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September 20, 2014

September 20, 2014

Posted by **Marie** on Thursday, November 11, 2010 at 10:40am.

The question is: A solution with a volume of one liter contains 0.45 mole HC2H3O2 and 0.48 mole C2H3O2 ion. (Ka for HC2H302 is 1.82 x 10-5) What is the [H+] of the solution and the pH of this solution?

I am just not sure where to start. The formula is pH = -log[H+]; the expression for Ka and Kb would be Ka = [H+][C2H3O2-]/[HC2H3O2]; Kb = [NH4+][OH-]/[NH3]

Thanks.

- College Chemistry 100 -
**Dr Russ**, Thursday, November 11, 2010 at 10:54amI am not sure where the Kb comes from, however, your expressions for Ka and pH are correct.

I assume the concentrations are the starting concentrations.

So at the start

[H+]=0

[C2H3O2-] = 0.48 M

[HC2H3O2]= 0.45 M

at equilibrium

[H+]= x (we are tying to find this)

[C2H3O2-] = 0.48 M + x

[HC2H3O2]= 0.45 M -x

(each mole of HC2H3O2 that dissociates give one mole of H+ and one mole of the ion)

So from the Ka expression

Ka = [x][0.48+x]/[0.45-x] = 1.82x10^-5 mole l^-1

if we assume that x is small wrt 0.48 and 0.45 then

[x][0.48]/[0.45] = 1.82x10^-5 mole l^-1

so x=1.7x10^-5 mole l^-1

alternatively you can solve the quadratic.

so pH=-log(1.7x10^-5) = 4.8

but please check the maths!

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