Question

Part A) A student measured 0.3363 of 99.2% pure sulfamic acid into a beaker and dissolved it in carbon dioxide free water. The solution was titrated with NaOH, where it took 28.20 to reach a pH of 4.5 and 28.31 to reach a pH of 9.2. Calculate the concentration of NaOH at the two pH values.

Part B) A student titrated 25.00 mL of an unknown acid mixture with NaOH that was standardized in the problem above. The unknown mixture required 11.40 mL to reach the first equivalence point at a pH of 4.5, and 36.10 mL to reach the second equivalence point at pH 9.2. Calculate the concentration (in mM) of H3PO4 and H2PO4 in the acid mixture unknown.

Answer 1

in part a) the units are not specified, I will assume that the mass is in grams and the volume in milliliters. The volume was converted to liters in all mathematical operations.

if with 28.20 volume of NaOH the pH was 4.5 and with 28.31 of NaOH the pH was 9.2, this jump in the pH indicates the equivalence point. Then, we can assume that 28.31 of NaOH contains the same moles that sulfamic acid.

the moles of sulfamic acid is calculated:

moles=(mass/molar mass)*(purity/100) = (0.3363g / 97.1 g/mole) (99.2%/100)=0.003436 moles

and we can calculate the concentration of NaOH:

Concentration of NaOH = moles of NaOH / volume of NaOH = 0.003436 moles / 0.0282 L = 0.122 M

b) In the first equivalence point we can calculate the moles added of NaOH:

moles added to the first equivalence point = concentration of NaOH* volume added of NaOH

moles=0.122 M* 0.0114 L = 0.0014 moles

the stoichiometric ratio is 1:1, so this number of moles is equal to H3PO4, we can calculate the concentration of H3PO4:

concentration=moles/volume = 0.0014 moles/0.025 L = 0.0556 M = 55.6 mM

For the second equivalence point we calculate the moles added:

moles = concentration of NaOH*volume added of NaOH = 0.122 M* (0.0361-0.0114 L) = 0.003 moles

these moles are the same that the moles of H2PO4, and we can calculate its concentration:

concentration= moles/volume = 0.003 moles/ 0.025 L = 0.120 M = 120 mM.