Solutions to Chemical Reactions Problem Set

This problem set was developed by S.E. Van Bramer for Chemistry 145 at Widener University.

  1. Titration/acid base problem. 0.4563 g of NaOH is weighed out and disolved in 50.00 mL of H2O. This is titrated against a solution of HNO3. The initial volume of the burette is 2.35 ml. The final volume of the burette is 47.98 ml. What can you determine?

    1. First give the balanced reactions
      NaOH (aq) + HNO3 (aq) -> H2O + NaNO3 (aq)
    2. Next find the moles of NaOH

      3. (0.4563 g)/(39.997 g/mole) = 0.01141 moles NaOH
    3. From the stoichiometry of the balanced reaction this gives 0.01141 moles HNO3
    4. The volume of HNO3 added is

      5. 47.98 ml - 2.35 ml = 45.63 mL
    5. Therefor the concentration of the nitric acid solution is

      6. (0.01141 moles)/(0.04563 L) = 0.2500 M

  2. Acid Base equations and reactions. A base solution is made by disolving 4.987 g of Potassium Hydroxide in 500.0 ml of water. 36.84 ml of this base solution is used to titrate 25.00 ml of sulfuric acid. Write the total, total ionic, and net ionic equations for these reactions. What can you calculate from this information? What are your results?

    1. Total Equation: 2 KOH (aq) + H2SO4 (aq) -> 2 H2O + K2SO4 (aq)
    2. Total Ionic Equation: 2 K1+ (aq) + 2 OH1- (aq) + 2 H1+ (aq) + SO42- (aq) -> 2 H2O + 2 K1+ (aq) + SO42- (aq)
    3. Net Ionic Equation: 2 OH1- (aq) + 2 H1+ (aq) -> 2 H2O
    4. Find the moles of KOH

      5. (4.987 g)/(56.10564 g/mole) = 0.08889 moles
    5. From the stoichiometry this gives 0.04445 moles H2SO4
    6. The concentration of the acid is

      7. (0.04445 moles)/(0.0250 L) = 1.78 M

  3. When the following solutions are mixed, does a precipitate form? Write out the total, total ionic, and net ionic equations.

    1. silver nitrate and rubidium chloride
    2. lead nitrate and potassium chloride
      1. Overall Equation:
        AgNO3 (aq) + RbCl(aq) -> AgCl(s) + RbNO3 (aq)
      2. Total Ionic Equation:
        Ag1+ (aq) + NO31- (aq) + Rb1+(aq) + Cl1-(aq) -> AgCl(s) + Rb1+(aq) + NO31- (aq)
      3. Net Ionic Equation:
        Ag1+ (aq) + Cl1-(aq) -> AgCl(s)
    3. mercury (I) nitrate and hydrochloric acid
      1. Overall Equation:
        Hg2(NO3)2 (aq) + 2 HCl (aq) -> Hg2Cl2 (s) + 2 HNO3 (aq)
      2. Total Ionic Equation:
        Hg21+(aq) + 2 NO31- (aq) + 2 H1+(aq) + 2 Cl1- (aq) -> Hg2Cl2 (s) + 2 H1+ (aq) + 2 NO31-
      3. Net Ionic Equation:
        Hg21+(aq) + 2 Cl1- (aq) -> Hg2Cl2 (s)
    4. calcium chloride and sodium carbonate
      1. Overall Equation:
        CaCl2 (aq) + Na2CO3 (aq) -> 2 NaCl(aq) + CaCO3 (s)
      2. Total Ionic Equation:
        Ca2+(aq)+ 2 Cl1- (aq) + 2 Na1+ (aq)+ CO32- (aq) -> 2 Na1+(aq) + 2 Cl1-(aq) + CaCO3 (s)
      3. Net Ionic Equation:
        Ca2+(aq) + CO32- (aq) -> CaCO3 (s)
    5. magnesium nitrate and calcium chloride
      1. Overall Equation:
        Mg(NO3)2 (aq) + CaCl2 (aq) -> Ca(NO3)2 (aq) + MgCl2 (aq)
      2. Total Ionic Equation:
        Mg2+ (aq) + 2 NO31- (aq) + Ca2+ (aq) + 2 Cl1- (aq) -> Ca 2+(aq) + 2 NO31- (aq) + Mg2+(aq) + 2 Cl1- (aq)
      3. Net Ionic Equation:
        No Reaction
    6. potassium sulfate and barium chloride
      1. Overall Equation:
        K2SO4 (aq) + BaCl2 (aq) -> BaSO4 (s) + 2 KCl (aq)
      2. Total Ionic Equation:
        2 K1+ (aq)+ SO42- (aq) + Ba2+ (aq) + 2 Cl1- (aq) -> BaSO4 (s) + 2 K1+(aq)+ 2 Cl1- (aq)
      3. Net Ionic Equation:
        SO42- (aq) + Ba2+ (aq) -> BaSO4 (s)

  4. Precipitation Reactions and Solubility.
    1. Step 1: 0.8765 g of silver (I) nitrate is placed in a 250 mL volumetric flask diluted to the mark with deionized water. Determine the concentration of each ion in solution.

      The reaction would be:

      AgNO3(s) -> Ag1+(aq) + NO3-1(aq)

      0.8765 grams of AgNO3(s), MW 169.87 g/mole, 0.0051598 moles
      Volume of solution is 250 mL, or 0.2500 L. Gives concentration of 0.02064 M. Stoichiometry is 1:1 so
      [Ag1+(aq)] = 0.02064 M
      [NO3-1(aq)]= 0.02064 M

    2. Step 2: 1.8793 g of potassium chloride is placed in a 250 mL volumetric flask diluted to the mark with deionized water. Determine the concentration of each ion in solution.

      The reaction would be:

      KCl(s) -> K1+(aq) + Cl-1(aq)

      1.8793 grams of KCl(s), 74.55MW g/mole, 0.02521 moles
      Volume of solution is 250 mL, or 0.2500 L. Gives concentration of 0.1008 M. Stoichiometry is 1:1 so
      [K1+(aq)] = 0.1008 M
      [Cl-1(aq)]= 0.1008 M

    3. Step 3: 50.0 mL of the silver (I) nitrate solution and 50.0 mL of the potassium chloride solution are mixed together in an erlenmyer flask. Determine the mass of any precipitate formed and the concentration of each ion in solution.

      The ionic reaction is:

      Ag1+(aq) + NO3-1(aq) + K1+(aq) + Cl-1(aq) -> AgCl (s) + NO3-1(aq)+ K1+(aq)

      The net ionic reaction is:
      Ag1+(aq) + Cl-1(aq) -> AgCl (s)

      Since only 50 mL of each solution is used, recalculate moles:
      Ag1+(aq) = 0.02064 M * 0.050 L = 0.001032 moles
      NO3-1(aq)= 0.02064 M * 0.0500 L = 0.001032 moles
      K1+(aq) = 0.1008 M* 0.050 L = 0.00504 moles
      Cl-1(aq) = 0.1008 M* 0.050 L = 0.00504 moles

      The ICF Table is as follows:
      Reaction Ag1+(aq) Cl-1(aq) -- AgCl (s)
      Initial 0.001032 moles 0.00504 moles
      Change 0.001032 0.001032 0.001032
      Final 0 0.004008 0.001032

      For the precipitate, AgCl, 0.001032 moles, 143.32 g/mole, 0.1479 g. For the solution:
      [Cl-1(aq)] = 0.004008 moles (unreacted) / 0.100 L = 0.0401 M
      [NO3-1(aq)] = 0.001032 moles / 0.100 L = 0.01032 M
      [K1+(aq)] = 0.00504 moles / 0.100 L = 0.0504 M

Please send comments or suggestions to svanbram@science.widener.edu

Scott Van Bramer
Department of Chemistry
Widener University
Chester, PA 19013

© copyright 1996, S.E. Van Bramer
This page has been accessed 13654 times since 1/5 /96 .

Last Updated: Friday, November 03, 2000 10:29:05 AM