Guide
1
Consider the example of the formation of sparingly soluble compounds.

Na2SO4 + BaCl2 = BaSO4 + 2NaCl

or variant in an ionic form:

2Na + + SO42- + Ba2 ++ 2Cl- = BaSO4 + 2Na + + 2Cl-
2
Note that reacted only ionsbarium and sulfate ions, the status of other ions have not changed, so this equation can be written in abbreviated form:

Ba2 + + SO42- = BaSO4
3
In addressing ionic equations, you must observe the following rules:

- the same ions of the two parts are excluded;

- it should be remembered that the amount of electric charges in the left side of the equation must be
equal to the amount of electrical charges on the right side of the equation.
4
Examples:

Write ionic equations of the reactions of interaction between the aqueous solutions of the following substances: a) HCl and NaOH;b) AgNO3 and NaCl;a) H2SO4 and K2SO3;g) SN3SOOH and NaOH.

decision.Write down the interaction of these substances in molecular form:

a) HCl + NaOH = NaCl + H2O

b) AgNO3 + NaCl = AgCl + NaNO3

in) K2CO3 + H2SO4 = K2SO4 + CO2 + H2O

g) SN3SOOH +NaOH = CH3COONa + H2O
5
Note that the interaction of these substances may, for as a result of the binding of ions to form a weak electrolyte (H 2 O) or soluble substance (AgCl), or a gas (CO2).
6
In the case of variant d) the reaction proceeds toward greater ion binding, i.e., formation of water even though there are two weak electrolyte (acetic acid and water).But this is because the water - a weak electrolyte.
7
Eliminating identical ions from the left and right sides of the equation (in the case of option a) - ions of sodium and chlorine in the case of b) - sodium ions and nitrate ions, in the case) - potassium ions andsulfate ions), d) - sodium ions, to receive these ionic equations:

a) H + + OH- = H2O

b) Ag + + Cl- = AgCl

in) CO32- + 2H + = CO2 + H2O

g) SN3SOOH + OH- = CH3COO- + H2O