[MitchTwitchita]

Hey guys, I was wondering if anyone could help me out with a few problems?

1. A sample contains an unknown amount of citric acid, H3C6H5O7.  If 0.6667 g of the sample requires 35.09 mL of 0.1000 M NaOH to neutralize the H3C6H5O7 completely, what is the percentage of H3C6H5O7 in the sample?  The molar mass of H3C6H5O7 is 192.13 g/mol.  The balanced equation for the reaction is:

H3C6H5O7 + 3NaOH > 3H2O + Na3C6H5O7

M=n/V
Therefore, n=MV
=0.1000 x 0.03509
=0.003509

0.003509 mol NaOH x 1 mol H3C6H5O7/3 mol NaOH x 192.13 g NaOH/1 mol NaOH
=0.2247 g

Therefore, 0.2247 g/0.6667g x 100% = 33.71%


2.  Sodium thiosulfate, Na2S2O3, is an important reagent for titrations.  Its solutions can be standardized by titrating the iodine released when a weighed amount of potassium hydrogen iodate, KH(IO3)2 (389.912 g/mol), is allowed to react with excess potassium iodide in acidic solutions.  The net ionic equations are:

production of iodine from KH(IO3)2: IO3- + 5I- + 6H+ > 3I2 + 3H2O
titration of iodine:  I2 + 2S2O3(2-) > 2I- + S4O6(2-)

What is the molarity of a sodium thiosulfate solution if 25.30 mL are required to titrate the iodine released from 0.1732 g of KH(IO3)2?

0.1732 g KH(IO3)2 x 1mol KH(IO3)2/389.912 g KH(IO3)2 = 4.442 x 10^-4 mol
4.442 x 10^-4 mol KH(IO3)2 x 3 mol I2/1 mol KH(IO3)2 x 2 mol Na2S2O3/1 mol I2
=2.665 x 10^-3 mol Na2S2O3
Therefore, M=n/V
=2.665 X 10^-3 mol/0.02530 L
=0.1053 mol/L

3.  You have 0.7790 g of an unknown monoprotic acid, HA, which reacts with NaOH according to the balanced equation:

HA + NaOH > NaA + H2O

If 45.89 mL of 0.1039 M NaOH is required to titrate the acid to the equivalence point, what is the molar mass of the acid?


n+MV
=0.1039 mol/L x 0.4589 L
=0.004768 mol naOH

4.768 x 10^-3 mol NaOH x 1 mol HA/1 mol NaOH x 0.7790 g HA/1 mol HA
=0.003714 g HA

It would be greatly appreciated if anyone can tell me if and where I'm going wrong.  Thanks a whole lot.

[RBF]

Just for part 1 of your questions -  You have the following as part of your calculation:

0.003509 mol NaOH x 1 mol H3C6H5O7/3 mol NaOH x 192.13 g NaOH/1 mol NaOH
=0.2247 g

Check your dimensional analysis.  Your units for the last factor should be for citric acid.  I believe your numbers are correct, though, so it may have been a typo rather than a conceptual disconnect.