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Topic: Acid Base Equilibrium  (Read 2612 times)

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Gravemind

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Acid Base Equilibrium
« on: March 05, 2008, 04:58:33 PM »
I'm having a little trouble understanding Acid Base Equilibrium, its probably from the fact that I didn't fully understand regular Equilibrium.  Ive asked my teacher and some fellow students to try to help me understand but the way they explain it just doesn't work for me.  Heres an example of the stuff I'm just not understanding:

Calculate the pH change produced by adding .1moles of solid NaOH to 500 ml of 1.00M HC2H3O2 and 1.00M NaC2H3O2

HC2H3O2->H+ + C2H3O2-

So we have 1mol/.5L=2
Ka = 1.8x10-5

1.8x10-5=[H+][C2H3O2-]/[HC2H3O2]

1.8x10-5=x(2+.1)/(2-.1)
x= 1.62x10-5 which makes the pH 4.78

Now i just flip the signs for the NaC2H3O2 so its:
1.8x10-5=x(2-.1)/(2+.1)
x=1.98x10-5 which makes the pH 4.70

But when I looked at the answer it says I should've got a pH of 4.75 and 4.92 to give me a difference of .18, but I only got a difference of .08


Offline enahs

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Re: Acid Base Equilibrium
« Reply #1 on: March 05, 2008, 09:13:38 PM »
You have a weak acid and it's conjugate base.
What does that mean?
What is it's initial pH?

You then have a strong base reaction with a weak acid; what happens?


Also note, always think about what is happening; do not get so focused on the numbers you miss the science. You are saying if you add a base the pH is going to go down? Does that make sense?
« Last Edit: March 05, 2008, 09:26:14 PM by enahs »

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