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美国高考SATII化学专题分析——Net Ionic Equations

来源:51SAT 2010-04-16   美国留学   免费评估   答疑
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SAT是Scholastic Aptitude Test的缩写,是申请几乎所有美国大学必须参加的考试。通常,希望继续接受高等教育的高中生需要参加SAT考试,并且SAT考试得分是获取奖学金的重要标准之一。大部分美国大学要求SAT考试作为录取的条件并根据SAT得分授予奖学金。

 Net ionic equations are equations that show only the soluble, strong electrolytes reacting (these are represented as ions) and omit the spectator ions, which go through the reaction unchanged. When you encounter net ionic equations on the SAT II Chemistry test, you’ll need to remember the following solubility rules, so memorize them! Also keep in mind that net ionic equations, which are the bare bones of the chemical reaction, usually take place in aqueous environments. Here are those solubility rules:

  1. Most alkali metal compounds and compounds are soluble.
  2. Cl-, Br-, I- compounds are soluble, except when they contain Ag+, , or Pb2+.
  3. F- compounds are soluble, except when they contain group 2A metals.
  4. , , , and CH3COO- compounds are soluble.
  5. compounds are soluble, except when they include Ca2+, Sr2+, Ba2+, Ag+, Pb2+, or .
  6. , , , , S2-, OH-, and O2- compounds are insoluble.
  7. Group 2A metal oxides are classified as strong bases even though they are not very soluble.
     
    The two solubility rules that you will use the most are numbers 1 and 4. You must memorize that all group 1A metal and ammonium compounds are soluble. As soon as you see a compound , Li, Na, K, Rb, Cs, or Fr, you should know that it’s soluble. Also, all nitrates are soluble—look at the end of the compound. If it ends in , you know that it’s soluble.
     What’s the big deal with solubility? Well, if the ion is soluble, it won’t form a precipitate, and this means it doesn’t react and should be left out of the net ionic equation. The key is first to write the compound’s chemical formula and then determine if it’s soluble. If it is soluble, then ionize it—if it isn’t, don’t ionize it; leave it as a molecule.
     Here are some additional rules about common reaction types that you should be familiar with for the exam:
  • If an insoluble precipitate or gas can be formed in a reaction, it probably will be.
  • Oxides (except group 1A) are insoluble, and when reacted with water, they form either acids (nonmetal oxides) or bases (metal oxides).
  • There are six strong acids that completely ionize: HCl, HBr, HI, HNO3, H2SO4, HClO4. All other acids are weak and are written together, as molecules.
  • The strong bases that ionize are oxides and hydroxides of group 1A and 2A metals. All other oxides and hydroxides are considered weak and written together, as molecules.
     
    Now try writing some net ionic equations, using the rules above.
     Example
     Write the net ionic equation for a mixture of solutions of silver nitrate and lithium bromide.
     Explanation
Ag+ + + Li+ + Br-
    This is a double replacement reaction. Both compounds are soluble, so everything ionizes. If anything is formed, it will come from recombining the “inside” two ions with the “outside” two ions to make LiNO3 and AgBr. If either of them is insoluble, a precipitate will be formed, and the ions that react to form it will be in our net ionic equation; the other ions are spectators and should be omitted! As we said, the two possible products are lithium nitrate and silver bromide. Since halides are soluble except those containing silver, mercury, or lead, we have a precipitate of silver bromide, and our net ionic equation looks like this:
Ag+ + Br-AgBr
    Example
     Hydrochloric acid and sodium hydroxide are mixed. Write the net ionic equation.
     Explanation
     This is a mixture of a strong acid and a strong base, so each ionizes completely.
H+ + Cl- + Na+ + OH-
    The two possible compounds formed are sodium chloride, which is soluble, and water, which is molecular; thus water is the only product in our net ionic equation.
H+ + OH-H2O
    Example
     Chlorine gas is bubbled into a solution of potassium iodide; write the net ionic equation.
     Explanation
     This one is a single replacement, so you need to consider the activity series. Since halogens are involved, you can determine their activity by using the periodic table: Cl is more active than I.
Cl2 + K+ + I-
    Remember that halogen is diatomic and that all potassium compounds are soluble. The resulting compound is also soluble, so K+ is a spectator and is left out of the final equation.
Cl2 + I-I2 + Cl-
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