College Chemistry 100

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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]

Thanks.

  • College Chemistry 100 -

    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

    [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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