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dentalstudent2021
hey guys, there is a trick to it that I can't seem to grasp.
the answer is 6.98 btw rather than 8.
the answer is 6.98 btw rather than 8.
Thanks, jjj7236!
Yup, it would work for bases too. Just make sure you're careful in terms of checking whether the are asking for pOH or pH.
Also, remember HCL is a strong acid and thus completely dissociates. With weak acids and weak bases the dissociated concentrations are usually much less than the initial acid and base concentration. Thus to find the [H] concentration of 1x10-8 [acetic acid (CH3OOH)] you would have to use to Ka value to determine the amount of [H] that dissociated. You cannot just add 1x10^-8 to 1x10^-7 like I did before... because the [H] would be even smaller than 1x10^-8... but just have in mind that this would be less additional [H] and thus the pH would be even closer/VERY close to 7
So with a strong base, you would still add 1x10^-7 because the H+/OH conc would be the same for water?
And for weak acid, would you do H+ = sqrt( HA*Ka) and add that to 1*10^-7?
Awesome thanks! Glad I found this out..wouldve missed it.Think you have the right idea, but might have just miss typed:
With a strong base, you would sum the 1x10^-7 concentration of [OH] within the water with the [OH] concentration produced from the base which as far as the DAT is concerned is equal to the concentration of the [Strong Base].
Yup!
Just to double check: you mean this works for bases as well when the BASE concentration is also less than 1*10^-7?
Thanks, jjj7236!
Yup, it would work for bases too. Just make sure you're careful in terms of checking whether the are asking for pOH or pH.
Also, remember HCL is a strong acid and thus completely dissociates. With weak acids and weak bases the dissociated concentrations are usually much less than the initial acid and base concentration. Thus to find the [H] concentration of 1x10-8 [acetic acid (CH3OOH)] you would have to use to Ka value to determine the amount of [H] that dissociated. You cannot just add 1x10^-8 to 1x10^-7 like I did before... because the [H] would be even smaller than 1x10^-8... but just have in mind that this would be less additional [H] and thus the pH would be even closer/VERY close to 7
that would be the ka = (H)(A-) / (HA) equation right ?
can you give a instance where they would give an acid and ask for POH ? if thats what you meant. or if they give base and ask for pH.
Yup, that is the Ka value I was talking about.{Equilibrium (K) expression for an acid (a)}
Sample Question...(if you want real exposure you need to work Destroyer and Chad's questions)
What is the pOH of an aqueous solution of 1x10^-8 HCL..
hey guys, there is a trick to it that I can't seem to grasp.
the answer is 6.98 btw rather than 8.
This applies when the OH concentration is less than 1x10^-7 right? Say OH= 1*10^-8. pOH=8, and you would think this seems odd for a base.The concentration of acid here is very low and therefore you need to consider the contribution of water.
Let's pretend you forgot to do this and you took the negative log of 1x10 exp -12. You would got an answer of 12, correct ? This can not be so, since an HCl solution is acidic and the pH must be under 7!!!. This should throw up a red flag signifying you have encountered a trick problem, in such a case the pH will be extremely close to 7, this trick will work perfectly for the DAT.
ALWAYS be careful of answers that make no sense.
Hope this helps
Dr. Romano
This applies when the OH concentration is less than 1x10^-7 right? Say OH= 1*10^-8. pOH=8, and you would think this seems odd for a base.
A basic solution would be done in a similar way.......If the base concentration was say...1 x10 exp -10........... You may have erroneously thought the pOH was 10...thus the pH would be 4
Clearly this is not possible......Again the answer would be approx 7 !!!!!
Keep hammering away !
Thanks Dr. Romano. so for this one, what would you estimate the pOH then if pH is around 7.