6. Ionic equations
Double replacement reactions and other reactions of ions in aqueous solution are usually represented by that are known as “net ionic equations”. To write a net ionic equation, one firstly writes an equation in which all strong electrolytes are shown as dissociated ions in solution (without steps, of course). For example:
1. Formation of slightly soluble substance:
CuSO4 + 2NaOH = Cu(OH)2 + Na2SO4
C
u2+
+ SO42-
+ 2Na+
+ 2OH-
= Cu(OH)2
+ 2Na+
+ SO42-
-
this is a complete ionic equation.
Cu2+ + 2OH- = Cu(OH)2 - this is net ionic equation.
2. Formation of weak electrolyte:
a) HCl + KOH = KCl + H2O,
H+
+ Cl-
+K+
+ OH-
= K+
+ Cl-
+ H2O,
H+ + OH- = H2O.
b ) СH3COONa + HCl = CH3COOH + NaCl,
CH3COO- + Na+ + H+ + Cl- = CH3COOH + Na+ + Cl-,
CH3COO- + H+ = CH3COOH.
3. Formation of gas:
а
)
K2CO3
+ 2HNO3
= 2KNO3
+ CO2
+ H2O,
2K++ CO32- + 2H+ + 2NO3- = 2K+ + 2NO3- + CO2 + H2O,
CO32- + 2H+ = CO2 + H2O.
b) FeS + 2HCl = FeCl2 + H2S.
I
ron
(II) Sulfide is slightly
soluble
and
Hydrogen
Sulfide is a weak electrolyte (gas). Therefore:
F eS + 2H+ + 2Cl- = Fe2++ 2Cl- + H2S,
FeS + 2H+ = Fe2++ H2S.
Interactions in the solutions of electrolytes are realized completely if in the result the following will be formed:
а) weak electrolyte;
б) sediments;
в) gases.
PRACTICE PROBLEMS
Write the dissociation equation for the ions of the following compounds:
H2SO4, HClO4, Ca(OH)2, AlCl3, Fe2(SO4)3, K2CO3, Bi(NO3)3, Na2HPO4, NH4HCO3, Co(NO3)3.
Write molecular, complete and net ionic equations:
BaCl2 + Fe2(SO4)3 →
Cu(NO3)2 + Na2CO3 →
Na2SiO3 + H2SO4 →
FeCl3 + KOH →
H3PO4 + Ba(OH)2 →
Cu(OH)2 + HCl →
Na2HPO4 + CaCl2 →
Which of two reactions does proceed in the solution up to the end and why? Draw the corresponding molecular, complete and net ionic equations:
Calcium Chloride + Magnesium Nitrate →
Sodium Carbonate + Zinc Chloride →
Lithium Sulfate + Potassium hydroxide →
Iron (II) Sulfate + Sodium hydroxide →
Iron (III) hydroxide + Nitric Acid →
Draw molecular equations of the reactions between substances interacted in aqueous solution according to the following scheme:
Ba2+ + SO42- → BaSO4;
Fe(OH)3 + 3H+ → Fe3+ + 3H2O;
CaCO3 + 2H+ → Ca2+ + CO2 + H2O;
3Mg2+ + 2PO43- → Mg3(PO4)2;
Ca2+ + 2 F- → CaF2.
Laboratory training
Experiment 1. Slightly soluble substances production
A) Put separately 5-6 drops of Na2SO4, (NH4)2CO3, Na3PO4 solutions into three test-tubes and add 3-4 drops of BaCl2 solution into each of them. What does happen? Draw molecular and net ionic equations.
B) Put 5-6 drops of CuSO4 and Cu(NO3)2 solutions into two test-tubes. Add 4-5 drops of NaOH solution into one test-tube and 4-5 drops of Ba(OH)2 into another. What does happen? Draw molecular and net ionic equations.
Experiment 2. Weak electrolytes production
Put 5-6 drops of CH3COONa and Pb(CH3COO)2 solutions into two test-tubes, add 5-6 drops of 2 N HCl or 2 N H2SO4 into each of them. Heat the mixture and identify by odor the acetic acid vapor isolation. Draw molecular and net ionic equations.
Experiment 3. Gases production
A) Put 5-6 drops of NH4Cl into the test-tube; add the same volume of NaOH. Heat the mixture on water bath. Identify by odor the kind of gas. Draw molecular and net ionic equations.
B) Put 0,01 - 0,02 g of CaCO3 or BaCO3 solid salt into the test-tube and add 4-5 drops of 2 N HCl solution. What does happen? Draw molecular and net ionic equations.
Experiment 4. Properties of amphoteric hydroxides and their production
Produce Zn(OH)2 precipitation in two test-tubes by adding drops of ZnSO4 solution to 1-2 drops of NaOH solution to forming of sediment. Add 5-6 drops of 2 N HCl into one test-tube and 5-6 drops of 2 N NaOH into another. What does happen with Zn(OH)2 precipitation? Draw molecular and net ionic reactions of Zn(OH)2 production and its dilution in acid and alkali.