[FirgeGobLin]

Hi there,

I am missing on little piece to my puzzle and would be very grateful if someone could help me out.
If you look to the initials of the ICE box I have set up for my reaction table, I do not understand how the H3O+ is 1E-7 as shown in my book's solution. It is this one little part that is hindering me from completing the problem.

Problem: Phenylacetic acid (C6H5CH2COOH simplified here to HPAc) builds up in the blood of persons with phenylketonuria, an inherited disorder that, if untreated, causes mental retardation and death. A study of the acid shows that pH of 0.12M HPAc is 2.62. What is the Ka of phenylacetic acid?

 Concentration (M)            HPAc(aq)     +      H2O(l)          H3O+          +        PAc-(aq)
      Initial                         0.12                    -----                       1E-7                        0
      Change
      Equilibrium

I was under the impression that the products start out with 0 for initial. I don't understand why there would be any amount under the products for initial.

Thanks!

Jaime

[Loyal]

The 10^-7 is the amount that is produced from water through this reaction

2 H₂O <-> OH^- + H₃O⁺

I would think though with a pH of 2.62 the amount produced by water would be very small.  Even though it is more accurate to include the amount from water I don't think it will impact the answer much.   

Though if your book says to include it then it may be a good idea to do so.

[nj_bartel]

You only need to take into account the autoionization of water when the concentration of the acid is on the order of 10^-6 M, I believe.

[Loyal]

You only need to take into account the autoionization of water when the concentration of the acid is on the order of 10^-6 M, I believe.

Technically you should do it all of the time, but you are able to do an approximation to save time.  It only becomes heavily mandatory when you are dealing with a high precision or a low concentration.

[nj_bartel]

There are always approximations    If you need an answer that accurate, you're likely going to have to take into account how the autoionization varies with temperature, pressure, concentration of deuterium, etc.

I can say right now that if you're using sig figs to arrive at your final answer, the autoionization of water is never going to come into play with a pH of 2ish and a concentration of acid of 0.1 M, with 2 sig figs.

[Loyal]

There are always approximations    If you need an answer that accurate, you're likely going to have to take into account how the autoionization varies with temperature, pressure, concentration of deuterium, etc.

I can say right now that if you're using sig figs to arrive at your final answer, the autoionization of water is never going to come into play with a pH of 2ish and a concentration of acid of 0.1 M, with 2 sig figs.

Very true.  I mean I have learned quite quickly in Phyiscal Chemistry you can drive yourself crazy with corrections.   Stupid universe doesn't like to keep it simple. 

[Borek]

Using 10^-7 as initial concentration, and ignoring water autodissociation equilibrium doesn't make much sense. Even when the acid concentration is very low, it won't give correct answer, as dissociating acid shifts water dissociation thus initial is no longer 10^-7. ICE table (at least in its simplest form, could be there are some more elaborate methods tjhat I am not aware off) is not able to deal with the problem.

For example (see attachement) 3e-7M solution of acetic acid contains 2.945e-7M of acetate ion - thus acetic acid is a source of 2.945e-7M H⁺. Total concentration of H⁺ is 3.253e-7M - so water autodissociation produced 3.254e-7-2.945e-7=3.09e-8M - over three times less than 10^-7.

So - when you have to account for water autodissociation you are out of the ICE table usability range, when you are inside of the ICE table usability range, using 10^-7 for initial is just a waste of time, as it can be safely ignored.