Determination of a partition coefficient Mass of mandelic acid: 6.01g [NaOH]: 0.

Question

Determination of a partition coefficient

Mass of mandelic acid: 6.01g

[NaOH]: 0.1036

Table 1: titration of a 10.0 ml aliquot of mandelic acid solution with standard NaOH

average of three titrations: 38.31ml

table 2: titration of the 5.0 ml sample of extracted water

Part A:

a) calculate the moles of sodium hydroxide used to tirate the 10.00 ml aliquot of mandelic acid.

b) using the mole ration between sodium hydroxie and the mandelic acid, calculate the moles of mandelic acid in the 10.0 ml aliquot

c) given the molar mass of mandelic acid is 152.15g/mol determine the mass of the mandelic acid in the 10.0 ml aliquot.

d) determine the concentration of mandelic acid in g/ml

burette reading trial 1 trial 2 trial 3 final reading 38.1ml 41.40ml 41.02ml initial reading 0.0 2.80 2.78 volume NaOH used 38.1ml 38.60ml 38.24ml

Explanation / Answer

Mandelic acid is a monoprotic weak acid and is denoted as HA.

a) Moles of sodium hydroxide, NaOH required to neutralize mandelic acid = (average volume of NaOH in L)*(concentration of NaOH in molarity) = (38.31 mL)*(1 L/1000 mL)*(0.1036 M) = 3.968916*10-3 mole ? 3.969*10-3 mole (ans, correct to 4 sig. figs).

b) Write down the balanced chemical equation for the reaction between mandelic acid, HA and NaOH as

HA (aq) + NaOH (aq) --------> NaA (aq) + H2O (l)

As per the stoichiometric equation,

1 mole HA = 1 mole NaOH.

Mole(s) of HA neutralized by 3.969*10-3 mole NaOH = (3.969*10-3 mole NaOH)*(1 mole HA/1 mole NaOH) = 3.969*10-3 mole (ans).

c) We have 10.0 mL aliquot of mandelic acid. The mass of mandelic acid present in 10.0 mL aliquot of mandelic acid = (3.969*10-3 mole)*(152.15 g/mol) = 0.60388 g ? 0.604 g (ans).

d) Concentration of mandelic acid in 10.0 mL aliquot = (mass of mandelic acid)/(volume of aliquot) = (0.604 g)/(10.0 mL) = 0.0604 g/mL (ans).

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