Chemistry

A 5.55g sample of a weak acid with Ka=1.3 x10^-4 was combined with 5.00 mL of 6.00 M NaOH and the resulting solution was diluted to 750 mL. The measured pH of the solution was 4.25. What is the molar mass of the weak acid?

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  1. ..............HA + NaOH ==> NaA + H2O
    ..............x.....0........0.....0
    add................30 mmols..........
    change.......-30...-30.....+30.....+30
    equil........x-30....0......+30....+30

    pH = pKa + log (base)/(acid)
    4.25 = 3.886 + log(30/(A)
    Solve for A. I found 12.98; therefore, the value for x, the initial HA, is 12.98 + 30 = 42.98 mmoles or 0.04298 moles.
    n = grams/molar mass. Solve for molar mass. You know g and n.

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