CHM 235 Quantitative Analysis
Spring 2007
Dr. S.A. Skrabal
SOLUTIONS TO PROBLEM SET 1
Concentration units, solutions, dilutions, stoichiometry
11 January 2007
Note: Correct significant figures are used throughout this solution set!
1. Consider a 50.0 ppm solution of scandium nitrate, Sc(NO3)3. (A) What is the molarity of Sc3+ in this solution?
(B) What is the concentration in ppm of NO3- in this solution?
(A) Remember 1 ppm = 1 mg/L
(50.0 mg Sc(NO3)3/L)(1 g/1000 mg)(1 mol Sc(NO3)3/230.97 g Sc(NO3)3)(1 mol Sc3+/1 mol Sc(NO3)3) =
2.16 x 10-4 mol Sc3+/L
(B) (50.0 mg Sc(NO3)3/L)(1 g/1000 mg)(1 mol Sc(NO3)3/230.97 g Sc(NO3)3)(3 mol NO3-/1 mol ScNO3)3)
(62.00 g NO3-/mol NO3-)(1000 mg/g) = 4.03 x 101 mg/L = 40.3 mg/L = 40.3 ppm
2. A volume of 3.5 ounces (1.0 oz = 29.6 mL) of an algicide containing 7.0% Cu (w/v) is added to 10,000 gallons
(1 gal = 3.79 L) of pool water. Assume the density of pool water is 1.00 g/mL. (A) What is the concentration of Cu
(in ppb) in the pool? (B) What is the molarity of Cu in the pool?
(A) Remember 1 ppb = 1 µg/L
(3.5 oz)(29.6 mL/oz)(7.0 g Cu/100 mL)(µg/10-6g)(1/10,000 gal)(1 gal/3.79 L) = 1.9 x 102 µg/L =
1.9 x 102 ppb
(B) (1.9 x 102 µg/L)(10-6 g/µg)(1 mol Cu/63.55 g Cu) = 3.0 x 10-6 mol/L
3. An aqueous solution consists of 95.0 g of sodium thiosulfate pentahydrate, Na2S2O3 5H2O, in a total solution
volume of 400.0 mL. (A) What is the molarity of the solution? (B) What is the concentration of sulfur in this
solution in ppm? (C) What is the molarity of Na+ in this solution? (D) What is the concentration of Na+ in this
solution on a %(w/v) basis?
(A) (95.0 g Na2S2O3 5H2O)(1 mol Na2S2O3 5H2O/248.18 g Na2S2O3 5H2O)(1/400 x 10-3 L) = 0.957 mol/L
(B) (0.957 mol Na2S2O3 5H2O/L)(2 mol S/1 mol Na2S2O3 5H2O)(32.06 g S/mol S)(1000 mg/g) = 6.14 x 104 mg/L
= 6.14 x 104 ppm
(C) (0.957 mol Na2S2O3 5H2O/L)(2 mol Na+/1 mol Na2S2O3 5H2O) = 1.91 mol Na+/L
(D) (1.91 mol Na+/L)(22.99 g Na+/mol Na+) = 43.9 g Na+/L = 4.39 g/0.100 L = 4.39 g/100 mL
%(w/v) = 100(4.39 g Na+/100 mL) = 4.39%
4. A student pipets 12.0 mL of a 1.50 M nitric acid solution into a 250.0 mL volumetric flask, adds deionized water
to the mark and mixes well. What is the molarity of the nitric acid solution in the flask?
CconcVconc = CdilVdil
(1.50 M)(12.0 mL) = Cdil(250.0 mL); Cdil = ((1.50 M)(12.0 mL))/250.0 mL = 7.20 x 10-2 M
5. As a result of research in my laboratory, we have found the concentration of dissolved zinc (Zn) in the lower
Cape Fear River to range between about 5 and 60 nM. (1 nM = 1 x 10-9 M). Express this range of concentrations
in parts per billion (ppb or µg/L).
(5 x10-9 mol Zn2+/L)(65.38 g Zn2+/mol Zn2+)(106µg/g) = 3 x 10-1 µg/L = 3 x 10-1 ppb
(60 x 10-9 mol Zn2+/L)(65.38 g Zn2+/mol Zn2+)(106µg/g) = 3.9 µg/L = 3.9 ppb
6. What is the molality (m) of a 5.00 % by weight aqueous solution of NaOH?
5.00% by wt. NaOH solution means 5.00 g NaOH/100.0 g soln. In 100.0 g of solution, 5.00 g is NaOH, 95.00
g is H2O. So, (5.00 g NaOH/95.0 g H2O)(1 mol NaOH/40.00 g NaOH)(1000 g/kg) = 1.32 mol NaOH/kg H2O =
1.32 m
7. Calculate the number of grams of each substance required to prepare the following solutions: (A) 250 mL of
0.100 M NaOH; (B) 1.00 L of 0.0275 M K2Cr2O7; (C) 500 mL of 0.0200 M AuCl3.
(A) (250 mL)(1.00 L/1000 mL)(0.100 mol NaOH/L)(40.00 g NaOH/mol NaOH) = 1.00 g NaOH
(B) (1.00 L)(0.0275 mol K2Cr2O7/L)(294.20 g K2Cr2O7/mol K2Cr2O7) = 8.09 g K2Cr2O7
(C) (500 mL)(1.00 L/1000 mL)(0.0200 mol AuCl3/L)(303.33 g AuCl3/mol AuCl3) = 3.03 g AuCl3
8. How many milliliters of concentrated hydrochloric acid (38.0% (w/w), density 1.19 g/cm 3) are required to
prepare 1.00 L of a 0.100 M solution? Remember that 1 mL = 1 cm 3.
(1.00 L)(0.100 mol HCl/L) = 0.100 mol HCl needed.
(38.0 g HCl/100 g soln)(1.19 g soln/mL soln)(1 mol HCl/36.46 g HCl) = 1.24 x 10-2 mol HCl/mL soln
(1 mL soln/1.24 x 10-2 mol HCl)(0.100 mol HCl) = 8.06 mL of conc. HCl
9. What mass of solid La(IO3)3 (663.6 g/mol) is formed when 50.0 mL of 0.250 M La3+(aq) is mixed with 75.0 mL
of 0.302 M IO3- (aq)? The reaction is: La3+ (aq) + 3 IO3- (aq) ---> La(IO3)3 (s).
First figure out which is limiting: La3+ or IO3-:
(50.0 mL)(L/1000 mL)(0.250 mol La3+/L) = 1.25 x 10-2 mol La3+
(75.0 mL)(L/1000 mL)(0.302 mol IO3-/L) = 2.26 x 10-2 mol IO3Since 1.25 x 10-2 mol of La3+ requires (3 mol IO3-/mol La3+) = 3.75 x 10-2 mol IO3-, IO3- is the limiting reagent.
(2.26 x 10-2 mol IO3-)(1 mol La3+/3 mol IO3-)(1 mol La(IO3)3/1 mol La3+)663.6 g La(IO3)3/mol La(IO3)3) =
5.00 g La(IO3)3
10. What is the molar concentration of Cl- in a solution prepared by mixing 25.0 mL of 0.025 M NaCl with 35.0 mL
of 0.050 M BaCl2? Assume the volumes are additive.
(25.0 mL)(1 L/1000 mL)(0.025 mol NaCl/L)(1 mol Cl-/1 mol NaCl) = 6.25 x 10-4 mol Cl(35.0 mL)(1 L/1000 mL)(0.050 mol BaCl2/ L)(2 mol Cl-/1 mol BaCl2) = 3.50 x 10-3 mol Cl(6.25 x 10-4 mol + 3.50 x 10-3 mol)/(25.0 x 10-3 L + 35.0 x 10-3 L) = 6.87 x 10-2 mol Cl-/L
11. To determine the concentration of ethanol in cognac, a 5.00 mL sample of cognac is diluted to
0.500 L. Analysis of the diluted cognac gives an ethanol concentration of 0.0844 M. What is the molar
concentration of ethanol in the undiluted cognac?
CconcVconc = CdilVdil
Cconc(5.00 x 10-3 L) = (0.0844 M)(0.500 L)
Cconc = 8.44 M
12. A solution has a Ag+ (aq) concentration of 7.5 ng/L. What is the molarity of Ag+ (aq)?
(7.5 ng Ag+/L)(g/109 ng)(1 mol Ag+/107.87g Ag+) = 7.0 x 10-11 mol/L
13. Treatment of a 0.4000 g sample of impure potassium chloride (KCl) with an excess of AgNO 3 (aq) resulted in
the formation of 0.7332 g of AgCl by way of the following reaction:
Ag+ (aq) + Cl- (aq) ---> AgCl (s).
Calculate the percentage of KCl in the sample.
(0.7332 g AgCl)(1 mol AgCl/143.32 g AgCl)(1 mol Cl-/1 mol AgCl)(1 mol KCl/1 mol Cl-)
(74.55 g KCl/1 mol KCl)(100/0.4000g) = 95.35%
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14. A 6.42% (w/w) Fe(NO3)3 (241.81 g/mol) solution has a density of 1.059 g/mL. Calculate (A) the molar
concentration of Fe(NO3)3 in this solution and (B) the mass in grams of Fe(NO3)3 in each liter of this solution.
(A) (6.42 g Fe(NO3)3/100 g soln)(1.059 g soln/mL soln)(1 mol Fe(NO3)3/241.81 g Fe(NO3)3)(1000 mL/L) =
0.281 M
(B) (0.281 mol Fe(NO3)3/L)(241.81 g Fe(NO3)3/mol Fe(NO3)3) = 67.9 g Fe(NO3)3/L
15. Calculate the grams of substance required to prepare the following solutions: (A) 250 mL of 5.00%(w:v)
NaNO3; (B) 500 mL of 1.00%(w:v) NH4NO3; (C) 1000 mL of 10.0%(w:v) AgNO3.
(A) 5.00%(w:v) = (5.00 g NaNO3/100 mL soln.)100. So (250 mL soln.)(5.00 g NaNO3/100 mL soln.) = 12.5 g
NaNO3
(B) 1.00%(w:v) = (1.00 g NH4NO3/100 mL soln.)100. So (500 mL)(1.00 g NH4NO3/100 mL soln.) = 5.00 g NH4NO3
(C) 10.0%(w:v) = 10.0 g AgNO3/100 mL soln.)100. So (1000 mL)(10.0 g AgNO3/100 mL soln.) = 100 g AgNO3
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