pH of a strong acid with a concentration of less that 1x10^-7

[Extirpator]

Sorry if someone already asked this, but I couldn't find it in the forum...

The example given states:

A 1x10^-8 solution of HCl would appear to be 8 unless you take into account the H+ ions from the dissociation of water... which you must take into account for acid/base concentrations below 1x10^-7.

I think the problem for me is that I haven't done this type of math in a very long time, but maybe someone can refresh my memory...

They set it up as:

Kw = (x + 1x10^-8)(x) = 1.0x10^-14

They solve for x and get x = 9.5x10^-8 <--- This is the part where I'm not sure how they got their answer.

they finally get a pH of 6.98

Any help is appreciated, thanks!

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[Cjc555777]

They tell you that the HCl that you add is 1x10^-8. If you take a container of water with nothing else in it the Kw of that contain is 1x10-14, and in that water you have equal amount of H+ and OH- so if you do the math [H+]=[OH-]=1x10-7. 1x10-7 is 10 times larger than the amount of HCl that you are adding, that means that the HCl litterally has no (almost no) effect on the pH of the solution because your concentration of H+ from the water is higher than what you are adding. Does that help, if not I can try to reword it if you need.

 

[Extirpator]

Sorry, I should have explained my question a little better. I understand why you need to do it this way, I'm just lost as to how they got x=9.5x10^-8 M... I'm essentially just looking for someone to explain the math to get that number. I know it's not that difficult ha, but for some reason I can't recall how to go about it.

So my question is how to get ---> x = 9.5x10^-8

by solving this eqauation ---> Kw = (x + 1x10^-8)(x) = 1.0x10^-14

Thanks

 

[Chowdder]

I actually have only come across one question that requires the use of a quadratic equation (what you're looking for to solve this) on a practice question. And that question wasn't from AAMC, but rather princeton review. Nevertheless, a quadratic equation is what you're looking for to solve this problem.

(x +1x10^-8)(x) = 1x10^-14
solve it to get a polynomial
x^2 + 1x10^-8x - 1x10^-14 = 0

use quadratic formula

(wikipedia)
where b = 1x10^-8
a = 1
c = -1x10^-14

You should come out with two values, a negative one and a positive one for the value of x. Negative one is eliminated because you can't have negative concentration. So the positive value is the correct one.

If you're not sure of how to solve this, let me know and I'll quickly solve it for you.

Cheers

 

[Extirpator]

got it

Thanks a lot for your help!

 

[BerkReviewTeach]

Sorry if someone already asked this, but I couldn't find it in the forum...

The example given states:

A 1x10^-8 solution of HCl would appear to be 8 unless you take into account the H+ ions from the dissociation of water... which you must take into account for acid/base concentrations below 1x10^-7.

I think the problem for me is that I haven't done this type of math in a very long time, but maybe someone can refresh my memory...

They set it up as:

Kw = (x + 1x10^-8)(x) = 1.0x10^-14

They solve for x and get x = 9.5x10^-8 <--- This is the part where I'm not sure how they got their answer.

they finally get a pH of 6.98

Any help is appreciated, thanks!

Because the test is multiple choice, you probably should avoid the quadratic equation and make a quality guess. The [H+] will be about 1.1 x 10exp-7 (10exp-7 from the water and 10exp-8 from HCl). If [H+] = 1.1 x 10exp-7, then the pH = 7 - log 1.1 = 6.9something. The best answer choice between 6.9 and 7.0 is the winner.

BTW, I used a BR math trick to go from [H+] = 1.1 x 10exp-7 to pH = 7 - log 1.1. The pH = -(power of 10) - log unit #. The derivation can be found in the BR book if you want to look it up, but I'd suggest just taking it at face value.

The point is that on a multiple choice exam where you get no calculator, approximate quickly and as close to the real number as possible.

 

[Extirpator]

Because the test is multiple choice, you probably should avoid the quadratic equation and make a quality guess. The [H+] will be about 1.1 x 10exp-7 (10exp-7 from the water and 10exp-8 from HCl). If [H+] = 1.1 x 10exp-7, then the pH = 7 - log 1.1 = 6.9something. The best answer choice between 6.9 and 7.0 is the winner.

BTW, I used a BR math trick to go from [H+] = 1.1 x 10exp-7 to pH = 7 - log 1.1. The pH = -(power of 10) - log unit #. The derivation can be found in the BR book if you want to look it up, but I'd suggest just taking it at face value.

The point is that on a multiple choice exam where you get no calculator, approximate quickly and as close to the real number as possible.

yeah, I realize that doing the quadratic equation without a calculator would slow me down. I just wanted to understand how they got there so that I could memorize it as something more than just a random equation.

Your way of doping it is obviously quite a bit faster ha. So thanks for the tip, I'll probably end up using this way if I get an actual question on it.

I think I'm all set on this, so thanks to all for the help

 

[Dr Gerrard]

Why would you not take this into consideration when you have concentrations greater than 10^-7? Is it because if that were the case, the hydrogen ion concentration of the acid would be much greater than that of water and thus would make the water's concentration negligible?

 

[Cjc555777]

That's exactly why!