pH of this solution?

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AggieJohn

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Here's a good question, that I was told appeared on a previous MCAT:

What is the pH of a 1.0 * 10^-8 M HCl solution?
 
Yes, it should be 8:

pH = -log[H+]
If we have 1*10^-8 as the [HCl], the [H+] = 1*10^-8

so, pH = -log (1*10^-8) = 8
 
You're telling me that you add HCl (a strong acid) to water, and the pH is basic? C'mon now... 😉
 
That seems wrong because the pH is way too high...plus it would be too easy to answer. I think the concentration of HCl is low so you have to factor in H+ from water. So -log(1.08*10^-8 +1*10^-7)=6.95? Could be wrong..dont remember chem at all....
 
the dissociation of water is x10^-7 so that means that the concentration of HCl that makes x10^-8 is VERY small. Thus its going to be only slight below neutral but IT WILL BE ACIDIC. I am not going to do any calculations right now but it will be around 6.9
 
yeah jfleong is right on. when the acid/base conc. is extremely low, you have to take into account the contribution from water. so since its normally pH 7, it has 1x10^-7 [H+] conc. and adding the HCl adds 1x10^-8 to this, which gives a total of 11x10^-8 for [H+]. this is a pH of 6.95
 
topdogg82 said:
yeah jfleong is right on. when the acid/base conc. is extremely low, you have to take into account the contribution from water. so since its normally pH 7, it has 1x10^-7 [H+] conc. and adding the HCl adds 1x10^-8 to this, which gives a total of 11x10^-8 for [H+]. this is a pH of 6.95


another way to rule out a pH of 8 is to think about what you have in the solution. if you have HCl, which is a strong acid, you cannot get a pH that's >6.99. so if you have a 10^-8 M HCl, you must factor in the contribution from water, and since the molarity of the HCl is so low, the pH should be around a 6-6.99 range.
 
I got this problem right from intuition. I knew it had to slightly less than 7. HOwever, in my gen chem class they never taught us to consider the acid dissociation of water when the acidic concentration is so low. If ph is 1e-7, we assumed it was 7. So in this case, 1e-8, based on our class the amount of [OH] would be 1 e-6 using using [H+][OH] = 1e-14. Consequently, since [OH] is greater one would think it would be basic. I feel cheated, i go to a large public big ten university and earned good grades and feel cheated. I mean we learned acid dissociation, but it was never used in the aforementioned situation. What book did you use, we used brown lemay bursten and it didn't cover this.
 
Will Hunting said:
I got this problem right from intuition. I knew it had to slightly less than 7. HOwever, in my gen chem class they never taught us to consider the acid dissociation of water when the acidic concentration is so low. If ph is 1e-7, we assumed it was 7. So in this case, 1e-8, based on our class the amount of [OH] would be 1 e-6 using using [H+][OH] = 1e-14. Consequently, since [OH] is greater one would think it would be basic. I feel cheated, i go to a large public big ten university and earned good grades and feel cheated. I mean we learned acid dissociation, but it was never used in the aforementioned situation. What book did you use, we used brown lemay bursten and it didn't cover this.

nevermind, we did I just forgot. Doh, and it was only a few months ago.
 
Ok, so it has been mentioned. I had not even thought of the calculations to be quite honest. I knew, as previously stated that it would be only slightly below 7, so I would choose 7 or 6.95 or something. Thats awesome that you got the calculations jfleong =)
 
actually if someone wanted this exact example is in Kaplan's book (ch. on acids/bases)
 
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