ENCE 433 Dr. Alba Torrents

ENVIRONMENTAL ENGINEERING ANALYSIS Fall 1996

Homework Set # 3 SOLUTION

1. Hydrochloric acid is an strong acid completely ionized in dilute aqueous solution. The self-ionization of water occurs at very slight extent in pure water, and in the presence of H3O+ from the HCl, the self-ionization of water is repressed. This means that the reaction:

2 H2O H3O+ + OH-

is driven to the left, because the reaction between HCl and H2O produces and excess of H3O+ . Thus

[H3O+ ] = contribution from HCl + contribution from self ionization of water

= 0.400 + <<< 0.0000001 0.400

Using Kw we can calculate [OH-]:



  • 2. (a) Ammonium carbonate is a basic salt. Its cation, NH4+, is a weak acid with Ka = 5.7x10-10. Its anion is a weak base with


    • Since Kb(CO3-2) > Ka (NH4+), solutions of ammonium carbonate are basic.

  • (b) Potassium perchlorate, KClO4, is a neutral salt. Neither K+ nor ClO4- has any tendency to donate or accept a proton in dilute aqueous solutions. The reaction between a strong base KOH and the strong acid HClO4 produces KClO4.
  • (c) Potassium hydrogen sulfate (also called potassium bisulfate) is an acidic salt. The anion, HSO4-, is a moderately strong acid with Ka = 1.2 x 10-2. The cation K+ is neither an acid not a base.
  • (d) Barium oxide, BaO, is a basic salt. Oxides of the alkaline earth metals are basic. The reaction that occurs when BaO dissolves in water is

  • BaO(s) + H2O Ba+2(aq) + 2 OH- (aq)
  • (e) The anion HSO3- is an ampholyte.


    • Since Ka > Kb, solution of NaHSO3 are acidic.

    1. Methylammonium nitrate is an acidic salt. Its cation, CH3NH3+, is a weak acid. Its anion, NO3-, has virtually no tendency to accept protons in dilute solutions.

    3.


    Species: H+, OH-, HNO2, NO2-

    MBE: 0.1 = [NO2-] + [HNO2]

    ENE: [H+] = [NO2-] + [OH-]

    PBE: [H+] = [NO2-] + [OH-]


    As we add acid we can assume [H+] >>> [OH-]

    then from PBE: [H+] = [NO2-]

    from MB : [HNO2] = 0.1 - [H+]

    In Ka [H+] 2 = 10-3.29 (0.1 - [H+])

    [H+]2 + 10-3.29[H+] = 10-4.29 = 0


    pH = 2.16

    Assumption [H+] >>> [OH-] is OK as [OH- ] = 10-11.84