Posted by johnny on Sunday, June 8, 2008 at 9:47pm.
A generic equation can be written as follows:
RCOOH + NaOH ==> H2O + RCOONa
mols NaOH = L x M
mols RCOOH = mols NaOH (since the acid is monoprotic).
mols RCOOH = grams/molar mass.
You have mols and grams. Solve for molar mass.
The pH of 4.92 at 17.54 mL NaOH is extraneous information unless you are asked to calculate Ka for the acid.
Yeah asking for Ka was the second part of the question, but it wasn't one of the review questions
moles OH- = (liters)(molarity)
(0.115 mol NaOH/L)(0.03383L) = 3.89x10-3 mol OH- used to reach the end point
Moles acid = moles OH-
moles of acid neutralized = 3.89x10-3 mol
0.288 g acid/0.03383L = 8.513g/L
(8.513g/L)/(0.115 mol/L) = ________g/mole
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