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Topic: pH?  (Read 14353 times)

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Offline Shea

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pH?
« on: August 22, 2006, 06:50:27 PM »
I need help with this question.  It is not from an assignment.  It is in my book, and it tries to explain how to do it, but I don't understand...

Determine the pH of a solution with a hydrogen concentration of 3.5 x 10^-4

This is what it says.

pH = -log [3.5 x 10^-4]
= -log 3.5 - log 10^-4
=  - .54 - (-4)
=  - .54 + 4
=  3.46

Where does the .54 come from?

And what Exactly does log mean in this?

Offline Shea

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Re: pH?
« Reply #1 on: August 22, 2006, 09:07:58 PM »
Never mind, I got it now.  I'm even more lost though on this.

Write an expression for Kp in the following:

2NO2 -> N2O4

and

Write an expression for the ionic product of water Kw

and

Write an expression for Kc for the dissociation reaction of butanoic acid

H(C4H7O2) <-> C4H7O2- + H+

I don't get the Kp, Kc, and Kw.  The book doesn't explain this, even though it says on the cover that it makes chemistry easy to learn with full explanations...

Offline Borek

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Re: pH?
« Reply #2 on: August 23, 2006, 02:56:13 AM »
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Offline Borek

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Re: pH?
« Reply #3 on: August 23, 2006, 02:59:22 AM »
I don't get the Kp, Kc, and Kw

http://en.wikipedia.org/wiki/Reaction_quotient

Kp uses pressures, Kc and Kw use concentrations. Kw is just Kc for water dissociation, although it is usually assumed that water concentration is constant.
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Offline Shea

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Re: pH?
« Reply #4 on: August 23, 2006, 06:10:08 PM »
Thanks for explaining that, but what do you think the questions mean by, "write an expression?"

Offline Borek

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Re: pH?
« Reply #5 on: August 23, 2006, 06:19:12 PM »
Look at this page:

http://www.chembuddy.com/?left=pH-calculation&right=introduction-acid-base-equilibrium#eq1.1

Equation 1.1 is the expression for the acid dissociation constant.
« Last Edit: August 24, 2006, 02:28:04 AM by Borek »
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Offline Shea

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Re: pH?
« Reply #6 on: August 23, 2006, 07:32:01 PM »
Um, that page wont open.  When I click on it, it goes to a site that offers some highly questionable search topics...

Offline Will

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Re: pH?
« Reply #7 on: August 23, 2006, 07:56:52 PM »
Thanks for explaining that, but what do you think the questions mean by, "write an expression?"

This wikipedia article should help explain the equilibrium constant.

The following is what you need to remember:

and

When they tell you to write an expression they want you to fill in the above expression (ignore kf and kb).

For Kp see this. The expression is similar, just only include gases.

As for Borek's site, just put the "." inbetween chembuddy and com:
http://www.chembuddy.com/?left=pH-calculation&right=introduction-acid-base-equilibrium#eq1.1
That'll explain Kw.

Offline Shea

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Re: pH?
« Reply #8 on: August 23, 2006, 08:10:15 PM »
Given that [H+]=[C4H7O2-] = 3.8 x 10-3 M and [HC4H7O2] = 1 M, calculate Ka for this equilibrium.

How do you calculate Ka?

Borek's site said that for "HA <-> H+ + A-," Ka = [H+][A-] / [HA]

How do I get Ka from what's given in the question above?

Offline Borek

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Re: pH?
« Reply #9 on: August 24, 2006, 02:27:40 AM »
Sorry for the typo in url, I'll correct it in just a moment.

You have everything given - you must identify now what is HA and A- in terms of the substance present in your solution. That's if you have already identified H+ ;)
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Offline Shea

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Re: pH?
« Reply #10 on: August 24, 2006, 05:34:20 PM »
H(C4H7O2) <-> C4H7O2- + H+

Is it, "Ka = [H+][C4H7O2-] / [H(C4H7O2)]."

Offline Borek

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Re: pH?
« Reply #11 on: August 24, 2006, 05:39:37 PM »
OK, although formatting looks lousy ;)
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Offline Shea

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Re: pH?
« Reply #12 on: August 24, 2006, 05:46:05 PM »
So I would plug the numbers into that?

Ka = [3.8 x 10-3 M][3.8 x 10-3 M] / [1]

That doesn't seem right... Is that it?

Offline Will

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Re: pH?
« Reply #13 on: August 24, 2006, 05:48:02 PM »
Is that it?

Yes :)
(although I think the square brackets mean "concn of", so you can just stick with normal brackets- ( & ) when doing the math!)

Offline Shea

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Re: pH?
« Reply #14 on: September 01, 2006, 05:25:11 PM »
What is the pH of a solution with [H+] = 6 x 10-6?

Can someone just please tell me what the pH of this is???

I think its 5.22, but I'm wrong.

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