questions abotu pH and pOH


by V0ODO0CH1LD
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V0ODO0CH1LD
V0ODO0CH1LD is offline
#1
Sep19-13, 05:15 PM
P: 207
The definition of the pH and pOH of a substance is the negative log base ten of the concentrations of hydronium and hydroxide, respectively, in one molar of that substance. Right? I know I can talk about the pH of 3 mols of a certain substance in a liter of water (the pH of 3 molar of that substance). But is it a convention that if I just ask "what is the pH of substance X?" I mean the negative log base ten of the concentration of hydronium in one liter of water after mixing one mol of that substance in, or is it just implied?

Because when someone tells me that the pH of tomato juice is 4, I directly think "if I put one mol of tomato juice in a liter of water and measure the concentration of hydronium in the resulting solution I will find 10^-4 mols of H3O". Is that right?

Also, if I measure the amount of hydronium and hydroxide in one liter of water I will find 10^-7 mols for both, right? But if I plug that in the equilibrium constant equation I get
[tex] \frac{10^{-7}10^{-7}}{[H_2O]}=K_{eq} [/tex]
which means either that [H2O] = 1, or that the equilibrium constant of auto-ionazition of water isn't 10^-14 like I've been always told, but [H2O]*Keq = 10^-14. But the number of mols of H2O in one liter of water is 55.5, not one. So I'm confused..

Another question, let's say I have the reaction:
[tex] HX_{(aq)}\rightleftharpoons H^+_{(aq)}+X^-_{(aq)} [/tex] or
[tex] HX+H_2O\rightleftharpoons H_3O^++X^- [/tex]
(They mean exactly the same thing, right?)

And after equilibrium is reached I measure the concentration of ##X^-## and ##HX## to be ##10^{-3}M## and ##10^{-2}M##, respectively. So
[tex] \frac{[H^+]10^{-3}}{10^{-2}}=K_{eq}. [/tex]
Which allows me to solve for the pH of ##HX## if I know the constant of equilibrium of that substance, right?

But I've seen people rush to the conclusion that the concentration of hydronium and ##X^-## is the same because each molecule of ##HX## will become one of ##H^+## and one of ##X^-##, but what about the concentration of hydronium in the water before mixing ##HX## in? Is it that the concentration of hydronium in the water that doens't come from ##HX## is negligible? Or am I missing something?
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mfb
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#2
Sep19-13, 05:37 PM
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"Concentration" always means "amount of X in amount of Y" - the concentration does not depend on the volume (or amount of water) you consider, so you can just talk about the concentration (or here: pH) of your water.

Because when someone tells me that the pH of tomato juice is 4, I directly think "if I put one mol of tomato juice in a liter of water and measure the concentration of hydronium in the resulting solution I will find 10^-4 mols of H3O". Is that right?
No. The tomato juice itself has 10^(-4) mole/liter H3O+ ions. If you have 1 liter of tomato juice, you have 10^(-4) mole H3O+ ions. If you have 10 liters, you have 10^(-3) mole H3O+ ions.
If you mix 1 liter of tomato juice with 1 liter of water, you still have 10^(-4) mole H3O+ ions (approximately), but now those are distributed over 2 liters, so you get 10^(-4) mole/(2 liter) = 5*10^(-5) mole/liter H3O+ ions or a pH of ~4.3.

Tomato juice is a mixture of different substances, "one mole of it" does not exist.

Also, if I measure the amount of hydronium and hydroxide in one liter of water I will find 10^-7 mols for both, right?
With pure water, right.


Keq is not the KW used in the pH-definition. They differ by that factor of 55.

but what about the concentration of hydronium in the water before mixing HX in? Is it that the concentration of hydronium in the water that doens't come from HX is negligible?
In many cases, this is true (e.g. if the final pH is 5, 99% comes from the acid. With a pH of 4, it is 99,9% and so on). If it is not, you have to do the full calculation.
V0ODO0CH1LD
V0ODO0CH1LD is offline
#3
Sep19-13, 06:32 PM
P: 207
Okay, so the pH and pOH of a substance is the amount, in mols, of hydronium and hydroxide, respectively, in one liter of that substance. Right?

But in that equation [tex] \frac{[H^+][X^-]}{[HX]}=K [/tex]

which comes from the reaction ##HX_{(aq)}\rightleftharpoons H^+_{(aq)}+X^-_{(aq)}##, ##[HX]## still represents the amount of the substance ##HX## in one liter of water, ##[H^+]## still represents the amount of hydronium in that same liter of water and ##[X^-]## still represents the amount of ##X^-## anions still in that same liter of water, all in mols and after equilibrium. Right?

But the fact that sometimes people assume that ##[H^+]=[X^-]## is that the amount of hydronium not coming from ##HX## is negligible. Otherwise I would have to perform the whole calculation without that assumption. Right? I didn't get your last paragraph with the percentages.. Is that what you meant?

EDIT: My bad, I got it. I read "pH is 5, 99% comes" as "pH is 5.99% comes".. :)
still, how did you get that if the pH is 5, 99% comes from the acid?

Borek
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#4
Sep20-13, 02:37 AM
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questions abotu pH and pOH


Quote Quote by V0ODO0CH1LD View Post
EDIT: My bad, I got it. I read "pH is 5, 99% comes" as "pH is 5.99% comes".. :)
still, how did you get that if the pH is 5, 99% comes from the acid?
If pH is 5, concentration of H+ is 10-5 M. Pure water has pH of 7, so the concentration of H+ from water autodissociation is 10-7 M. That means H+ from water autodissociation is around

[tex]\frac {10^{-7}}{10^{-5}} 100\% = 1\%[/tex]

and around 99% comes from the acid.

That's a very crude approximation, as the acid presence shifts water autodissociation left, and in fact concentration of H+ from water autodissociation will be much lower. Easy to check - water autodissociation produces OH- and H+, and as pOH+pH=14, pOH=9 and [OH-] = 10-9 M. As water autodiossociation produces equal amounts of H+ and OH- concentration of H+ from autodissociation is 10-9 M as well. Compare that to 10-5 M in total and you will see that acid produced 99.99% of the H+ present.


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