BACK TITRATION SOLVING

Question 1

Given:

25 mL aliquot - 5.00 mL brandy / 1 Liter

brandy distilled into 50.00 mL K2Cr2O7 (0.0200 M)

20.00 mL Fe2+ (0.1253 M) PIPETTED
EXCESS FE2+ REMOVED WITH 7.46 mL K2CrO7

LOOK FOR PERCENT C2H5OH (46.07 g/mol) in brandy

Answer 1

Given:

25 mL aliquot - 5.00 mL brandy / 1 Liter

brandy distilled into 50.00 mL K2Cr2O7 (0.0200 M)

20.00 mL Fe2+ (0.1253 M) PIPETTED
EXCESS FE2+ REMOVED WITH 7.46 mL K2CrO7

LOOK FOR PERCENT C2H5OH (46.07 g/mol) in brandy

Question 2

A

Answer 2

A
HOW MANY MOL of FE 2+ OVERALL

(L of all Fe2+)(M of original Fe2) =A

Question 3

B

Answer 3

B
MOL OF Fe2+ EXCESS

(L of K2Cr2O7 used for backtitrate) (M of K2Cr2O7) (Stoic convert to Fe2+) = B

Question 4

A-B

Answer 4

A-B
Amount of FE2+ Reacted with MOL

A-B = Fe2+ mol that reacted

Question 5

C

Answer 5

C
MOL OF K2CR2O4 that reacted with Fe2+

(Fe2 that reacted)(STOICH CONVERT) = C

Question 6

D

Answer 6

D
MOL of K2CR2O4 OVERALL

(Original L of K2Cr3O4)(Original mol K2Cr2O4) = D

Question 7

C-D

Answer 7

C-D
MOL IN SOL’N K2CrO7 LEFT

C-D = to be compared left

Question 8

E

Answer 8

E
g/mL of C2H5OH diluted in 1000 mL

(K2CrO7 mol Left) (STOICH CONVERT to C2H5OH) (MM C2H2O5/1 mol C2H2O5)

= PARTIAL ANSWER/25 mL (aliquot) = concentration C2

Question 9

F

Answer 9

F
Final Concentration
C1V1 (sample) = C2V2 (aliquot)

C1 = C2V2/V1

C1 = g/ml C2H5OH (1000 mL) / 5.00 mL

= ans