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Topic: Acid base titration  (Read 1547 times)

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Offline Fish200398

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Acid base titration
« on: January 05, 2020, 02:14:11 AM »
Buffer of HC3H2O2 45 mL 0,2 M (ka = 1,3 10^-5) with NaC3H5O2 0,10 M 55 mL. Then added NaOH 40 mg. what is the ratio of HC3H5O2/C3H5O2- ?

NaOH will react with HC3H2O2, both quantity is 9 moles, so it will react completely become salt NaC3H5O2? and using [C3H5O2-]^2 = kw/ka . [NaC3H5O2] ? i am not sure about this, since my answer not in the options. how to find the ratio of  HC3H5O2/C3H5O2- ?

Offline Borek

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Re: Acid base titration
« Reply #1 on: January 05, 2020, 03:56:32 AM »
NaOH will react with HC3H2O2

yes

Quote
both quantity is 9 moles, so it will react completely

no
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Offline AWK

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Re: Acid base titration
« Reply #2 on: January 05, 2020, 03:58:25 AM »
Quote
NaOH will react with HC3H2O2, both quantity is 9 moles
9 milimoles
Quote
Then added NaOH 40 mg
How many milimoles?
Write down a balanced reaction and do simple stoichiometry.
AWK

Offline Fish200398

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Re: Acid base titration
« Reply #3 on: January 05, 2020, 05:28:34 AM »
1 mmol NaOH react with 9 mmol HC3H5O2. form 1 mmol NaC3H5O2. left 8 mmol HC3H5O2. total of C3H5O2^{-} = 6,5 mmol?
HC3H5O2 / C3H5O2^{-} = 8/6,5 = 16/13 ? is it true?

Offline AWK

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Re: Acid base titration
« Reply #4 on: January 05, 2020, 05:46:53 AM »
CH3COOH + NaOH = CH3COONa + H2O
18 mmol     9 mmol                     
after reaction
9 mmol       0 mmol     9 mmol    it does not matter
now addition
              +1 mmol NaOH 
After addition of NaOH
9 ±..           0              9 ±.. ???
AWK

Offline Fish200398

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Re: Acid base titration
« Reply #5 on: January 05, 2020, 11:22:54 AM »
noooo HC3H5O2 is 9 mmol at first, and there is 1 mole of NaOH

Offline AWK

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Re: Acid base titration
« Reply #6 on: January 05, 2020, 11:40:46 AM »
You have 9 mmol of salt and propanoic acid at the beginning. I formed this salt from an additional 9 mmol of acid and 9 mmol of NaOH.
AWK

Offline Fish200398

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Re: Acid base titration
« Reply #7 on: January 06, 2020, 05:20:49 AM »
there is 1 mole of NaOH so, produce 1 mole salt

Offline AWK

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Re: Acid base titration
« Reply #8 on: January 06, 2020, 06:26:00 AM »
Quote
Then added NaOH 40 mg
this is 1 millimole
Read the text carefully.
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Offline Fish200398

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Re: Acid base titration
« Reply #9 on: January 06, 2020, 07:46:38 AM »
so the aswer is 16/13?

Offline AWK

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Re: Acid base titration
« Reply #10 on: January 06, 2020, 08:05:40 AM »
No.
The acid is a little less and the amount of salt has increased the same but more.
Starting numbers are 9/9
AWK

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