
0.100M NaOH solution. Does the sample exceed the minimum limit ?
(density of vinegar is 1.01g/mL )
Show solution
#Post#: 1456--------------------------------------------------
Re: Help me on titration problem
DIR By: Michel
Date: December 8, 2012, 3:28 am
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In this forum, we like to help you solve your question yourself
(see forum rules)
start by writing the balanced reaction between acetic acid and
NaOH.
Then find the amount of NaOH consumed in the reaction (molar
concentration=mole/volume). Using the balanced chemical
reaction, find the corresponding amount of CH3COOH consumed (in
mole). Then find the mass of CH3COOH by multiplying the amount
(in mole) by the molar mass of acetic acid.
You can now find the % by mass.
Post your attempt here and we'll see if you got it right.
#Post#: 1462--------------------------------------------------
Re: Help me on titration problem
DIR By: jamesjpb
Date: December 8, 2012, 8:03 am
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i got an answer of 4 but my instructor said that 4.53% is the
correct answer . I cant do it please help me please ? :) can you
show the solution then i'll learn it by myself
#Post#: 1465--------------------------------------------------
Re: Help me on titration problem
DIR By: jamesjpb
Date: December 8, 2012, 8:22 am
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Can you help me PLEASE ? Show me how to solve it . I need it
tomorrow for my assignment :)
#Post#: 1469--------------------------------------------------
Re: Help me on titration problem
DIR By: Michel
Date: December 8, 2012, 10:35 am
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Okay,
lets go...
First, write the balanced chemical reaction for the reaction
between CH3COOH and NaOH.
CH3COOH + NaOH => CH3COONa + H2O
concentration of NaOH=0.1M, volume of NaOH solution=38.08ml,
therefore amount of NaOH in mole=?
Find this.
Since in the eqn of reaction, 1mol of NaOH reacts with 1mol of
CH3COOH, whatever you get as the mole of NaOH is the mole of
CH3COOH in the solution. Then mole= mass of CH3COOH/Molar mass
of CH3COOH.
Plug in your values and find the mass of CH3COOH.
You are given density of solution as 1.01 g/ml
density= mass/volume
you know the volume of the solution as 5ml,
solve for mass of the acetic acid solution, 1.01=mass/5,
mass=5x1.01=5.05g
You can then find the mass % by dividing the mass of acetic acid
by the mass of solution and multiply by 100
#Post#: 1475--------------------------------------------------
Re: Help me on titration problem
DIR By: jamesjpb
Date: December 8, 2012, 6:01 pm
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how can i get the mass of solution ? the mass of solution that i
computer is 60 ? im not sure with my answer
#Post#: 1476--------------------------------------------------
Re: Help me on titration problem
DIR By: jamesjpb
Date: December 8, 2012, 6:27 pm
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Can you show how mass solution is being computed ? i cant
understand how to get the mass of solution ..
#Post#: 1483--------------------------------------------------
Re: Help me on titration problem
DIR By: Michel
Date: December 9, 2012, 12:25 am
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You titrated 5ml of vinegar,
density of solution=mass of solution/volume of solution
1.01=mass of solution/5
mass=5x1.01
=5.05g
#Post#: 1484--------------------------------------------------
Re: Help me on titration problem
DIR By: jamesjpb
Date: December 9, 2012, 1:28 am
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how is the solution for the mass solvent ??
#Post#: 1489--------------------------------------------------
Re: Help me on titration problem
DIR By: Michel
Date: December 9, 2012, 6:41 am
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--- Quote from: jamesjpb link ---
>
> how is the solution for the mass solvent ??
>
>
--- End Quote ---
The Vinegar is the solution, and in my earlier post, I
calculated it's mass since we're given it's volume and density.
Mass=volume x density
=5 x 1.01
=5.05g
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