I am very stuck on how to figure out these with the data I have. I need each fro
ID: 536458 • Letter: I
Question
I am very stuck on how to figure out these with the data I have.I need each from Trial 1-3
I need the 1) Moles of Absorbic Acid, 2) Mass of Absorbic Acid, 3) % Difference
PLEASE SHOW CALCULATIONS I am very confused. Determination of Ascorbic Acid Molarity of the potassium iodate (KIO) solution 0.01 00 Mass of Ascorbic Trial I Trial 2 Trial 3 log 19 Initial Buret reading L 19. ml 0.00 ml Final Buret reading Volume of KIO3 titrant added 30, mL Calculations Calculate the mass of ascorbic acid in each sample determined by the titration and compare with your mass for each sample. Trial Moles of of Ascorbic Mass of Ascorbic %difference Mass ascorbic acid acid Acid (as weighed (determined by (determined by out) titration) titration 10 Show calculations:
Explanation / Answer
The balanced chemical reactions for the titration are
KIO3 + 5 KI + 6 H+ -------> 3 I2 + 6 K+ + 3 H2O ……(1)
C6H8O6 + I2 -------> C6H6O6 + 2 I- + 2 H+ ……(2)
As per the stoichiometric equation (1),
1 mole KIO3 = 3 moles I2
1 mole I2 = 1 mole C6H8O6
Therefore, 1 mole KIO3 = 3 moles C6H8O6
We shall fill up the tables as below:
Data
Trial 1
Trial 2
Trial 3
Mass of ascorbic acid (g)
0.10
0.11
0.10
Initial Buret Volume (mL)
0.00
19.31
0.00
Final Buret Volume (mL)
19.31
49.38
20.91
Volume of KIO3 titrant added (mL) = (final buret volume) – (initial buret volume)
19.31
30.07
20.91
Volume of KIO3 titrant added (L) = (volume of KIO3 titrant in mL)*(1 L/1000 mL)
0.01931
0.03007
0.02091
Next fill up the results table. Add an extra column as shown.
Trail
Moles of KIO3 = (volume of KIO3 in L)*(molarity of KIO3)
Moles of ascorbic acid (determine by titration) = 3*(moles of KIO3)
Mass of ascorbic acid (determined by titration) = (moles of ascorbic acid)*(molar mass of ascorbic acid) [Molar mass of ascorbic acid = 176.12 g/mol]
Mass of ascorbic acid (as weighed out)
% difference = (mass obtained by titration) – (mass weighed out)/(mass obtained by titration)*100
1
(0.01931 L)*(0.0100 mol/L) = 0.0001931 mole
3*0.0001931 mole = 0.0005793 mole
(0.0005793 mole)*(176.12 g/mol) = 0.1020 g
0.10 g
(0.1020 g) – (0.10 g)/(0.1020 g)*100 = 1.96
2
(0.03007 L)*(0.0100 mol/L) = 0.0003007 mole
3*0.0003007 mole = 0.0009021 mole
0.1589 g
0.11 g
(0.1589 g) – (0.11 g)/(0.1589 g)*100 = 30.77
3
(0.02091 L)*(0.0100 mol/L) = 0.0002091 mole
3*0.0002091 mole = 0.0006273 mole
0.1105 g
0.10 g
(0.1105 g) – (0.10 g)/(0.1105 g)*100 = 9.50
Data
Trial 1
Trial 2
Trial 3
Mass of ascorbic acid (g)
0.10
0.11
0.10
Initial Buret Volume (mL)
0.00
19.31
0.00
Final Buret Volume (mL)
19.31
49.38
20.91
Volume of KIO3 titrant added (mL) = (final buret volume) – (initial buret volume)
19.31
30.07
20.91
Volume of KIO3 titrant added (L) = (volume of KIO3 titrant in mL)*(1 L/1000 mL)
0.01931
0.03007
0.02091
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