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limp32 limp32
wrote...
11 years ago
You have a solution of 25.0 mL 1.0M sodium acetate + 10.0 mL 1.0 M acetic acid.  The observed pH of the solution is 5.25, which means the [H+]=5.62 x 10-6. I'm not sure how to complete this problem involving two chemicals in a solution.
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#1
11 years ago
The equilibrium is HAc Leftwards Arrow> H+ + Ac-
The moles and concentration of acetate is: 0.025 L X 1 mol/L = 0.025 mol / 0.035L = 0.714 mol/L

THe moles and concentration of acetic acid (HAc) is: 0.010 L X 1 mol/L / 0.035 L = 0.285 mol/L

Since you know the [H+], you can calculate the ionization constant:

Ka = [H+][Ac-]/[HAc] = 5.62X10^-6 X 0.714 / 0.285 = 1.4 X 10^-5
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