what is the molarity of 20.0 ml of a KCl solution that reacts completely with 30.0 ml of a 0.400 M Pb(NO3)2 solution?

The answer is `1.2M`.

First, start with the formula equation

`Pb(NO_3)_2(aq) + 2KCl(aq) -> PbCl_2(s) + 2KNO_3(aq)`

The complete ionic equation is

`Pb^(2+)(aq) + 2NO_3^(-)(aq) + 2K^(+)(aq) + 2Cl^(-)(aq) -> PbCl_2(s) + 2K^(+)(aq) + 2NO_3^(-)(aq)`

The net ionic equation, obtained after eliminating spectator ions (ions that can be found both on the reactants, and on the products' side), is

`Pb^(2+)(aq) + 2Cl^(-)(aq) -> PbCl_2(s)`

According to the solubility rules, lead (II) chloride can be considered to be insoluble in water.

Notice that we have a `1:2` mole ratio between `Pb(NO_2)_2` and `KCl` in the formula reaction; this means that 2 moles of the latter are needed for every 1 mole of the former for a complete reaction.

We know that `Pb(NO_3)_2`'s molarity, which is defined as the number of moles of solute divided by the liters of solution, is `0.400M`. THis means that

`n_(Pb(NO_3)_2) = C * V = 0.400M * 30.0 * 10^(-3)L = 0.012` moles

The number of `KCl` moles will then be

`n_(KCl) = 2 * n_(Pb(NO_3)_2) = 2 * 0.012 = 0.024` moles

This makes the potassium chloride's molarity equal to

`C_(KCl) = n/V = (0.024 mol es)/(20.0 * 10^(-3)L) = 1.2M`