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Chemistry

40.0 ml of an acetic acid of unknown concentration is titrated with 0.100 M NaOH. After 20.0 mL of the base solution has been added, the pH in the titration flask is 5.10. What was the concentration of the original acetic acid solution? (Ka(CH3COOH) = 1.8 × 10−5)

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  1. let X = M CH3COOH.
    (CH3COOH) = (40mL*XM/60 mL)
    (NaOH) added = (20 mL x 0.1M/60mL)

    ........CH3COOH + NaOH ==> CH3COO^- H2O
    I.......(40X/60)...0........0..........0
    add...............2/60...............
    change...-2/60... -2/60.....2/60.....2/60
    equilib (40X-2/60)..0........2/60....2/60
    Substitute into the Henderson-Hasselbalch equation and solve for X.
    5.10 = 4.74 + log(2/(40x-2) and solve for X

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  2. dkls,45

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