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Jules18

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Equilibrium question
« on: December 09, 2009, 05:18:16 PM »

Hey guys, I think I've done the better half of this question, but I'm stuck at the end:

Calculate the concentration of the acetate ion (M) in a solution prepared by dissolving 1.00×10-3 mol of HCl(g) in 1.00 L of 1.90 M aqueous acetic acid. Ka = 1.7E-5

.   CH3COOH --> CH3COO- + H+
initial:        1.90 M                             0                                         0
change:       - x                                +x                                       +x
equlbm:      ~1.9                                x                                        x

blah blah blah, calculations ... x = 5.68E-3 .   So that would be the acetate concentration before you add the 1E-3 mol H+.  

And at that point I'm not sure what to do.
« Last Edit: December 09, 2009, 05:38:15 PM by Jules18 »
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savy2020

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Re: Equilibrium question
« Reply #1 on: December 09, 2009, 07:26:23 PM »

.   CH3COOH --> CH3COO- + H+
initial:           1.90 M                             0                                         0
change:          - x                                +x                                       +x
equilibrium:      ~1.9                                x                                        x
Well before adding HCl is as above. But after that It would be...
.   CH3COOH <--> CH3COO- + H+
initial:           1.90 M                             0                                         0
change:          - x                                +x                                       +x
equilibrium:      ~1.9                                x                               10-3+x  ~10-3
You know why?
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AWK

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Re: Equilibrium question
« Reply #2 on: December 09, 2009, 07:56:30 PM »

.   CH3COOH --> CH3COO- + H+
initial:           1.90 M                             0                                         0
change:          - x                                +x                                       +x
equilibrium:      ~1.9                                x                                        x
Well before adding HCl is as above. But after that It would be...
.   CH3COOH <--> CH3COO- + H+
initial:           1.90 M                             0                                         0
change:          - x                                +x                                       +x
equilibrium:      ~1.9                                x                               10-3+x  ~10-3
You know why?

Should be
equilibrium:     1.9-x ~1.9                           x                               10-3+x  ~10-3
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