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Offline alextu

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pH calculations?
« on: October 09, 2006, 09:51:37 PM »
I'm having trouble solving pH problems...

I have an example that needs to be worked through, any help would be great, Thank you!
 
The example is as follows:
What is the pH of a 0.030 M pyridine solution that has a Kb= 1.9 X 10-9? The equation for the dissociation of pyridine is
C5H5N(aq) + H20(l) <---> C5H5NH+(aq) + OH-(aq)

Another example I need to be worked through is:
Para-Aminobenzoic acid (PABA), p-H2NC6H4, is used in some sunscreens and hair conditioning products.  Calculate the pH of an aqueous solution with [PABA] = .30 M and Ka=2.2 X 10-5.

Offline enahs

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Re: pH calculations?
« Reply #1 on: October 09, 2006, 11:07:03 PM »
What it help you to point out that Kw = Ka ? Kb, and so Ka = Kw/Kb ?

Do you know what Ka and Kb? They are the acid dissociation constant and base hydrolysis constant.

Ka = [H+] [A-]
             [HA]

Kb = [BH+][OH-]
               

Where HA is the concentration of the un-dissociated acid and B is the concentration of the base. That is what you have in your cases, the concentration of the original solutions (HA and/or B).
You know the HA -> H+ and A-. There for Ka = [X][X]
                                                                           [HA – X]
Simple solving for x, once you know x you know the concentration of the H+ and therefore the pH (once you "-log it) .

Using the one with Kb (base hydrolysis) you get the concentration of the OH-, so use the Kw=Ka*Kb and convert to Ka to  H+ concentration. Or pH + pOH = 14 (you have to be careful with this one though, because very dilute or very concentrated solutions can throw this off)


Note: The calculations will change some if it is a polyprotic acid.
« Last Edit: October 09, 2006, 11:27:33 PM by enahs »

Offline alextu

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Re: pH calculations?
« Reply #2 on: October 10, 2006, 12:38:38 AM »
I figured out the pH for the two I asked about but now I have been working on for a few hours and haven't been able to figure it out, please *delete me*

Calculate the pH of a .300M solution of methylammonium chloride, CH3NH3Cl, The Kb for methylamine, CH3NH2 is 3.7X10-4.

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Re: pH calculations?
« Reply #3 on: October 10, 2006, 02:50:31 AM »
CH3NH3Cl is methylammonium hydrochloride, salt of weak base and strong acid.

http://www.chembuddy.com/?left=pH-calculation&right=toc
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info

Offline alextu

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Re: pH calculations?
« Reply #4 on: October 10, 2006, 12:24:32 PM »
Can anyone calculate the methyammonium chloride problem through??  I keep coming up with an answer of 12.02, which is wrong, any help would be appreciated. Thanks

Offline enahs

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Re: pH calculations?
« Reply #5 on: October 10, 2006, 10:51:12 PM »
Can anyone calculate the methyammonium chloride problem through??  I keep coming up with an answer of 12.02, which is wrong, any help would be appreciated. Thanks

I get 5.54

Convert Kb to Ka, then it is just like the other problems.

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Re: pH calculations?
« Reply #6 on: October 11, 2006, 03:34:40 AM »
Salt of a weak base (methylammonium) contains this base protonated - CH3NH3+. This protonated base is a weak acid. So what you have to do is to calculate pH of the weak acid solution.

http://www.chembuddy.com/?left=pH-calculation-questions&right=pH-weak-base-hydrolysis-q1
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info

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