Experiment 9 Post-lab Questions 1) A student completes a titration of a sample o
ID: 546289 • Letter: E
Question
Experiment 9 Post-lab Questions 1) A student completes a titration of a sample of KHP using 20.23 mL of 0.1101 M NaOH. If the KHP sample was diluted with 28.21 mL of water, how many grams of KHP were in the solution? 2) If a 0.2423 g sample of malic acid is diluted to a volume of 17.52 mL of water using 21.04 mL of 0.1733 M KOH as per the reaction below, what is the experimental molar mass of the malic acid? H2A represents malic acid, a diprotic acid. H2A (aq) + 2 KOH (aq) KA (aq) + 2 H2O (l) What mass of citric acid would you need to make 75 mL of a 0.25 M solution? What is the solute and solvent in this solution? In terms of "like dissolves like" why does the solute dissolve? Would the solute dissolve in hexane? Explain. )Explanation / Answer
Q1
KHP + NaOH = H2O + NaKP
mol of NaOH = MV = (0.1101)(20.23*10^-3) = 0.002227323 mol of base
mol of KHP = 1:1 ratio implies 0.002227323mol of acid
mass of KHP = mol*MW = 0.002227323*204.22 = 0.4548 g of KHP
Q2.
molar mass of acid:
MM = mass/mol
mol of acid = 1/2*mol of base
mol of base = MV = (0.1733)(21.04*10^-3 ) = 0.003646 mol of base
mol of acid = 1/2*0.003646 = 0.001823 mol of acid
MW = mass/mol = 0.2423/0.001823 = 132.91 g/mol
Q3
M = mol/V
mol = MV = 0.25*0.075 = 0.01875*192.1235 = 3.60 g of citric acid
the solut edissolves because it is polar, and water is polar...
it will not dissolve in hexane, since it is nonpolar