# 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?

Dec 25, 2014

The answer is $1.2 M$.

$P b {\left(N {O}_{3}\right)}_{2} \left(a q\right) + 2 K C l \left(a q\right) \to P b C {l}_{2} \left(s\right) + 2 K N {O}_{3} \left(a q\right)$

The complete ionic equation is

$P {b}^{2 +} \left(a q\right) + 2 N {O}_{3}^{-} \left(a q\right) + 2 {K}^{+} \left(a q\right) + 2 C {l}^{-} \left(a q\right) \to P b C {l}_{2} \left(s\right) + 2 {K}^{+} \left(a q\right) + 2 N {O}_{3}^{-} \left(a q\right)$

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

$P {b}^{2 +} \left(a q\right) + 2 C {l}^{-} \left(a q\right) \to P b C {l}_{2} \left(s\right)$

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 $P b {\left(N {O}_{2}\right)}_{2}$ and $K C l$ 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 $P b {\left(N {O}_{3}\right)}_{2}$'s molarity, which is defined as the number of moles of solute divided by the liters of solution, is $0.400 M$. THis means that

${n}_{P b {\left(N {O}_{3}\right)}_{2}} = C \cdot V = 0.400 M \cdot 30.0 \cdot {10}^{- 3} L = 0.012$ moles

The number of $K C l$ moles will then be

${n}_{K C l} = 2 \cdot {n}_{P b {\left(N {O}_{3}\right)}_{2}} = 2 \cdot 0.012 = 0.024$ moles

This makes the potassium chloride's molarity equal to

${C}_{K C l} = \frac{n}{V} = \frac{0.024 m o l e s}{20.0 \cdot {10}^{- 3} L} = 1.2 M$