Topic: pH of Buffer (Read 10255 times)

Crashley2154 · « **on:** April 06, 2005, 11:09:12 PM »

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 · « **Reply #1 on:** April 07, 2005, 02:19:23 AM »

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 · « **Reply #2 on:** April 07, 2005, 04:20:42 AM »

Quote from: Crashley2154 on April 06, 2005, 11:09:12 PM

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

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

Quote

.5/82.03379=0.006095

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

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

Topic: pH of Buffer (Read 10255 times)

Crashley2154

pH of Buffer

Garneck

Re:pH of Buffer

Borek

Re:pH of Buffer

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