Question

Lead ions can be removed from solution by precipitation with sulfate ions. Suppose a solution contains lead(II) nitrate.Write a complete ionic equation to show the reaction of aqueous lead(II) nitrate with aqueous potassium sulfate to form solid lead(II) sulfate and aqueous potassium nitrate.

Answer 1

The equation:
Pb(NO3)2 + K2SO4 = PbSO4 + 2KNO3
The ionic equation:
$Pb^{2+} + SO4^{2-} = PbSO4$

Answer 2

Answer: The net ionic equation is written below.

Explanation:

Net ionic equation of any reaction does not include any spectator ions.

Spectator ions are defined as the ions which does not get involved in a chemical equation. They are found on both the sides of the chemical reaction when it is present in ionic form.

The chemical equation for the reaction of lead (II) nitrate and potassium sulfate is given as:

$Pb(NO_3)_2(aq.)+K_2SO_4(aq.)\rightarrow PbSO_4(s)+2KNO_3(aq.)$

Ionic form of the above equation follows:

$Pb^{2+}(aq.)+2NO_3^-(aq.)+2K^+(aq.)+SO_4^{2-}(aq.)\rightarrow PbSO_4(s)+2K^+(aq.)+2NO_3^-(aq.)$

As, potassium and nitrate ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and are spectator ions.

The net ionic equation for the above reaction follows:

$Pb^{2+}(aq.)+SO_4^{2-}(aq.)\rightarrow PbSO_4(s)$

Hence, the net ionic equation is written above.