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pH of Buffer

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Crashley2154:
A buffer solution was prepared by mixing 50.00 mL of a 0.100 molar solution of CH3COOH with 0.500 grams of NaCH3COO. The resultion mixture is Diluted to 100.00 mL. What is the pH of the solution?

pKa of acetic acid is 1.75 x 10^-5
CH3COOH FM=60.05196
NaCH3COO FM=82.03379

(.05L x 0.1M)/60.05196=8.326 x 10^-5

.5/82.03379=0.006095
1.7(1.75 x 10^-5)+log((8.32612 X 10^-5)/0.006095)=-1.8645

This answer can not be correct because pH can not be negitive.

Garneck:
You got it all wrong:

Acetic acid:
C=0,1 M
V=0,05 L

n= 0,1*0,05 = 0,005 moles

dilluting to 0,1 L

C[CH3COOH] = 0,005/0,1 = 0,05 M

Sodium acetate:
m = 0,5 g
M[CH3COONa] = 82 g/mole
n= 0,5/82 = 0,0061 moles

C[CH3COONa] = 0,0061/0,1 = 0,062

pH=pKa + log([CH3COONa]/[CH3COOH]) = - log (1,75*10^-5) + log (0,062/0,050) = 4,757 + 0,093 = 4,85

Borek:

--- Quote from: Crashley2154 on April 06, 2005, 11:09:12 PM ---(.05L x 0.1M)/60.05196=8.326 x 10^-5
--- End quote ---

You are dividing number of moles by molar mass, I have no idea why.


--- Quote ---.5/82.03379=0.006095
--- End quote ---

This is number of moles of CH3COONa - but it is not concentration yet, so you can't put it into HH equation.


--- Quote ---1.7(1.75 x 10^-5)+log((8.32612 X 10^-5)/0.006095)=-1.8645
--- End quote ---

And that's total mess - HH equation requires pKa in the first postion, not Ka. As for the rest see my comments above.

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