Physical chemistry 2

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50 ml of 0.02M acetic acid is titrated with 0.1M NaOH.Calculate the pH of the solution when 10ml of NaOH is added

  • Physical chemistry 2 -

    50 mL x 0.02M = 1 mmol HAc.
    10 mL x 0.1M = 1 mmol NaOH.

    HAc + NaOH ==> H2O + NaAc
    1 mmol 1mmol = e.p.

    So you have a solution of Ac^- in aqueous solution. The (Ac^-) is 1 mmol/60 mL = about 0.017M but you nee to do it more accurately.
    ........Ac^- + HOH ==> HAc + OH^-
    I........0.017...........0.....0
    C........-x.............x......x
    E......0.017-x..........x......x

    Kb for Ac^- = (Kw/Ka for HAc) = (HAc)(OH^-)/(Ac^-) and solve for x = OH^-, then convert to pH. I expect you will need to use the quadratic.

  • Physical chemistry 2 -

    0.0012

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