.5 g unknown acid in water in 250 ml beaker 2.4 initial pH Standard 0.10 M NaOh
ID: 558644 • Letter: #
Question
.5 g unknown acid in water in 250 ml beaker 2.4 initial pH Standard 0.10 M NaOh solution Possible unknowns: benzoic acid, sulfanilic acid, or potassium hydrogen phthalate Titrahn 3.75 4. 5. 6. Determine the volume of NaOH needed to reach the midpoint. The pH at this point should equal the pKa for your unknown acid. What is the pKa of the unknown acid From the titration data, calculate the moles of acid. Remember, the unknown acid is monoprotic. Calculate the molar mass of the unknown acid. Identify the unknown acid. 7.Explanation / Answer
0.5 gm unknown acid = mass / molar mass = 0.5 / M mole.
at equivalence point , volume of NaOH nedded = 108 * 2 = 216 ml.
216 ml of 0.1 M NaOH = 0.216 * 0.10 = 0.0216 mole.
therefore,
0.0216 = 0.50 / M
M = 23.15 gm /mole.
molar mass of unknown acid is 23.15 gm / mole.
to confirm the unknown acid, please provide the volume of NaOH at equivalence point.
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