[unsavedhero]

My chemistry professor (PhD from harvard; telling you this so you know he's not a moron) said "The answer is not 8 or 7 or 6.96 or 6.959. Come to class to find out the answer"
I'm really confused apparently -log(1.0x10^-8) is not enough to solve this problem. anyone know the answer? A lot of people tried and failed even though giving seemingly correct answers. any ideas guys? please and thank you

[Kate]

Hmmm, since the HCl concentration is so low, you probably have to take into consideration the autoprotolysis of water.

The H⁺ concentration will be 10^^-7 plus 10^^-8. But the -log of this sum gives me 6.96 and you just said it's wrong, so I don't know.

Btw, he could be a moron with a PhD from Harvard. 

[unsavedhero]

hmm yes i've asked around and everyone is getting the same answer. I'm starting to think his answer is different maybe to like the hundreths place or maybe he's just talking about significant figures? idk he's just being a little wise guy is all. but one quick question about the autoprotolysis of water. you said there will be 1.0 x 10^-7 H+ atoms from the water and 1.0 x 10^-8 from HCl right? so does water ALWAYS form 1.0 x 10^-7 H+ and 1.0 x 10^-7 OH at 25 celsius? oh and when do we take into account the autoprotolysis of water? is it only when acid/base concentrations are small? and what is considered small? thank you

[Big-Daddy]

Ok ...

Anything which does not take into account the self-ionization of water is an approximation. It becomes more valid as the action of the water becomes less relevant, which occurs as you have an increasingly acidic or basic solution (this can come about due to using acids/bases with higher Ka and Kb values respectively, or high concentrations).

Now, look into exact acid-base calculations. You will find the equations you need there. This - http://www.chem1.com/acad/pdf/c1xacid2.pdf - is the best introductory online tutorial I've ever seen and the case you're looking for is covered early.

You'll find the solution is even closer to 7 than your guess. Think about what that means: since we assumed all the acid dissociated, 10^-8 M must come from the acid; this means that less than 10-7 M H⁺ came from the water. (How much? Must be equivalent to [OH^-] of course, since water's self-ionization is the only equilibrium acting as a source of hydroxide ions!) In other words, adding more acid and base suppresses the equilibrium and effect of water's ionization, so that it contributes less. It contributes the full 10^-7 M to each, only when there are no acids or bases present.

[Kate]

so does water ALWAYS form 1.0 x 10^-7 H+ and 1.0 x 10^-7 OH at 25 celsius?

No. The autoprotolysis of water is an equilibrium, so if there's another "source" of H⁺ ions, like the complete dissociation of a strong acid such as HCl, then by Le Chatelier's principle the equilibrium of water will be displaced towards the formation of water and the concentration of H⁺ will be lower than 1x10^-7 M.

oh and when do we take into account the autoprotolysis of water? is it only when acid/base concentrations are small? and what is considered small? thank you

Depends on whether it's a strong acid/base or weak acid/base.

I went through my notes and it says that for a strong acid/base, as long as the concentration of acid or base is superior to 10^-5, then you can ignore the autoprotolysis of water, because water will contribute with even less than 10^-7 M of H⁺ ions and the error you're making is small and negligible.

For weak acids and bases there are 3 possible approximations:

1) If the acid/base is weak (k_a x 1000 > C_HA) and the concentration relatively high (k_a x C_HA > 1000 x k_w), then you can ignore the autoprotolysis of water and so the concentration of H₃O⁺ comes as the square root of the product k_a x C_HA. When I say you can ignore the ionization of the acid I mean that the concentration of HA at equilibrium is equal to the initial concentration of acid. The H⁺ concentration would still come from the ionization of the acid, but it's just a way to simplify the equation that you'd get.

2) If the acid is not that weak (k_a x 1000 > or approximately C_HA) and the concentration relatively high (K_a x C_HA > 1000 x K_w), then you can't ignore the ionization of the acid but you can ignore the autoprotolysis of water.

3) If the acid/base is weak (k_a x 1000 > C_HA) but the concentration relatively low (k_a x C_HA < or approximately 1000 x k_w), then you can't ignore the autoprotolysis of water but you can ignore the ionization of the acid.

[DanChen2014]

I got 6.978.

What is the answer?

[Bublik]

I got 6.975, and I most likely rounded less strictly than above, so the above answer is probably more correct

[Borek]

Correct answer is in another question: what is the temperature?

You all assumed pKw of 14. Not bad, but not necessarily correct either.

http://www.chembuddy.com/?left=pH-calculation&right=water-ion-product

Then, he can be hinting at a pretty small effect of the non-zero ionic strength of the solution.

[Borek]

so does water ALWAYS form 1.0 x 10^-7 H+ and 1.0 x 10^-7 OH at 25 celsius?

No. The autoprotolysis of water is an equilibrium, so if there's another "source" of H⁺ ions, like the complete dissociation of a strong acid such as HCl, then by Le Chatelier's principle the equilibrium of water will be displaced towards the formation of water and the concentration of H⁺ will be lower than 1x10^-7 M.

I suppose you mean "concentration of H⁺ from water autodissociation will be lower than 1x10^-7 M". If there is an acid present, solution has definitely more than 10^-7 M H⁺.

[Big-Daddy]

Correct answer is in another question: what is the temperature?

You all assumed pKw of 14. Not bad, but not necessarily correct either.

http://www.chembuddy.com/?left=pH-calculation&right=water-ion-product

Then, he can be hinting at a pretty small effect of the non-zero ionic strength of the solution.

We're looking at a very dilute acid here, and I thought the percentage effect of activity corrections tends to be less for more dilute salts (because ionic strength is smaller etc.) so this correction would be very small indeed for this case?

As for pK_w=14, I think we kind of have to assume 25°C if not given a temperature. Then Kw = 1.008*10^-14, as opposed to 1*10^-14 as we were using before. I feel (but have not checked) this won't make any difference in the first few decimal places of pH.

[Borek]

this correction would be very small indeed for this case?

Yes.

As for pK_w=14, I think we kind of have to assume 25°C if not given a temperature. Then Kw = 1.008*10^-14, as opposed to 1*10^-14 as we were using before. I feel (but have not checked) this won't make any difference in the first few decimal places of pH.

Yes again. But this is kind of question where we have to catch at straws.