Back Titration: Definition and Problem Solving
Back titration is a technique that use in analytical chemistry which this procedure allows us to determine the concentration of an unknown concentration of substance by reacting it with an excess volume of another reactant of known concentration. The remaining solution that not reacted with the substance is then titrated back with other reactant, taking into account the molarity of the excess which was added
Back titrations can be used for many reasons, including: when the sample is not soluble in water, when the sample contains impurities that interfere with forward titration, or when the end-point is more easily identified than in forward titration.
This is the examples of problem that involved back titration:
Chemistry Tutor Site Problem 1
Iodoform (CHI3) is used in some antiseptic preparations. Iodoform reacts with Ag+ ions to form AgI and CO gas:
CHI3 + 3AgNO3 + H2O 3AgI(s) + CO(g) + 3HNO3
Suppose that a 1.000 g sample of an antiseptic preparation is treated with 25.00 mL of 0.02964 M AgNO3 and then the excess Ag+ is back-titrated with 3.22 mL of 0.03542 M KSCN to a red FeSCN2+ end point (Volhard method).
KSCN + AgNO3 AgSCN(s) + KNO3
What is the weight percent of CHI3 (393.7 g/mol) in the antiseptic?
Anwer:
How much total Ag was in the 25 ml of solution? 25.00 ml * .02964 mmol / ml = .741 mmol
How much excess Ag was present in the 25 ml? Assuming your chemical equation is correct (I’m a little puzzled by the mention of the Fe), it reacted with the thiocyanate in a 1-to-1 ratio, so find the moles of thiocyanate:
= 3.22 ml * .03542 mmol / ml = .114 mmol
How much Ag reacted with the antiseptic?
.741 mmol - .114 mmol = .627 mmol
How much CHI3 would be equivalent to that much Ag?
.627 mmol Ag * mmol CHI3 / 3 mmol Ag = .209 mmol CHI3
How much would that weigh?
.209 mmol * 393.7 mg/mmol = 82.3 mg
What per cent of 1 g is 82.3 mg? 82.3 mg/1000 mg = 8.23%
Chemistry Tutor Site Problem 2
A 0.4755 g sample containing ammonium oxalate (NH4)2C2O4 and inert compounds has dissolved in water and made alkaline with KOH. The liberated NH3 was distilled into 50 mL of 0.1007 N H2SO4. The excess H2SO4 was back titrated with 11.13 mL of 0.1214 N NaOH. Calculate the %N and %(NH4)2C2O4 in the sample..
Anwer:
Molarity H2SO4 = 0.1007 / 2 = 0.05035
Moles H2SO4 = 50 x 0.05035 / 1000 = 0.00252
Moles NaOH = 11.13 x 0.1214 / 1000 = 0.00135
2NaOH + H2SO4 = Na2SO4 + 2H2O
Moles H2SO4 titrated by NaOH = 2 x 0.00135 = 0.00270
Moles H2SO4 in excess = 0.00270 - 0.00252 = 0.00018
Moles NH3 = 2 x 0.0018 = 0.00036
the ratio between ammonium oxalate and ammonia is 2 : 1
moles ammonium oxalate = 2 x 0.00036 = 0.00072
Molar mass = 124 g/mol
Grams ammonium oxalate
= 0.00072 mol x 124 g/mol = 0.0893
= 0.0893 x 100 / 0.4755 = 18.8 % ( % ammonium oxalate in the sample)
Moles N = 0.00072 x 2 = 0.00144
Grams N = 0.00144 x 14 = 0.02016
% = 0.02016 x 100 / 0.4755 = 4.24 % ( % N in the sample)
Chemistry Tutor Site Problem3
A 0.4376 g aspirin tablet was heated with 50.00 mL of 0.196 M sodium hydroxide solution. The aspirin reacts according to the equation:
C6H4(OCOCH3)COOH (aq) + 2NaOH (aq) —-> C6H4(OH)COONa (aq) + CH3COONa (aq) + H2O (l)
After cooling, the resulting solution was titrated against 0.298 M HCl to determine the amount of unreacted NaOH. A titre of 18.64 mL was obtained.
ok then it asks for amount of NaOH initially added to the asprin, and then the amount of HCl used, which i CAN DO. It’s just the next questions that I can’t figure out for the life of me:
Calculate the amount of NaOH is excess after reaction with the aspirin
Calculate the amount of NaOH that reacted with the asprin
Answer
you are given the number of moles of OH
ie moles = 0.196 mol/L x 0.05000 L
excess OH (after reaction) = no moles HCl
ie excess = 0.298 mol/L x 0.01864L
(can convert to mass)
Amount reacted = initial moles OH - excess OH
this will lead you (divide by 2) to moles aspirin
Should be of some help. I’ve done some very nasty ones of these.
Add comment April 10th, 2008