Perk Learning Lab, MGCCC

Created by Tara L. Moore

 

GENERAL CHEMISTRY,  5th ed., Whitten, Davis & Peck

 

Chapter 11 Review Quiz

 

MATCHING

  1. Buret                             F. Normality                 K. Secondary Standard
  2. End Point                      G. Oxidation                 L. Standard Solution
  3. Equivalence Point         H. Oxidizing Agent      M. Standardization
  4. Half-reaction                 I. Primary Standard      N. Titration
  5. Indicator                        J. Reducing Agent        O. Reduction

 

  1. A solution that has been titrated against a primary standard.
  2. Number of equivalent weights of solute per liter of solution
  3. Chemically equivalent amounts of reactant have reacted.
  4. Loss of electrons
  5. Gain of Electrons
  6. Change in color and titration is stopped
  7. a solution of accurately known concentration
  8. Volume of a standard solution required to react with a specific amount of a substance is determined.

 

PROBLEM SOLVING

  1. Calculate the mass of NaOH required to prepare 3.40L of 2.18M solution.

 

  1. Calculate the molarity and normality of a solution that is 22.3% (NH2)2SO4. The specific gravity of the solution is 1.35.

 

  1. Balance the following equation. Then calculate the normality and molarity of an HCl solution if 55.7mL of the solution reacts with 2.46 g of Na2CO3.

                     HCl  +  Na2CO3 à  NaCl  +  CO2  +  H2O

 

  1. If 400 mL of 0.600 M  KOH solution is added to 150 mL of 0.0200 M H2CO3 solution, the resulting solution will  be ____  molar in ____ (excess reactant) and ____ molar in K2CO3.

 

  1. A vinegar solution is 9.37% acetic acid, HC2H3O2 . Its density is 1.007 g/mL. What is the normality?

 

  1. Calculate the normality and molarity of a solution that contains 43.6 g of H3AsO4 in enough H2O to make 350mL of solution.

 

  1. What is potassium hydrogen phthalate?

 

  1. Why can sodium carbonate be used as a primary standard for solutions of acids?

 

  1. What are the properties of an ideal primary standard?

 

  1. An impure sample of (COOH)2· 6H2O that had a mass of 3.44g was dissolved in water and titrated with standard NaOH solution. The titration required 63.7 mL of 0.342 M  NaOH solution. Calculate the percentage of (COOH)2· 6H2O in the sample.     2NaOH  +  (COOH)2 à  Na2(COO)2  + 2H2O

 

Balance the following ionic equations.

  1.  Cu (s)  +  NO3- (aq) à  Cu+2 (aq) +   NO2 (g)

 

  1.   Cl2 (g) à  ClO3- (aq)  +  Cl- (aq)

 

  1.   Al(s)  +  NO3- (aq) +  OH- (aq) +  H2O  à  Al(OH)4- (aq) +  NH3 (g)

 

  1.  CrO4-2 (aq)  +  H2O (l) +  HSnO2- (aq)  à  CrO2- (aq)  + OH- (aq)  +  HSnO3- (aq)

 

  1. C2H4 (g) +  MnO4- (aq)  +  H+ (aq) à  CO2 (g) +  Mn+2 (aq)  +  H2O (l)

 

 

 

 

 

 

 

 

 

 

ANSWERS

  1. K
  2. F
  3. C
  4. G
  5. O
  6. B
  7. L
  8. N
  9. ?g NaOH = 3.4 L (2.18mol/1L) (40g NaOH/1 mol) = 296.48g NaOH

 

  1. ?g/L (NH2)2SO4= (1.35g/1 mL) (1000 ml/ 1L)(0.223)= 301.05g/L (NH2)2SO4

       ?M = (301.05g/L)[1mol/ 128g (NH2)2SO4] = 2.35 mol/L (NH2)2SO4

       ?N = (2.35 mol/L)(4)= 9.4 N  (NH2)2SO4

 

  1. 2 HCl  +  Na2CO3 à 2 NaCl  +  CO2  +  H2O

        ? mol HCl = 2.46 g Na2CO3 [1mol/ 106g Na2CO3][2 mol HCl/ 1 mol Na2CO3] = 0.046 mol HCl

        ?M = (0.046 mol HCl/ 0.0557 L) = 0.825 M  HCl

        ?N = (0.825M) (1)= 0.825 N  HCl

 

  1. 2KOH  +  H2CO3  à  K2CO3  +  2H2O

? mol H2CO3 have = 0.15 L (0.02mol/L) = 0.003 mol H2CO3 (actually have)

? mol KOH have = (0.4L) (0.6mol/L) = 0.24 mol KOH (actually have)

? mol H2CO3 needed = 0.24 mol KOH (1 mol H2CO3/ 2 mol KOH) = 0.12 mol H2CO3 needed

H2CO3 is the limiting reactant used to determine the amount of products produced.

      

       ? mol KOH needed = 0.003 mol H2CO3 (2 mol KOH/ 1 mol H2CO3) = 0.006 mol KOH needed

? mol excess KOH = 0.24 mol KOH (have) – 0.006 mol KOH (needed) = 0.234 mol KOH remain

? mol K2CO3 = 0.003 mol H2CO3 (1mol K2CO3/1 mol H2CO3) = 0.003 mol K2CO3 produced

 

  1. ? g/L = (1.007g HC2H3O2/ 1 mL)(1000 mL/ 1L) ( 0.0937) = 94.35 g/ L  HC2H3O2

? M = (94.35g/L)(1 mol/ 60 g HC2H3O2) = 1.572 M HC2H3O2

? N = (1.572 M) (3) = 4.716 N  HC2H3O2

 

  1. ? M = (43.6 g H3AsO4/ 350mL) (1000mL/ 1L) (1 mol/ 142 g H3AsO4) = 0.877 M  H3AsO4

? N = (0.877 M) (3) = 2.63 N  H3AsO4

 

  1. KHP is a primary standard for bases.

 

  1. Sodium carbonate is a salt. However, because a base can be broadly defined as a substance that reacts with hydrogen ions, in this reaction Na2CO3 can be thought of as a base.

 

  1. a. must not react with or absorb the components of the atmosphere, such as water vapor, oxygen, and carbon dioxide.

b. must react according to one invariable reaction

c. must have a high percentage purity

d. should have a high formula weight to minimize the effect of error in weighing

e. must be soluble in the solvent of interest

f. should be nontoxic

g. should be readily available (inexpensive)

h. should be environmentally friendly

 

  1. ? mol NaOH = 0.0637L ( 0.342 molNaOH/L) = 0.021 mol NaOH

? mol (COOH)­2= 0.021 mol NaOH [1 mol (COOH)2/ 2 mol NaOH]= ).0108 mol (COOH)­2

? g (COOH)­2· 6H2O = 0.0108 mol [ 198 g (COOH)­2· 6H2O/ 1 mol] = 2.138 g (COOH)­2· 6H2O

? % purity = (2.138 g/ 3.44 g) * 100% = 62.16% (COOH)­2· 6H2O

 

19.   Cu  +  NO3 -  à  Cu+2  +   NO2    Use Acid Rules

 

Cu0  +  (N+5 O3-2) -  à  Cu+2  +  N+4O2-2

             +5 -6 = -1

 

Red: 2H+  +  (N+5 O3) -  +  1 e  à  N+4O2  +  H2O

                       +5            minus       +4 =    +1 electron

Ox:                         Cu0  - 2 e à  Cu+2

                                  0       minus      +2    = - 2 electrons

 

Red:  x2 [2H+  +  (N+5 O3) -  +  1 e  à  N+4O2  +  H2O]

Ox:                         Cu0  - 2 e à  Cu+2

 

Red:  4H+  +  2(N+5 O3) -  +  2 e  à  2N+4O2  + 2 H2O]

Ox:                         Cu0  - 2 e à  Cu+2

 

Combine:  4H+  +  2 NO3-  + Cu  à  2NO2  +  2H2O  + Cu+2

 

20. Cl2  à  ClO3 -  +  Cl-   Use Acid Rules

 

Cl20  à  (Cl+5O3-2) -  +  Cl-

 

Red:               Cl20  +  2e à  2Cl-

                         0     minus     -2 = +2 electrons

Ox:  6 H2O  +  Cl20  - 10 e  à  2 (Cl+5 O3) -  + 12H+

                         0        minus      10 = - 10 electrons

 

Red:      x5   [Cl20  +  2e à  2Cl -]

Ox:  6 H2O  +  Cl20  - 10 e  à  2 (Cl+5 O3) -  + 12H+

 

Red:               5Cl20  +  10e à  10Cl -

Ox:  6 H2O  +  Cl20  - 10 e  à  2 (Cl+5 O3) -  + 12H+

 

Combine:  6 H2O  + 6Cl2  à  10Cl-  +  2ClO3-  +  12H+

 

 

 

21. Al(s)  +  NO3- (aq) +  OH- (aq) +  H2O  à  Al(OH)4- (aq) +  NH3 (g)    Use Basic Rules

 

Al0  +  (N+5 O3 -2)- +  OH- +  H2O  à  [Al+3 (OH)4 ] -  + N-3 H3 +1

 

Red:   3H2O +  3H2O  +  (N+5 O3 )-  +  8 e  à  N-3 H3  +  6OH-  +  3OH-

                                            +5              minus     -3 =  +8 electrons

Ox:  4 H2O  +  8OH-  + Al0 – 3 e  à [ Al+3(OH)4] -  +  4H2O  +4OH-

                                       0             minus  +3  = -3 electrons

 

Red:  x3 [3H2O +  3H2O  +  (N+5 O3 )-  +  8 e  à  N-3 H3  +  6OH-  +  3OH- ]

Ox:  x8 [ 4 H2O  +  8OH-  + Al0 – 3 e  à [ Al+3(OH)4] -  +  4H2O  +4OH- ]

 

Red:  9H2O +  9H2O  + 3 (N+5 O3 )-  +  24 e  à  3N-3 H3  +  18OH-  +  9OH-

Ox:  32H2O  +  64 OH-  +   8Al0 – 3 e  à 8 [ Al+3(OH)4] -  +  32 H2O  +  32 OH-

 

Combine: 18 H2O  + 3NO3  +  5OH-  +  8Al  à  3NH3  +  8Al(OH)4 -

 

22. (Cr+6 O4-2 )-2  +  H2O  +  (H+1 Sn+2 O2-2) -  à  (Cr+3 O2 -2) -  + OH-  +  (H+1 Sn+4 O3 -2) -   Use Basic Rules

 

Red: 2H2O  +   Cr+6O4  +  3e  à  Cr+3O2-  +  4OH-

                         +6   minus       +3 = +3 electrons

Ox:  2OH-  +  HSn+2O2-  -  2 e  à  HSn+4O3-  +  H2O

                         +2       minus     + 4 = -2 electrons

 

Red: x2 [2H2O  +   Cr+6O4  +  3e  à  Cr+3O2-  +  4OH-]

Ox:  x3  [2OH-  +  HSn+2O2-  -  2 e  à  HSn+4O3-  +  H2O]

 

Red: 4H2O  +   2Cr+6O4  +  6e  à  2Cr+3O2-  +  8 OH-

Ox:   6OH-  +  3 HSn+2O2-  -  6 e  à 3  HSn+4O3-  + 3 H2O

 

Combine:  H2O  +  2CrO4-2  +  3HSnO2-  à  2CrO2-  +  2OH-  +  3HSnO3-

 

23. C2 -2 H4 +1 +  (Mn+7 O4-2 )-  +  H+ à  C+4 O2-2 +  Mn+2  +  H2O      Use Acid Rules.

 

Red:  8 H+  +  Mn+7O4-  +  5e  à  Mn+2  +  4 H2O

                            +7     minus        +2 = +5 electrons

Ox:  4H2O  + C2-2H4  -  12e à  2C+4O2  +  12H+

                     - 4       minus     +8  = - 12 electrons

 

Red:  x12 [8 H+  +  Mn+7O4-  +  5e  à  Mn+2  +  4 H2O]

Ox: x5  [4H2O  + C2-2H4  -  12e à  2C+4O2  +  12H+ ]

 

Red:   96H+  +  12Mn+7O4-  +  60e  à  12Mn+2  +  48 H2O

Ox:    20H2O  + 5 C2-2H4  -  60e à  10C+4O2  +  60H+

 

Combine:  36H+  +  12MnO4-  +  5C2H4  à  12 Mn+2  +  28H2O  +  10 CO2