[xshadow]

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

[AWK]

This is a buffer solution.

[xshadow]

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

[AWK]

Buffer solution always contains mixture od two compounds; salt and weak acid (or base).
Two wise men prepared a special formula for such calculations.

[Borek]

Assume neutralization went to completion.

[xshadow]

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!

[xshadow]

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!

[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. 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.

[magician4]

@ 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

[Borek]

@ 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.

[Stronzio2]

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