Asked by Trista Nigh
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?
Answers
Answered by
DrBob222
..............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.
..............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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