Precipitation Rxns
Example 1
Reaction of aqueous solutions of aluminum hydroxide and nitric acid
Step 1: Write out formulas keeping in mind that aqueous solution means the substance is soluble and gets an (aq) label.
molecular: Al(OH)3 (aq) + HNO3 (aq)
Step 2: Decide what will combine with what. Cation from first takes anion from second (metathesis or exchange). Al3+ takes NO3- and H+ takes OH-. Because Al has three plus charge, need three nitrate ions. When H+ takes OH-, you get water.
Al(OH)3 (aq) + HNO3 (aq) Al(NO3)3 + H2O
Step 3: Decide if the products are soluble, or if they ppt using solubility chart on page 134.
Rule 2 says that all nitrates are soluble, so aluminum nitrate remains aqueous. Now add labels to molecular equation.
Al(OH)3 (aq) + HNO3 (aq) Al(NO3)3 (aq) + H2O (l)
Step 4: balance the equation. Start with metal and polyatomic first, then move to hydrogen and oxygen (H+ and OH-). One Al on reactant, need one Al on products. 3 nitrates on products need three nitrates on reactant (add 3 before HNO3 (aq)). Now do hydrogen and oxygen. 6 hydrogen in reactants, so I need three waters. 3 oxygen in hydroxide, 3 oxygen from water. The equation is balanced.
Al(OH)3 (aq) + 3 HNO3 (aq) Al(NO3)3 (aq) + 3 H2O (l)
Step 5: Write a complete ionic equation. Split everything that is aq into ions-leave solids and liquids together as molecules.
Al3+ (aq) + 3 OH- (aq) + 3 H+ (aq) + NO3-(aq) → Al3+ (aq) + 3 NO3- (aq) + 3 H2O (l)
Step 6: Spectator ions are those ions that are found as ions on both sides of the equation. They do not participate in the reaction. These can be removed from the CIE-when they are, you are left with a NIE. In this example, Al3+ and NO3- ions are the spectators. The NIE becomes
3 H+ (aq)+ 3 OH- (aq) → 3 H2O (l) or H+ (aq)+ OH- (aq) → H2O (l)
Example 2
Reaction of aqueous solutions of ammonium sulfate with calcium iodide
Step 1: write out formulas. Ammonium is NH4+ and sulfate is SO42-. So you need two ammoniums for every sulfate. Calcium is Ca2+ and iodide is I-. So you need two iodide ions for each calcium ion.
Molecular: (NH4)2SO4 (aq) + CaI2 (aq) →
Step 2: cation from the first combines with the anion from the second and vice versa, ammonium with iodide and calcium with sulfate. Ammonium is plus one, iodide is minus one, so they combine in a one to one ratio. Calcium is plus two, sulfate is minus two, so they combine in a one to one ratio.
(NH4)2SO4 (aq) + CaI2 (aq) → NH4I + CaSO4
Step 3: decide whether they form a ppt or if they remain soluble and add labels.
Rule 1: all ammonium salts are soluble.
Rule 4: all sulfates are soluble except with calcium
(NH4)2SO4 (aq) + CaI2 (aq) → NH4I (aq) + CaSO4 (s)
Step 4: Balance the equation. There are two ammoniums as reactants, need two ammonium as product. Put a two in front of ammonium iodide. This takes care of balancing.
(NH4)2SO4 (aq) + CaI2 (aq) → 2 NH4I (aq) + CaSO4 (s)
Step 5: Write a complete ionic equation. Separate all aq species into ions and leave solids/liquids together.
2 NH4+ (aq) + SO42- (aq) + Ca2+ (aq) + 2 I- (aq) → 2 NH4+ (aq) + 2 I- (aq) + CaSO4 (s)
Step 6: Remove spectators to write NIE.
Ca2+ (aq) + SO42- (aq) → CaSO4 (s)
Example 3
Reaction of aqueous solutions of sodium hydroxide with barium carbonate
NaOH (aq) + BaCO3 (aq) → Na2CO3 (aq) + Ba(OH)2 (aq)
Because everything stays soluble, they are all spectator ions and there is no reaction, NR.
Acid-Base reactions
Lets look at the reaction of sulfuric acid with sodium hydroxide.
H2SO4 (aq) + NaOH (aq) →
Step 2: Again, the cation from the first takes the anion from the second and vice versa. The cation in the first is H+, which takes the anion from the second, OH-. This makes water. The cation in the second is Na+, which takes the anion from the second, SO42-. This makes sodium sulfate in a two to one ratio. Notice-this is a salt and water=Neutralization!
H2SO4 (aq) + NaOH (aq) → H2O + Na2SO4
Step 3: Decide if anything precipitates or not and add labels.
Rule 1: all first row cations are soluble. So, sodium sulfate is aq.
H2SO4 (aq) + NaOH (aq) → H2O (l) + Na2SO4 (aq)
Step 4: Balance. There is one sulfate on each side of the rection-good. One sodium on reactant side, two on product. Need a two in front of sodium hydroxide. Now balance H and O. 4 hydrogen on reactant, need two water on product side. Two oxygen from hydroxide on reactant side and 2 oxygen from water on product side-good.
H2SO4 (aq) + 2 NaOH (aq) → 2 H2O (l) + Na2SO4 (aq)
Step 5: Write CIE-separate all aq things, leave solids and liquids together.
2 H+ (aq) + SO42- (aq) + 2 Na+ (aq) + 2 OH- (aq) → 2 H2O (l) + 2 Na+ (aq) + SO42- (aq)
Step 6: Remove spectators to write NIE. The spectators are sodium and sulfate ions.
2 H+ (aq)+ 2 OH- (aq) → 2 H2O (l) or H+ (aq)+ OH- (aq) → H2O (l)
Molarity:
Question: How would you concentrate 100 mL of 5 M HCl to get a 2 M HCl solution?
5M(100mL) = 2M(x mL)
x mL = 250 mL
You would take 100 mL of the 5M solution and dilute to 250 mL.
This equation only works when you want the concentration or volume of the same solution. If you need to convert between two different solutions, you must consider mol:mol ratio-
Question: How many mL of 0.25 M HCl are needed to neutralize 30.0 mL of 0.50 M Ba(OH)2?
First, write a balanced equation: 2 HCl (aq) + Ba(OH)2(aq) → 2 H2O (l) + BaCl2(aq)
NIE: H+ (aq) + OH-(aq) → H2O (l)
This is a neutralization rxn where you produce a salt and water.
0.50 mol Ba(OH)2 0.030 L 2 mol HCl L______1000 mL
L 1 mol Ba(OH)2 0.25 mol HCl 1 L
IV. Redox Reactions
R1. Determine the oxidation number of sulfur in (a) S8 (b) SCl2 (c) Na2SO3(d) SO42-
(a)0 (b) +2 (c) +4 (d) +6
R2. Write a balanced ME and NIE for the reaction of aluminum with hydrobromic acid.
ME: 2 Al (s) + 6 HBr (aq) 2 AlBr3 (aq) + 3 H2 (g)
NIE: 2 Al (s) + 6 H+ (aq) 2 Al+3 (aq) + 3 H2 (g)
R3. Which of the following metals will be oxidized by Pb(NO3)2: Zn, Cu, Fe?
Any metal > than Pb on the activity series
R4. Mn2+ (aq) + NaBiO3 (s) → Bi3+ (aq) + MnO4 (aq). Which species is oxidized and which is reduced?
Since Mn goes from +2 to +7, Mn+2 is oxidized
Since Bi goes from +5 to +3, NaBiO3 is reduced