Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: objectivist on April 03, 2005, 04:19:43 PM
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Hello. Help would be much appreciated!
What is the pH of a solution if 100.0 mL of the solution is 0.21 M with respect to acetic acid and contains 1.59 g of sodium acetate, NaCH3CO2?
i found out the Molarity of sodium acetate to be 0.1939 but from there I am lost. i tried adding the -log of the two concentrations to get the pH but that is wrong.
Can anyone help?
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Do I do an ICE Table?: the
Initial
Change
Equilibrium
thing?
and use "Ka" or something?
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Okay...i think i was looking at the wrong chapter and may have figured it out.
but i am not sure if it is right:
CH3CO2H + H20 -- H30+ + CH3CO2-
I 0.21 0 0.1939
C -x +x +x
E 0.21 - x x 0.1939 + x
Ka = (x)(0.1939 + X)
0.21 + x
but then cuz "x" is small u take it out....
The Ka of acetic acid is (1.8 e-5) (thats the one i use, right?)
and solve for "x" and get 1.72e-6
-log(1.722e-6) = a pH of 5.76 ?!???
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Well...i just found out that that answer is wrong.
Can anyone tell me what I did wrong?
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Use the Henderson-Hasselbalch Equation...
pH = pK_a - log([base]/[acid])
The base in this case is CH3CO2- and the acid is HCH3CO2. The pK_a is the same as -log(1.8e-5).
So your final answer is
pH = -log(1.8e-5) - log(.1939/.21)
Tah dah!
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:D Thank you so much PkoThari13!!!
I dont know what the hell i was doing...