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#Post#: 1256--------------------------------------------------
Confusing buffer question
DIR By: Chris
Date: November 27, 2012, 11:36 am
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Hello, this my homework and i've to submit it latest friday.
I've been attempting them but i'm confused:-(
Can someone here help do them!!!!
1.) Calculate the change in pH obtained on addition of 5x10^-3
mole NaOH to a 0.1L buffer solution of 0.12M Na2HPO4 and 0.10M
KH2PO4 solutions. Assuming no volume change (Ka2=6.2x10^-8
2.)Calculate the conc of sodium propanoate that must be added to
a 0.14M solution of propanoic acid to prepare a buffer soln of
pH=4.65
Ka=1.3x10^-5
If the concs of the two solns are equal, In whaat ratio of
concentrations will you mix the two solutions to prepare a
buffer solution of pH=6.00
please help me answer. Thanx
#Post#: 1273--------------------------------------------------
Re: Confusing buffer question
DIR By: Michel
Date: November 28, 2012, 7:07 am
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Remember, We don't solve your hw for you...we help you solve it
(forum rules) For the 2nd Question.
First find Pka of C2H5COOH.
Then find the concentration of the C2H5COO- in 0.14M propanoic
acid, using the eqn for Ka.
When C2H5COONa is added, it ionizes completely, thus adding its
molar amount of C2H5COO- into the solution.
You can find the amount of C2H5COONa needed using the HH eqn.
pH= pKa + log[(conj.b ase/acid)]
then C2H5COO- will be the only unknown.
Remember there's already C2H5COO- in the solution(from the
propanoic acid) You have to consider this when plugging values
into the HH eqn.
The 2nd part is simply using HH again. pKa is constant, so just
find the ratio of the conj. Base to acid, probably safe to
assume all C2H5COO- comes from the salt, but you can also be
more precise by considering the C2H5COO- from the acid.
#Post#: 1275--------------------------------------------------
Re: Confusing buffer question
DIR By: Michel
Date: November 28, 2012, 8:28 am
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Not so sure about your 1st question...hope someone here comes up
to help, or else I might read up polyprotic acids:O
Not exactly something I feel like reading now :P
#Post#: 1352--------------------------------------------------
Re: Confusing buffer question
DIR By: Chemist@
Date: December 2, 2012, 10:33 am
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For the first one, only KH2PO4 will react with NaOH as Na2HPO4
is very very weak acid (its Kb should be bigger than Ka).
Calculate the number of moles of KH2PO4 that reacted with NaOH
and the number of moles that left. Then use the
Henderson-Hasselbalch equation to calculate the pH of the
H2PO4-/HPO42- buffer.
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