Answer
a. $Ba^{2+}(aq) + SO_4^{2-}(aq) => BaSO_4(s)$
b. $Pb^{2+}(aq) + 2Cl^-(aq) => PbCl_2(s)$
c. No reaction occurs
d. No reaction occurs
e. $Cu^{2+}(aq) + 2OH^-(aq) => Cu(OH)_2(s)$
Work Step by Step
a.
Examining our solubility rules and swapping the cations which each other, we see that the only precipitate that would form would be $BaSO_4$. Hence, we know that the barium ions and the sulfate ions would combine to form it making our net ionic equation.
b.
We do the same thing in part a and realize that $PbCl_2$ is insoluble and is the precipitate. 2 chloride ions and one lead ion would form the solid, hence making the net ionic equation with $Cl^-$ having a coefficient of 2.
c.
From our solubility rules, we see that all of the salts are soluble in water. Hence, there would be no net ionic equation because no precipitate is formed.
d.
From the previous question c, the same thing occurs with no precipitate forming. There will be no reaction.
e.
Using our solubility rules, copper(II) hydroxide is the only insoluble salt. It requires 2 moles of hydroxide ions for every 1 mole of copper(II) ions. Hence, the net ionic equation looks like that with a coefficient of 2 in front of $OH^-$.
