Posted by Marie on Thursday, November 11, 2010 at 10:40am.
The topic is about acids, bases, and salts. The two equations are given: H20 + HCL -> H30 + CL; H20 + HC2H3O2 <-> H30 + C2H3O2
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]
- College Chemistry 100 - Dr Russ, Thursday, November 11, 2010 at 10:54am
I 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
[C2H3O2-] = 0.48 M
[HC2H3O2]= 0.45 M
[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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