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a 5.55g sample of a weak acid with ka=1.3*10^-4 was combined with 5.00ml of 6.00 M NAOH and the resulting solution was diluted to 750mL. The measured pH of the solution was 4.25. what is the molor mass of the weak acid.
if used the formula
i don't know how to find the molar mass

  • chemistry -

    oops..i posted this question twice

  • chemistry -

    I assume the unknown acid is a monoprotic one.
    ..HA...+...NaOH ==> NaA....+ ... H2O
    5.55 g..0.005*6.00..0.03 moles..0.03 mols.

    0.03 moles/0.750 L = 0.04 M
    Ka = 1.3E-4 = (H^+)(A^-)/(HA)
    1.3E-4 = 5.62E-5)(0.04)/(X)
    Solve for X = molarity HA.
    Then M x L = M x 0.750 = moles HA
    moles HA = g/molar mass.
    You have moles HA and g HA, solve for molar mass HA.
    Post your work if you get stuck.

    There is nothing wrong with the equation you used but the log value skews the answer a little. I can go through that with you if you wish but post your work if you want that done so I can use your values.
    I obtained an answer for molar mass of about 430 (that isn't an exact number).

  • chemistry -

    mol HA + NaOH --> NaA + H2O
    i x 0.030
    f x-0.030 0 0.030

    pH = pKa + log(0.030/x-0.030)

    x = 0.042 mol
    PM=5.5g/0.042mol = 129.1 g/mol

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