# Question #993ab

Apr 29, 2015

SIDE NOTE I'll start wy saying that your solubility product constants should be the other way around - the ${K}_{s p}$ for lead (II) chloride should be $1.19 \cdot {10}^{- 4}$ and the ${K}_{s p}$ for silver chloride should be $1.79 \cdot {10}^{- 10}$.

Since the other answer used them as they were given to you, I'll solve again in more detail using the correct solubility product constants.

Here's how you'd go about solving this problem. Take a look at the three equilibrium reactions you have to work with

$\textcolor{b l u e}{\left(1\right)} : P b C {l}_{2} r i g h t \le f t h a r p \infty n s P {b}^{2 +} + 2 C {l}^{-}$, ${K}_{3} = 1.19 \cdot {10}^{- 4}$

$\textcolor{b l u e}{\left(2\right)} : A g C l r i g h t \le f t h a r p \infty n s A {g}^{+} + C {l}^{-}$, ${K}_{4} = 1.79 \cdot {10}^{- 10}$

$\textcolor{b l u e}{\left(3\right)} : P b C {l}_{2} + 2 A {g}^{+} r i g h t \le f t h a r p \infty n s 2 A g C l + P {b}^{2 +}$

Notice that equilibrium $\textcolor{b l u e}{\left(2\right)}$ has the ions on the products' side, but equilibrium $\textcolor{b l u e}{\left(3\right)}$ has 2 silver cations on the reactants' side. This means that you must reverse the second equilibrium and multiply it by 2.

$2 A {g}^{+} + 2 C {l}^{-} r i g h t \le f t h a r p \infty n s 2 A g C l$

The solubility product constant will be

${K}_{\text{1/4}} = \frac{1}{K} _ {4}^{2} = \frac{1}{1.79} ^ 2 \cdot {10}^{20} = 3.12 \cdot {10}^{19}$

Add the two equilibrium reactions to get

$P b C {l}_{2} + 2 A {g}^{+} + \cancel{2 C {l}^{-}} r i g h t \le f t h a r p \infty n s P {b}^{2 +} + \cancel{2 C {l}^{-}} + 2 A g C l$

When you add equilibrium reactions together, the resulting ${K}_{s p}$ will be the product of the solubility product constants of the two separate reactions.

As a result, you'll get

${K}_{\text{final" = K_3 * K_"1/4}}$

${K}_{\text{final}} = 1.19 \cdot {10}^{- 4} \cdot 3.12 \cdot {10}^{19} = \textcolor{g r e e n}{3.7 \cdot {10}^{15}}$