# Chemical Forums

## Chemistry Forums for Students => High School Chemistry Forum => Topic started by: Shea on August 22, 2006, 06:50:27 PM

Title: pH?
Post by: Shea 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?
Title: Re: pH?
Post by: Shea 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...
Title: Re: pH?
Post by: Borek on August 23, 2006, 02:56:13 AM
And what Exactly does log mean in this?

http://en.wikipedia.org/wiki/Logarithm
Title: Re: pH?
Post by: Borek 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.
Title: Re: pH?
Post by: Shea on August 23, 2006, 06:10:08 PM
Thanks for explaining that, but what do you think the questions mean by, "write an expression?"
Title: Re: pH?
Post by: Borek on August 23, 2006, 06:19:12 PM

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.
Title: Re: pH?
Post by: Shea 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...
Title: Re: pH?
Post by: Will 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 (http://en.wikipedia.org/wiki/Equilibrium_constant) 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 (http://en.wikipedia.org/wiki/Partial_pressure#Equilibrium_constants_of_reactions_involving_gas_mixtures). 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.
Title: Re: pH?
Post by: Shea 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?
Title: Re: pH?
Post by: Borek 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+ ;)
Title: Re: pH?
Post by: Shea on August 24, 2006, 05:34:20 PM
H(C4H7O2) <-> C4H7O2- + H+

Is it, "Ka = [H+][C4H7O2-] / [H(C4H7O2)]."
Title: Re: pH?
Post by: Borek on August 24, 2006, 05:39:37 PM
OK, although formatting looks lousy ;)
Title: Re: pH?
Post by: Shea 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?
Title: Re: pH?
Post by: Will 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!)
Title: Re: pH?
Post by: Shea 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.
Title: Re: pH?
Post by: Borek on September 01, 2006, 05:30:43 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.

You are calculating it right, although you may be using too much significant figures in your answer.
Title: Re: pH?
Post by: Shea on September 01, 2006, 06:03:30 PM
Really?  I told my teacher that I used a calculator, and did it the same way as they show in my book, but she said, "I assure you, it's incorrect."

Should I just say 5.2?
Title: Re: pH?
Post by: Borek on September 01, 2006, 06:19:07 PM
Really?  I told my teacher that I used a calculator, and did it the same way as they show in my book, but she said, "I assure you, it's incorrect."

Should I just say 5.2?

Or even 5. As long as you have not missed something. What is exact wording of the question?
Title: Re: pH?
Post by: Shea on September 01, 2006, 06:23:40 PM
That is the exact wording.
Title: Re: pH?
Post by: english on September 01, 2006, 08:22:43 PM
Don't forget that pH is a measure of the acidity/basicity of a solution.

pKa/pKb is a measure of the acidity/basicity of a compound.
Title: Re: pH?
Post by: Shea on September 06, 2006, 04:30:40 PM
How would I find the pH of a 0.6M solution of NH3?

I am utterly perplexed by this question.  :-\
Title: Re: pH?
Post by: Borek on September 06, 2006, 04:43:58 PM
This is weak base.

http://www.chembuddy.com/?left=pH-calculation&right=pH-weak-acid-base

Title: Re: pH?
Post by: Donaldson Tan on September 13, 2006, 09:29:50 AM
How would I find the pH of a 0.6M solution of NH3?

pH = 14 - pOH
pOH = - lg [ OH- ]

NH3 (aq) + H2O (l) <-> NH4+ (aq) + OH- (aq)

Kb = [ NH4+ ] [ OH- ] / [ NH3 ]

let concentration of OH- be x (unit M)
Kb = x2/(0.6 - x)
This is a quadratic expression. Solve for x.

pOH = - lg x
pH = 14 - pOH = 14 + lg x