CHM 235 Quantitative Analysis
Spring 2007
Dr. S.A. Skrabal
SOLUTIONS TO PROBLEM SET 5
Volumetric analysis
20 February 2007
NOTE: THERE ARE ADDITIONAL QUESTIONS ON THE BACK.
1. How many moles of solute are present in each of the following titrant solutions? (A) 19.34 mL of
0.04118 M KSCN (potassium thiocyanate); (B) 52.19 mL of 0.09886 M K2Cr2O7 (potassium
dichromate); (C) 16.52 mL of 5.97 x 10-2 M CH3CO2H (acetic acid).
(A) (19.34 x 10-3 L)(0.04118 mol KSCN/L) = 7.964 x 10-4 mol KSCN
(B) (52.19 x 10-3 L)(0.09886 mol K2Cr2O7/L) = 5.160 x 10-3 mol K2Cr2O7
(C) (16.52 x 10-3 L)(5.97 x 10-2 mol CH3CO2H/L) = 9.86 x 10-4 mol CH3CO2H
2. How many milliliters of a 0.09997 M KMnO4 (potassium permanganate) solution would be required
to exactly titrate 50.00 mL of 0.05201 M Na2C2O4 (sodium oxalate), given the following reaction?
2MnO4- (aq) + 5C2O42- (aq) + 6H+ (aq)  2Mn2+ (aq) + 10CO2 (g) + 8H2O (l)
(50.00 x 10-3 L)(0.05201 mol Na2C2O4/L)(2 mol KMnO4/5 mol Na2C2O4)(x L/0.09997 mol KMnO4) =
1.041 x 10-2 L = 10.41 mL
3. A wastewater was analyzed for Br- using a precipitation titration employing AgNO3 with a starchfluorescein indicator:
Ag+ (aq) + Br- (aq)  AgBr (s).
A volume of 1.25 mL of the wastewater was taken, diluted to a volume of 10.00 mL with deionized H2O,
then titrated to the endpoint with 5.01 mL of 0.1049 M AgNO3. What is the Br- concentration in the
sample?
(5.01 x 10-3 L)(0.1049 mol AgNO3/L)(1 mol Ag+/1 mol AgNO3)(1 mol Br-/1 mol Ag+)(1/1.25 x 10-3 L) =
0.420 M
4. Thiourea, {(NH2)2CS; FW = 76.12} in a 0.8767 g sample of organic material was quantitatively
extracted into a dilute sulfuric acid solution, then titrated with 29.95 mL of a 8.764 x 10-3 M solution of
Hg2+ according to the following reaction:
4(NH2)2CS (aq) + Hg2+ (aq)  [(NH2)2CS]4Hg2+ (aq).
What is the weight percent of thiourea in the sample?
(29.95 x 10-3 L)(8.764 x 10-3 mol Hg2+/L)(4 mol (NH2)2CS/1 mol Hg2+)(76.12 g (NH2)2CS/mol
(NH2)2CS))(100/0.8767 g sample) =9.116%
5. A 100.00 mL sample of groundwater was treated to convert all iron to Fe2+. A volume of 12.50 mL of
2.095 x 10-3 M K2Cr2O7 (potassium dichromate), resulting in the following reaction:
6Fe2+ (aq) + Cr2O72- (aq) + 4H+ (aq)  6Fe3+ (aq) + Cr3+ (aq) + 7H2O (l).
The excess K2Cr2O7 was back-titrated with 6.99 mL of 5.055 x 10-3 M Fe2+ solution. What is the
concentration of Fe2+ (in both molarity and ppm) in the groundwater sample?
Total Cr2O72- = (12.50 x 10-3 L)(2.095 x 10-3 mol Cr2O72-/L) = 2.619 x 10-5 mol Cr2O72Excess Cr2O72- = (6.99 x 10-3 L)(5.055 x 10-3 mol Fe2+/L)(1 mol Cr2O72-/6 mol Fe2+) = 5.89 x 10-6 mol
Cr2O72Cr2O72- reacted with Fe2+ = 2.619 x 10-5 mol – 5.89 x 10-6 mol = 2.030 x 10-5 mol Cr2O72-.
(2.030 x 10-5 mol Cr2O72-)(6 mol Fe2+/1 mol Cr2O72-)(1/100.00 x 10-3 L) =1.218 x 10-3 mol/L
(1.218 x 10-3 mol Fe2+/L)(55.85 g Fe2+/mol Fe2+)(1000 mg/g) = 68.03 mg/L or 68.03 ppm
6. A 0.4980 g sample that contains KBr (potassium bromide) is dissolved in 50 mL of deionized water
and titrated with 22.33 mL of 0.04332 M AgNO3. A blank titration requires 0.49 mL to reach the
endpoint. Calculate the weight percent KBr in the sample. The analysis reaction is:
Ag+ (aq) + Br- (aq)  AgBr (s).
Corrected volume = 22.33 mL – 0.49 mL = 21.84 mL
(21.84 x 10-3 L)(0.04332 mol AgNO3/L)(1 mol Ag+/1 mol AgNO3)(1 mol Br-/1 mol Ag+)(1 mol KBr/1 mol
Br-)(119.0 g KBr/1 mol KBr)(100/0.4980 g sample) = 22.61% KBr
7. A solution of EDTA (ethylenediaminetetraacetic acid) was standardized using a solution of Ca2+
prepared from primary standard grade CaCO3. A 0.3899 g sample of CaCO3 (FW = 100.09) was
placed in a 500.0 mL volumetric flask, dissolved in a small amount of 6 M HCl, then diluted to volume
with deionized H2O. A 50.00 mL volume of this solution was then titrated to the endpoint with 38.48 mL
of EDTA solution, according to the following reaction:
Ca2+ (aq) + EDTA4- (aq)  Ca(EDTA)2- (aq).
What is the molarity of the EDTA solution?
(0.3899 g CaCO3)(1 mol CaCO3/100.09 g CaCO3)(1/500.0 x 10-3 L)(1 mol Ca/1 mol CaCO3) = 7.791 x
10-3 mol Ca2+/L
(50.00 x 10-3 L)(7.791 x 10-3 mol Ca2+/L)(1 mol EDTA/1 mol Ca2+)(1/38.48 x 10-3 L) = 1.012 x 10-2 mol
EDTA/L
8. A Kjeldahl analysis is employed to determine the nitrogen content of a sample of cheddar cheese. A
0.8991 g sample of cheese is digested, the nitrogen is oxidized to NH4+, converted to NH3 by the
addition of NaOH, then distilled into a flask containing 50.00 mL of 0.1059 M HCl. The excess HCl was
back-titrated with 21.80 mL of 0.1097 M NaOH. (A) What is the weight percent of N in the cheese? (B)
What is the weight percent protein in the cheese, assuming that there is 6.38 g of protein per gram of
nitrogen in the cheese? The reactions are:
Digestion: organic N  NH4+ (aq)
NH4+ neutralization: NH4+ (aq) + OH- (aq)  NH3 (g) + H2O (l)
Distillation of NH3 into standardized HCl: NH3 (g) + H+ (aq)  NH4+ (aq)
Titration of excess HCl with standardized NaOH: H+ (aq) + OH- (aq)  H2O (l)
Total HCl = (50.00 x 10-3 L)(0.1059 mol HCl/L) = 5.295 x 10-3 mol HCl
Excess HCl = (21.80 x 10-3 L)(0.1097 mol NaOH/L)(1 mol HCl/1 mol NaOH) = 2.391 x 10-3 mol HCl
HCl reacted with NH3 = 5.295 x 10-3 mol – 2.391 x 10-3 mol = 2.904 x 10-3 mol HCl
(A) (2.904 x 10-3 mol HCl)(1 mol NH3/1 mol HCl)(1 mol NH4+/1 mol NH3)(1 mol N/1 mol NH4+)(14.01 g
N/1 mol N)(100/0.8991 g sample) = 4.525% N
(B) (2.904 x 10-3 mol HCl)(1 mol NH3/1 mol HCl)(1 mol NH4+/1 mol NH3)(1 mol N/1 mol NH4+)(14.01 g
N/1 mol N)(6.38 g protein/g N)(100/0.8991 g sample) = 28.9% protein
9. Disinfectant solutions are sold that contain approximately 3% hydrogen peroxide (H2O2) by weight.
A mass of 1.30 g of disinfectant solution is dissolved in 75 mL of deionized H2O, a small amount of
dilute H2SO4 is added, and the solution titrated with 36.79 mL of 0.01259 M KMnO4 according to the
following reaction:
5H2O2 (aq) + 2MnO4- (aq) + 6H+ (aq)  5O2 (aq) + 2Mn2+ (aq) + 8H2O (aq).
A blank titration required 0.19 mL to reach the endpoint. What is the weight percent H2O2 in the
disinfectant solution?
Corrected volume = 36.79 mL – 0.19 mL = 36.60 mL
(36.60 x 10-3 L)(0.01259 mol KMnO4/L)(5 mol H2O2/2 mol KMnO4)(34.02 g H2O2/1 mol H2O2)(100/1.30 g
sample) = 3.01% H2O2 (by wt.)
10. A 25.0 mL portion of a solution containing Tl(I) ion was treated with K2CrO4 (potassium chromate)
to form Tl2CrO4 (thallium(I) chromate). The solid Tl2CrO4 was filtered, washed, and dissolved in dilute
H2SO4 to form Cr2O72- (dichromate ion). The Cr2O72- thus produced was titrated with 39.72 mL of
0.1042 M Fe2+ solution. What is the mass of Tl(I) in the sample? The reactions are:
2 Tl+ (aq) + CrO42- (aq)  Tl2CrO4 (s)
2 Tl2CrO4 (s) + 2 H+ (aq)  4 Tl+ (aq) + Cr2O72- (aq) + H2O (l)
Cr2O72- (aq) + 6 Fe2+ (aq) + 14 H+ (aq)  6 Fe3+ (aq) + 2 Cr3+ (aq) + 7 H2O (l).
(39.72 x 10-3 L)(0.1042 mol Fe2+/L)(1 mol Cr2O72-/6 mol Fe2+)(2 mol Tl2CrO4/1 mol Cr2O72-)(2 mol Tl+/1
mol Tl2Cr2O4)(204.38 g Tl+/1 mol Tl+) = 0.5639 g Tl+