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Binding Energy

The K sp for ...

Question

The $K_{s p}$ for the following dissociation is $= 1.6 \times 10^{- 5}$ .
$P b C l_{2} (s) ⇌ P b^{2 +} (a q) + 2 c l^{-} (a q)$
Which of the following choices is correct for a mixture of 300 mL 0.134 M $P b (N O_{3})_{2}$ and 100 mL 0.4 M $N a C l$ ?

A

Q > K_{s p}

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B

Q < K_{s p}

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C

Q = K s p

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D

Not enough data provided

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Solution

The correct option is A $Q > K_{s p}$
Given $K_{s p} o f P b C l_{2} = 1.6 \times 10^{- 5}$
$P b (N O_{3})_{2} : m m o l e s = 300 m L \times 0.134 M = 40.2$
$N a C l : m m o l e s = 100 m L \times 0.4 M = 40$
This implies. $[P b]^{2 +} = \frac{40.2}{400} \approx 0.1 M$
$[C l]^{-} = \frac{40}{400} = 0.1 M$
$Q_{s p} = [P b^{2 +}] [2 C l^{-}]^{2} = 4 \times 10^{- 3} > K_{s p}$

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