Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: Fish200398 on January 05, 2020, 02:14:11 AM
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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- ?
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NaOH will react with HC3H2O2
yes
both quantity is 9 moles, so it will react completely
no
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NaOH will react with HC3H2O2, both quantity is 9 moles
9 milimoles
Then added NaOH 40 mg
How many milimoles?
Write down a balanced reaction and do simple stoichiometry.
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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?
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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 ±.. ???
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noooo HC3H5O2 is 9 mmol at first, and there is 1 mole of NaOH
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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.
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there is 1 mole of NaOH so, produce 1 mole salt
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Then added NaOH 40 mg
this is 1 millimole
Read the text carefully.
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so the aswer is 16/13?
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No.
The acid is a little less and the amount of salt has increased the same but more.
Starting numbers are 9/9