Chemistry

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A solution is made by mixing 500 ml of 0.172 M NaOh with exactly 500 ml of 0.100 M CH3COOH. Calculate the equilibrium concentration of the species below.
Ka of CH3COOH is 1.8 x10^-5

[H+]=?
{OH-]=?
[CH3COOH]=?
[Na+]=?
[CH3COO-]=?

  • Chemistry -

    mols NaOH = M x L = approx 0.0860
    molw CH3COOH = approx 0.05 = HAc

    ........NaOH + HAc ==> NaAc + H2O
    I....0.0860.....0.......0.......0
    Add...........0.050..............
    C.....-0.050.-0.050....0.050...0.050
    E......0.036....0......0.050

    So you have 0.0360 mols NaOH
    0 mols HAc
    0.050 mols NaAc = CH3COONa
    all in 500 + 500 = 1000 mL H2O(assuming the volumes are additive).

    M = mols/L each except OH^-
    You find that from the hydrolysis of the Ac^- which is done as follows:
    (Ac^-) = 0.05mols/L = 0.05
    .........Ac^- + HOH ==> HAc + OH^-
    I........0.05.............0....0
    C.........-x..............x.....x
    E.......0.05-x............x.....x

    Kb for Ac^- = (Kw/Ka for HAc) = (x)(x)/(0.05-x) and solve for x = (OH^-). From that you can obtain H^+.
    Post your work if you get stuck.

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