What is the PH of....

[ezsanche]

a [10^-8] molar solution of HCL?

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[BerkReviewTeach]

a [10^-8] molar solution of hcl?

6.91

 

[ezsanche]

How did you get that number?

 

[Longshanks]

6.91

How did you get that number?

How BRT got that number:

Based on what you wrote it is an aqueous solution, and you're given the molarity of HCL. So in terms of HCL the [H+] = 10^-8. However, the pH is not 8! It is an aqueous solution so the [H+] from HCL is less than the contribution of [H+] from water itself (which is 10^-7). Therefore, due to the contribution of the dilute HCL, the pH should be slightly more acidic than 7. On the MCAT you won't need to calculate exact numbers so you should be able to logically reason that the answer choice must be 6.5<x<7, as the answer choices wouldn't typically be that nit picky.

 

[loveoforganic]

edit that last sentence out. it doesn't make sense.

 

[ezsanche]

Well i worked out the problem and I end up getting two different answers based on two different reasoning.
(assuming standard condition)

Answer #1: Since most of the H+ concentration in the solution is due to the autoionization of water; the H+ concentration due to water is 10^-7 molar. The HCL donates a very little amount of H+ ,10^-8 molar. The sum of the total H+ concentration in solution should be 1.1 x 10^-7 (10^-7 + 10^-8). The negative log of 1.1 x 10^-7 is 6.9586. pH = 6.9586.

Answer #2: Because they are way more H+ ion due the autoionization of water, the very small of HCL doesn't dissociate fully and doesn't contribute to the total H+ concentration as much as it would be from the predicted value (le chatelier principle). So the pH will remain closer to 7..

 

[BerkReviewTeach]

Answer #1: Since most of the H+ concentration in the solution is due to the autoionization of water; the H+ concentration due to water is 10^-7 molar. The HCL donates a very little amount of H+ ,10^-8 molar. The sum of the total H+ concentration in solution should be 1.1 x 10^-7 (10^-7 + 10^-8). The negative log of 1.1 x 10^-7 is 6.9586. pH = 6.9586.

Yeah, that's it exactly. I guessed at the log of 1.1 as being around 0.09, because I had no calculator nearby. But yeah, the HCl should fully dissociate, just as it would at even higher concentrations, resulting in a solution with 1.1 x 10-7 M H+.

But Longshanks is 100% correct that the answer choices on the MCAT would give you such a huge hint that you wouldn't need to calculate the answer.

The choices would probably be something like:

• A. 6.95
  B. 8.00
  C. 6.00
  D. 7.05

You know the solution is slightly acidic, so it must be less than 7. But it can't be as low as 6, which would require 10-6 M HCl. Only choice A fits.

Longshanks method is brilliant and insightful, and beats our method of doing math by a good margin.

 

[loveoforganic]

I'm blanking out on the method to actually do the math. Would you mind, if you remember BRT?

 

[boaz]

This would be the most rigorous solution to the question:

We have two 'equilibria' happening:

1) H2O ==> H+ + OH-
2) HCl ==> H+ +Cl-

K1=[H+][OH-]=10^-14
K2=[H+][Cl-]=10^7

Now, the charges in solution must be balanced (otherwise the beaker will go flying ):

[H+]=[OH-] + [Cl-]

Rearranging the above gives:

[OH-]=[H+] - [Cl-]

If we plug that into the equilibrium expression K1, we get:

K1=[H+][OH-] = [H+]([H+]-[Cl-])

which is in the form of a quadratic equation.

Solving it shows that [H+]=1.051*10^-7

pH = 6.98.

Obviously, on the MCAT you're not going to see anything that needs this kind of calculation. You won't see anything more than what BRT shows above.

 

[happyfellow]

So does that mean we could effectively have HCl by itself and have a basic pH if the concentration was (10^-9)? Would the pH then be 9?

I know this questions is silly but I always though that if we had a strong acid by itself that the pH would be acidic.

 

[loveoforganic]

Thanks ishch, apparently I forgot everything from quant!

Happyfellow - did you read the thread? Did anyone suggest that?

 

[BerkReviewTeach]

I'm blanking out on the method to actually do the math. Would you mind, if you remember BRT?

The eqilibrium method employed by ishchayill is absolutely exact, but as they mentioned, it's a bit math heavy.

For the MCAT, we can assume that the amount of H+ in the solution is a sum of the 10-7 M from the autoionization of water and the 10-8 M from the full dissociation of HCl. This results in [H+] = 1.1 x 10-7.

Once you know the [H+], you normally plug into pH = -log [H+] = -log(1.1 x 10-7). But the shortcut is to work that equation to the following:

-log (1.1 x 10-7) = - (log 1.1 + log 10-7) = - (log 1.1 + (-7)) = 7 - log 1.1. This is the short cut equation.

• Short cut equation: pH = -(power of 10) - mantissa

The log of 2 is 0.3 and the log of 1 is 0, so the log of 1.1 is about 0.05. This means that the pH will be 7 - 0.05 = 6.95.

In terms of a non-math solution, we know that water itself is pH = 7, so adding the small amount of acid from 10-8 M HCl will make it slightly more acidic, lowering the pH slightly. This eliminates choices B and D. In order to be pH = 6, the [H+] would need to be 10-6 M, which it's not. This eliminates choice C and leaves only choice A.

I'm not sure I answered your question, but I did write a whole bunch of words.

 

[loveoforganic]

I was looking for the exact exact answer that ishch gave, and I've already taken the MCAT, but I'm sure your explanation will be beneficial to someone else, so

 

[BerkReviewTeach]

Thanks ishch, apparently I forgot everything from quant!

Isn't that a good thing, because remembering quant or p chem is just not natural.

BTW, I really appreciate your posts. Same thing with ishchayill. I'm really glad the two of you help so much around this site.

 

[loveoforganic]

Isn't that a good thing, because remembering quant or p chem is just not natural.

BTW, I really appreciate your posts. Same thing with ishchayill. I'm really glad the two of you help so much around this site.

I can agree with the sentiment about quant wholeheartedly Haven't experienced p. chem yet.

And aw shucks, right back at you!

 

[Longshanks]

Longshanks method is brilliant and insightful, and beats our method of doing math by a good margin.

Why thank you! Couldn't have done it without TBR though... before that I was just trudging through the math and doing things straight 'by the book' thinking I didn't get chem, but TBR helped me gain confidence in using intution if you understand what's going on.

So does that mean we could effectively have HCl by itself and have a basic pH if the concentration was (10^-9)? Would the pH then be 9?

I know this questions is silly but I always though that if we had a strong acid by itself that the pH would be acidic.

Exactly, it would be strongly acidic. I was just being foolish when I wrote that other statement. See what happens when one rushes and don't stop to think for a second? If it was HCL by itself it would have to be acidic, and HCL is pretty darn strong.

 

[boaz]

Thanks ishch, apparently I forgot everything from quant!

Going into quant I thought I'd hate it (I'm more of an organic thinker ) but then I came to really appreciate it (at least most of it) when I really understood on a conceptual level what's going on.

Isn't that a good thing, because remembering quant or p chem is just not natural.

BTW, I really appreciate your posts. Same thing with ishchayill. I'm really glad the two of you help so much around this site.

The "aha!" moment is one of the things I live for, and when I can I try to help others experience such moments.

The eqilibrium method employed by ishchayill is absolutely exact, but as they mentioned, it's a bit math heavy.

For the MCAT, we can assume that the amount of H+ in the solution is a sum of the 10-7 M from the autoionization of water and the 10-8 M from the full dissociation of HCl. This results in [H+] = 1.1 x 10-7.

Once you know the [H+], you normally plug into pH = -log [H+] = -log(1.1 x 10-7). But the shortcut is to work that equation to the following:

-log (1.1 x 10-7) = - (log 1.1 + log 10-7) = - (log 1.1 + (-7)) = 7 - log 1.1. This is the short cut equation.

• Short cut equation: pH = -(power of 10) - mantissa

The log of 2 is 0.3 and the log of 1 is 0, so the log of 1.1 is about 0.05. This means that the pH will be 7 - 0.05 = 6.95.

In terms of a non-math solution, we know that water itself is pH = 7, so adding the small amount of acid from 10-8 M HCl will make it slightly more acidic, lowering the pH slightly. This eliminates choices B and D. In order to be pH = 6, the [H+] would need to be 10-6 M, which it's not. This eliminates choice C and leaves only choice A.

I'm not sure I answered your question, but I did write a whole bunch of words.

This paragraph sounds like it could have come straight out of TBR chem! Yea, this is definitely the way to approach it on the actual MCAT. I used TBR physics and gen chem review books and it did it for me .

 

[BerkReviewTeach]

This paragraph sounds like it could have come straight out of TBR chem! Yea, this is definitely the way to approach it on the actual MCAT. I used TBR physics and gen chem review books and it did it for me .

I swear, teaching for them completely molded my mind. I've fielded questions in office hours for so long that I speak like the books sometimes. My girlfirend calls it my Berkeley mode whenever I answer a simple question that way.

 

[ezsanche]

This would be the most rigorous solution to the question:

We have two 'equilibria' happening:

1) H2O ==> H+ + OH-
2) HCl ==> H+ +Cl-

K1=[H+][OH-]=10^-14
K2=[H+][Cl-]=10^7

Now, the charges in solution must be balanced (otherwise the beaker will go flying ):

[H+]=[OH-] + [Cl-]

Rearranging the above gives:

[OH-]=[H+] - [Cl-]

If we plug that into the equilibrium expression K1, we get:

K1=[H+][OH-] = [H+]([H+]-[Cl-])

which is in the form of a quadratic equation.

Solving it shows that [H+]=1.051*10^-7

pH = 6.98.

Obviously, on the MCAT you're not going to see anything that needs this kind of calculation. You won't see anything more than what BRT shows above.

That is what I was thinking of in my second method but I was unsure on how to do the math. Thanks everyone!