The molar absorptivities of an indicator HIn (K_a = 2.0 times 10^-5) and its con
ID: 526485 • Letter: T
Question
The molar absorptivities of an indicator HIn (K_a = 2.0 times 10^-5) and its conjugate base In^- at 540 nm were determined as epsilon_HIn = 250 and epsilon_In^- = 1250, respectively. Calculate the absorbance (b = 1.00 cm) for the buffered solution (pH = 5.00) having total indicator concentration of 1.00 times 10^-5 M. The KCIO_3 in a 0.1279 g sample of an explosive was determined by reaction with 50.00 mL of 0.08930 M Fe^2+: ClO^-_3 + 6Fe^2+ + 6H^+ rightarrow 6Cl^- + 3H_2O + 6Fe^3+ When the reaction was complete, the excess Fe^2+ was back-titrated with 14.93 mL of 0.08361 M Ce^4+. Calculate the percentage of KClO_3 (Mw: 122.6 g/mole) in the sample.Explanation / Answer
Answer to Q5)
ClO3- + 6Fe2++ 6H+ ---> Cl- + 6Fe3++ 3H2O
Fe2++ Ce4+ ------> Fe3++ Ce3+
Total number of moles of Fe2+= concentration of Fe2+XVolume of Fe2+ in L
= 0.08930 mol L-1 X 50 X 10-3 L= 4.465 X 10-3 mol
Number of moles of Ce4+= concentration of Ce4+XVolume of Ce4+ in L
= 0.08361 mol L-1 X 14.93 X 10-3 L= 1.248 X 10-3 mol
From equation 2, Excess number of moles of Fe2+= Number of moles of Ce4+
= 1.248 X 10-3 mol
Number of moles of Fe2+ reacted with KClO3 = Total number of moles of Fe2+ - Excess number of moles of Fe2+
=4.465 X 10-3 mol -1.248 X 10-3 mol
= 3.2167 X 10-3 mol
From equation 1, we know that 1 mol of KClO3 consumes 6 mol of Fe2+
Number of moles of KClO3 = Number of moles of Fe2+ reacted with KClO3 /6
= 3.2167 X 10-3 mol /6 = 5.3612 X 10-4 mol
Mass of KClO3 = Number of moles of KClO3 X Molar mass of KClO3
= 5.3612 X 10-4 mol X 122.6 g mol-1
= 0.0657 g
% KClO3 = Mass of KClO3 X 100 / Total mass of sample
= 0.0657 g X 100 / 0.1279 g = 51.4 %
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