Writing Chemical Equations

Transcription

1 Writing Chemical Equations Chemical equations for solution reactions can be written in three different forms; molecular l equations, complete ionic i equations, and net ionic equations. In class, so far, we have been writing molecular equations. This tutorial will show you the differences between these types of equations, and give you some practice in writing equations in the different forms.

2 The three types of equations are: I. The molecular equation that gives the overall reaction stoichiometry, but not necessarily the actual forms of the reactants and products in solution. II. The complete ionic equation that represents all strong electrolyte products and reactants in their ionic form. III. The net ionic equation that gives only those reactant and product components that undergo change in the reaction, and does not mention other components, called spectator ions, that do not undergo change.

3 The Molecular Equation Let s look at the reaction where aqueous silver nitrate and aqueous sodium chloride react together to form a silver chloride precipitate and sodium nitrate that remains in solution. First let s connect names with molecular formulas: aqueous silver nitrate = AgNO 3 (aq) aqueous sodium chloride = NaCl(aq) silver chloride precipitate = solid silver chloride = AgCl(s) sodium nitrate in solution = aqueous sodium nitrate = NaNO 3 (aq)

4 The Molecular Equation If we combine these molecular formulas into a reaction we get the molecular equation: AgNO 3 (aq) + NaCl(aq)=AgCl(s) + NaNO 3 (aq) Fortunately for you this simple example is already balanced y y p p y as written, so you don t have to put any extra effort into balancing the equation.

5 The Complete Ionic Equation The molecular equation contained three different ionic compounds in aqueous solution. All three of these ionic compounds are strong electrolytes and would ionize, so in reality the molecules AgNO 3 (aq), NaCl(aq), and NaNO 3 (aq) don t exist! The better way to represent the true chemistry in this reaction to break all strong electrolytes lt into it their thicomponent ions. Thus, AgNO 3 (aq) = Ag + (aq) + NO 3- (aq) NaCl(aq)= Na + (aq) + Cl - (aq) NaNO 3 (aq) = Na + (aq) + NO 3- (aq)

6 The Complete Ionic Equation When we break all strong electrolyes into their component ions we get the complete ionic equation. For this reaction the compete ionic equation is: Ag + (aq) + NO 3- (aq) + Na + (aq) + Cl - (aq)= AgCl(s) + Na + (aq) + NO 3- (aq)

7 The Complete Ionic Equation Notice the only thing you can separate into ions are aqueous ionic compounds, emphasis on aqueous, emphasis on ionic. Never try to separate covalent compounds into ions. Never try to separate a solid(s), liquid (l) or a gas(g) into ions. In this equation AgCl(s) is ionic, but it doesn t get changed into ions because it is a solid. Don t forget this. It will trip you up if you don t remember, and that will mean points off.

8 The Net Ionic Equation The complete ionic equation, Ag + (aq) + NO 3- (aq) + Na + (aq) + Cl - (aq)= AgCl(s) + Na + (aq) + NO 3- (aq) contains two ions that appear on both sides of the chemical equation, Na + (aq) and NO 3- (aq). Things that appear unchanged on both sides of a chemical reaction are called spectators. Essentially the showed up and watched the reaction, but they didn t actually take part in it.

9 The Net Ionic Equation In the Net Ionic reaction we remove the spectator ions. This helps us to emphasize the important molecules l and ions that are undergoing the reaction, and generally makes the equation shorter and easier to write down. The net ionic equation for this reaction is: Ag + (aq) + Cl - (aq)=agcl(s)

10 Example problem Write the balanced molecular, complete ionic and net ionic equations for the reaction of barium nitrate and potassium chromate in solution, where they react to form a barium chromate precipitate and aqueous potassium nitrate.

11 Write the equations for the reaction of barium nitrate t and potassium chromate to form solid barium chromate and aqueous potassium nitrate. First, what are the molecules? Barium nitrate = Ba(NO 3 ) 2 Potassium chromate = K 2 CrO 4 Barium chromate = BaCrO 4 Potassium nitrate = KNO 3

12 Write the equations for the reaction of barium nitrate t and potassium chromate to form solid barium chromate and aqueous potassium nitrate. Combining the molecules and putting in the physical forms to get an unbalanced molecular equation: Ba(NO 3 ) 2 (aq) + K 2 CrO 4 (aq)=bacro 4 (s) +KNO 3 (aq) Ba and CrO 4 are balanced, K and NO 3 will balance if we make it 2 KNO make it 2 KNO 3 Balanced molecular equation: Ba(NO 3 ) 2 (aq) + K 2 CrO 4 (aq) = BaCrO 4 (s) +2 KNO 3 (aq)

13 Write the equations for the reaction of barium nitrate t and potassium chromate to form solid barium chromate and aqueous potassium nitrate. Breaking all aqueous ionic compounds into their component ions to get the complete ionic equation we have: Ba(NO 3 ) 2 (aq) = Ba 2+ (aq)+ 2 NO 3- (aq) K 2 CrO 4 (aq) = 2 K + (aq) + CrO 2-4 (aq) 2 KNO = 2K + 3 (aq) (aq) + 2NO 3- (aq) and our complete ionic equation becomes: Ba 2+ (aq) + 2 NO 3- (aq) + 2 K + (aq) + CrO 4 2- (aq) = BaCrO 4 (s) +2 K + (aq) + 2NO 3- (aq)

14 Write the equations for the reaction of barium nitrate t and potassium chromate to form solid barium chromate and aqueous potassium nitrate. In making the complete ionic equation: Ba 2+ (aq) + 2 NO CO 3- (aq) + 2 K + (aq) + CrO 2-4 (aq) = BaCrO 4 (s) +2 K + (aq) + 2NO 3- (aq) Remember that you can t break the solid into its component ions. In addition, notice that this equation contained the covalent ions NO 3- and CrO 4-. Covalent ions cannot be broken into smaller ionic i pieces, so you have to remember what your covalent ions are, and never break them down into smaller pieces.

15 Write the equations for the reaction of barium nitrate t and potassium chromate to form solid barium chromate and aqueous potassium nitrate. Finally we remove the spectator ions that occur on both sides of the complete ionic equation to get the net ionic equation: Ba 2+ (aq) + CrO 2-4 (aq) = BaCrO 4 (s)

16 One last problem. Write the molecular, complete ionic and net ionic equations for the reaction of barium hydroxide and nitric acid. I stepped it up a notch here. I told you the reactants, but I didn t give you re the products, that is up to you. Actually this is not as bad as it sounds. The hydroxide part of the name barium hydroxide tells you that this compound is a base. The name nitric acid tells you the other reactant is an acid. So you have an acid base reaction, that means you should be looking for H + and OH - to be reacting to make H 2 O somewhere in this reaction.

17 Write the molecular, complete ionic and net ionic equations for the reaction of barium hydroxide and nitric acid. Associating molecular formula s with names Reactants: t Barium hydroxide = Ba(OH) 2 Nitric Acid = HNO 3 Products:? H 2 O (product of an acid-base reaction)

18 Write the molecular, complete ionic and net ionic equations for the reaction of barium hydroxide and nitric acid. Let s write the reactants and the only known product to see if we can figure out the missing gproduct Ba(OH) 2 + HNO 3 =? + H 2 O In acid base reactions like this, the OH of one molecule reacts with the H of the other molecule to make water. Let s combine the other parts of each reactant (Ba and NO 3 ) to make the other product. BaNO 3

19 Write the molecular, complete ionic and net ionic equations for the reaction of barium hydroxide and nitric acid. But BaNO 3 doesn t look right. Ba should be a +2 ion, and NO 3 is a -1 ion, so the charges don t balance. The proper product should be: Ba(NO 3 ) 2 The unbalanced molecular equation then is : Ba(OH) 2 + HNO 3 = B(NO) Ba(NO H 2 O

20 Write the molecular, complete ionic and net ionic equations for the reaction of barium hydroxide and nitric acid. Balancing we get: Ba(OH) HNO 3 = Ba(NO 3 ) H 2 O And adding physical forms to get our molecular equation we have: Ba(OH) 2 (aq) (q) + 2 HNO 3 (aq) (q) = Ba(NO 3 3) 2 (aq) (q) + 2 H 2 O(l) ()

21 Write the molecular, complete ionic and net ionic equations for the reaction of barium hydroxide and nitric acid. Breaking the aqueous ionic compounds into their ions we get: Ba(OH) 2 (aq) = Ba 2+ (aq) + 2OH - (aq) 2 HNO 3 (aq) = 2H + (aq) + 2 NO 3- (aq) Ba(NO 3 ) 2 (aq) = Ba 2+ (aq) + 2 NO 3- (aq) (And I hope you didn t try to break the H 2 O down. It is NOT ionic and it is NOT aqueous) The complete ionic equation then is: Ba 2+ (aq) + 2OH - (aq) + 2H + (aq) + 2 NO 3- (aq) = Ba 2+ (aq) + 2 NO 3- (aq) + 2H 2 O(l)

22 Write the molecular, complete ionic and net ionic equations for the reaction of barium hydroxide and nitric acid. Removing the spectators we have the net ionic equation: 2OH - (aq) + 2H + (aq) = 2H 2 O(l) Taking all terms to the lowest common denominator we have the final net ionic equation: OH - (aq) + H + (aq) = H 2 O(l) (Which is a true acid-base reaction)

23 Practice Problems 1. Write the molecular, complete ionic and net ionic equations that describe the solution reaction of sodium sulfate with lead nitrate. The solid product of this reaction is lead sulfate. 2. Write the molecular, complete ionic and net ionic equations that describe the solution reaction of sulfuric acid with the aluminum hydroxide. (Try them first, before you see my answers)

24 Write the equations that t describe the solution reaction of sodium sulfate with lead nitrate. The solid product of this reaction is lead sulfate. Molecules involved: sodium sulfate = Na 2 SO 4 lead nitrate = Pb(NO 3 ) 2 lead sulfate = PbSO 4 Some other unnamed product? 1 st try at molecular equation: Na 2 SO 4 (aq) + Pb(NO 3 ) 2 (aq) = PbSO 4 (s) +? Combining i the other halves of the reactants to make the other aqueous product Na 2SO 4 4( (aq) + Pb(NO 3 3) 2 (aq) (q) = PbSO 4 (s) + NaNO 3 3( (aq) Balancing: Na 2 SO 4 (aq) + Pb(NO 3 ) 2 (aq) = PbSO 4 (s) + 2 NaNO 3 (aq)

25 Write the equations that t describe the solution reaction of sodium sulfate with lead nitrate. The solid product of this reaction is lead sulfate. Starting guess: Na 2 SO 4 (aq) + Pb(NO 3 ) 2 (aq) = PbSO 4 (s) +? Combining the halves of the reactants that weren t used in Making PbSO 4 you predict that missing product is NaNO 3. Next guess Na 2SO 4 4( (aq) + Pb(NO 3 3) 2 (aq) (q) = PbSO 4 (s) + NaNO 3 3( (aq) Balancing for final molecular equation: Na 2 SO 4 (aq) + Pb(NO 3 ) 2 (aq) = PbSO 4 (s) + 2 NaNO 3 (aq)

26 Write the equations that t describe the solution reaction of sodium sulfate with lead nitrate. The solid product of this reaction is lead sulfate. Breaking up the aqueous ionic compounds: Na 2 SO 4 (aq) = 2Na + (aq) + SO 2-4 (aq) Pb(NO - 3 ) 2 (aq) = Pb 2+ (aq) + 2 NO 3 (aq) 2 NaNO 3 (aq) = 2 Na + (aq) + 2 NO 3- (aq) Complete ionic equation: 2Na + (aq) + SO 2-4 (aq) + Pb 2+ (aq) + 2 NO 3- (aq) = PbSO 4 (s) + 2 Na + (aq) (q) + 2 NO 3- (aq) (q) Net ionic equation: SO (aq) + Pb (aq) = PbSO 4 (s)

27 Write the equations that describe the solution reaction of sulfuric acid with the aluminum hydroxide. Reactants: Sulfuric acid = H 2 SO 4 Aluminum hydroxide = Al(OH) 3 Products:? Acid and base, so H 2 O Combing the other halves of the molecules Al 3+ + SO 2-4 But the charges don t work for a 1:1 complex so Al 2 (SO 4 ) 3 is better

28 Write the equations that describe the solution reaction of sulfuric acid with the aluminum hydroxide. Unbalanced molecular equation: H SO + Al(OH) =HO+Al(SO ) H 2 SO 4 + Al(OH) 3 = H 2 O + Al 2 (SO 4 ) 3 Balancing: 3 H 2 SO Al(OH) 3 = 6 H 2 O + Al 2 (SO 4 ) 3 Adding physical forms to get our final molecular equation: gp y g q 3 H 2 SO 4 (aq) + 2 Al(OH) 3 (aq) = 6 H 2 O(l) + Al 2 (SO 4 ) 3 (aq)

29 Write the equations that describe the solution reaction of sulfuric acid with the aluminum hydroxide. Separating aqueous ions: 3HSO (aq) = 6H (aq) + 3 SO 42 (aq) 2 Al(OH) 3 (aq) = 2Al 3+ (aq) + 6 OH - (aq) Al 2 (SO 4 ) 3 (aq) = 2Al 3+ (aq) + 3 SO 2-4 (aq) Putting into complete ionic equation: 6H + (aq) (q) + 3 SO 2-4 (aq) (q) + 2Al 3+ (aq) (q) + 6 OH - (aq) (q) = 6H 2 O(aq) + 2Al 3+ (aq) + 3 SO 2-4 (aq)

30 Write the equations that describe the solution reaction of sulfuric acid with the aluminum hydroxide. Removing spectators for net ionic: 6H + (aq)+ 6 OH - (aq) = 6H 2 O(aq) Taking to lowest common denominator to get our final net ionic equation: 1 H + (aq)+ 1 OH - (aq) = 1 H 2 O(aq)