finding pH please help!!!
Started by joonkimdds
The square root of 5e-14 is equal to the individual [H30+] and [OH-]. So simply square root it and you will get [H30+]. Then -log that you will find the pH. I think that is it.
how about pOH, can someone help me finding it without using 14-pH?
Advertisement - Members don't see this ad
The only other way you can find pOH, is to have Kb. You could use kw = ka x kb to solve for kb, but then you would need to know ka. So, I don't think you have enough info here to solve for the pOH using this method.
Hmmmm. Now I'm doubting my own solution.
Our solution is based on the idea that [H+] = [OH-], so that we can assume:
Kw = x^2
So, when we solve for x, we have:
5x10E-14 = x^2 = 2.24E10-7 = [H+] = [OH-]
So...
pH = -log[H+] = -log (2.24E-7) = 6.65
pOH = -log[OH-] = -log (2.24E-7) = 6.65
However, if I use pOH = 14 - pH, then pOH comes out to be a different number:
pOH = 14 - 6.65 = 7.35
Why the discrepancy? I'm Confused!!
Our solution is based on the idea that [H+] = [OH-], so that we can assume:
Kw = x^2
So, when we solve for x, we have:
5x10E-14 = x^2 = 2.24E10-7 = [H+] = [OH-]
So...
pH = -log[H+] = -log (2.24E-7) = 6.65
pOH = -log[OH-] = -log (2.24E-7) = 6.65
However, if I use pOH = 14 - pH, then pOH comes out to be a different number:
pOH = 14 - 6.65 = 7.35
Why the discrepancy? I'm Confused!!
That's exactly what I think and why I am confused
Kw is the autoionization constant....at 298k it is 10^-14....
The reaction is endothermic, thus the autoionization constant is elevated at higher temperature, as is the case at 37 celsius...
Since the problem says water, we may assume the solution is neutral. By definition, pOH = pH for a neutral solution. pH or pOH = pKw/2 for a neutral solution since the hydroxide and hydronium concentrations are equal.
oooh! I think I got it...That's exactly what I think and why I am confused
pOH = 14 - pH only when Kw = 10E-14, so in our case:
pOH = -log(5E-14) - pH
= 13.3 - 6.65 = 6.65
pOH = 6.65 = pH as expected!!
