Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: xshadow on November 19, 2016, 12:45:13 PM
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HI :)
How can I calculate the pH of a solution that i get mixing :
1000ml NH3 10-2 M
1000ml CH3COOH 10-3 M
Now I have this acid-base reaction:
NH3 + CH3COOH ----> NH4CH3COO
But NH4CH3COOH is an ampholytes because in H2O gives two reaction:
1) CH3COO- + H2O ---> CH3COOH + OH-
2)NH4+ + H2O ---> NH3 + H3O+
So in the reaction environment I should have:
a) [NH4CH3COOH] = 10-3M
b) some remaining NH3 (it was in excess) (I think 10^-2 - 10^-3)
How can I calculate the pH of this solution?
Thanks
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This is a buffer solution.
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This is a buffer solution.
mmhhh yes...
I should have NH4+/NH3 ...but I have also CH3COO- that I think it gives the reaction:
CH3COO- + H2O ---> CH3COOH + H3O+
SO how can I calculate pH fot this "solutuion" ?
(buffer solution of NH4+/NH3 but also some CH3COO- (coming from the salt) which gives hydrolysis- coming from the salt)
Thanks :)
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Buffer solution always contains mixture od two compounds; salt and weak acid (or base).
Two wise men prepared a special formula for such calculations.
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Assume neutralization went to completion.
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Buffer solution always contains mixture od two compounds; salt and weak acid (or base).
Two wise men prepared a special formula for such calculations.
yes...
But usually is:
Salt : NH4Cl
weak base : NH3
So I have a buffer solution of NH4+/NH3 and a strong base Cl- that doesn't hydrolize
BUT I have,in this case , a buffer solution of a chemical compound "A" (NH3/NH4+) and a weak base CH3COO- that reacts with water (hydrolize)
So how can I calculate pH considering also the protonation of CH3COO- in my buffer solution?
Thanks!
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Buffer solution always contains mixture od two compounds; salt and weak acid (or base).
Two wise men prepared a special formula for such calculations.
yes...
But usually is:
Salt : NH4Cl
weak base : NH3
So I have a buffer solution of NH4+/NH3 and a strong base Cl- that doesn't hydrolize
BUT I have,in this case , a buffer solution of a chemical compound "A" (NH3/NH4+) and a weak base CH3COO- that reacts with water (hydrolize)
So how can I calculate pH considering also the protonation of CH3COO- in my buffer solution?
Thanks!
I can assume this because I have a buffer solution and a WEAK base CH3COO-??
So I can neglect the hydrolisis of CH3COO- in order to give acetic acid?
BUT if I have a LOT of CH3COO- (much salt) isn't there a formula in order to calculate the pH of this system?
(I usually use formulas obtained with mass and charge balance...)
Thanks!
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For buffer solution a working ratio Bronsted acid/conjugated Bronsted base is from 1/10 to 10/1.
Your data fails in this range. So you can use the most approximated formula for calculation pH of solution (after correct stoichiometry of neutralization - your first post showed the correct way of this stage).
You should write textbook chapter concerning buffer solution. All these informations you can find there.
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@ AWK:
For buffer solution a working ratio Bronsted acid/conjugated Bronsted base is from 1/10 to 10/1.
Your data fails in this range.
assuming that with the excess of ammonia given, the acetic acid becomes more or less completely deprotonized, the resulting ratio of NH3/NH4+ should be like ~ (10-2 - 10-3) / 10-3 = 9 : 1 , which is well within named range
@ xshadow:
you could approximate the situation to be an ammonia / ammonium-buffer of above named composition, and use the calculation belonging to
for higher precision, I'd recommend to use a simplified charge balance - but this might be slightly overdoing here
regards
Ingo
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@ AWK:
For buffer solution a working ratio Bronsted acid/conjugated Bronsted base is from 1/10 to 10/1.
Your data fails in this range.
assuming that with the excess of ammonia given, the acetic acid becomes more or less completely deprotonized, the resulting ratio of NH3/NH4+ should be like ~ (10-2 - 10-3) / 10-3 = 9 : 1 , which is well within named range
I think AWK meant the ratio FALLS in the range and you are both saying exactly the same.
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K for this reaction = 31578 = ka(ch3cooh)/ ka(nh4)
So
Nh3= 0,0045 M
Ch3cooh= 0,0005 M
Nh4+ = 0,0005 M
Ch3cooh = 0 ( k is quantitative)
H+ = ka (nh4) * 0,0005 / 0.0045
pH=11-0.80=10.20