Acid-base question H-H equation

[moto_za]

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I am having some trouble figuring out part two of this question. Can someone please help?

For the equilbrium, CH3COOH <---> CH3COO- + H+, the pKa is 4.7.

1. What is the ratio of [CH3COO- ]/[ CH3COOH] at pH = 6.7?

100/1

2. If the concentration of [CH3COOH] + [CH3COO-]= 3.50 mM, what is the concentration of [CH3COO-] and the concentration of [CH3COOH] when the pH = 6.7?

 

[loveoforganic]

You have the ratio.

(100/101)*(3.5 mM)
(1/101)*(3.5 mM)

I'm moderately sure that's right. If not, I'm sure someone will correct me!

 

[PreMedDude]

I agree. Since you have the ratio of acetate to acetic acid, and the molar concentration. All you have to do is multiply the whole molar concentration by the ratio of each species. Always remember that it works like a percent....that is....multiply a percent by the total and you get the amounts of each molecule. Cool!

-JP

 

[Beta Cell]

2. If the concentration of [CH3COOH] + [CH3COO-]= 3.50 mM, what is the concentration of [CH3COO-] and the concentration of [CH3COOH] when the pH = 6.7?

6.7= 4.7 + log([A-]/[HA])

2= log([A-]/[HA])
[A-]/[HA]= 100
[HA]=[CH3COOH]; [A-]=[CH3COO-]

[HA] + 100[HA]= 3.50 mM

3.50 mM/101= [HA]=~ .035M =[HA] and 3.465M=[A-]