Answer:
Pb^2+(aq) + 2F-(aq) → PbF2(s)
Explanation:
Step 1: Data given
sodium fluoride = NaF
lead(II)nitrate Pb(NO3)2
Step 2: The unbalanced equation
NaF(aq) + Pb(NO3)2(aq) → PbF2(s) + NaNO3(aq)
Step 3: Balancing the equation
NaF(aq) + Pb(NO3)2(aq) → PbF2(s) + NaNO3(aq)
On the left side we have 2x NO3 (in Pb(NO3)2), on the right side we have 1x NO3 (in NaNO3). To balance the amount of NO3 we hvae to multiply NaNO3 on the right side by 2.
NaF(aq) + Pb(NO3)2(aq) → PbF2(s) + 2NaNO3(aq)
On the left side we have 1x Na (in NaF), on the right side we have 2x Na (in 2NaNO3). To balance the amount of Na we have to multiply NaF on the left side by 2. Now the equation is balanced.
2NaF(aq) + Pb(NO3)2(aq) → PbF2(s) + 2NaNO3(aq)
Step 4: Calculate net ionic equation
The net ionic equation, for which spectator ions are omitted - remember that spectator ions are those ions located on both sides of the equation - will , after canceling those spectator ions in both side, look like this:
2Na+(aq) + 2F-(aq) + Pb^2+(aq) + 2NO3-(aq) → PbF2(s) + 2Na+(aq) + NO3-(aq)
Pb^2+(aq) + 2F-(aq) → PbF2(s)
