# Will a precipitate form when we mix Pb(NO3)2(aq) with NaBr(aq) if the concentrations after mixing are 0.0150 M and 0.0350 M respectively?

Feb 2, 2016

No

#### Explanation:

If the two solutions are mixed a precipitation reaction occurs:

$P b {\left(N {O}_{3}\right)}_{2} \left(a q\right) + 2 N a B r \left(a q\right) \rightarrow P b B {r}_{2} \left(s\right) + 2 N a N {O}_{3} \left(a q\right)$

The ionic equation is:

$P {b}_{\left(a q\right)}^{2 +} + 2 B {r}_{\left(a q\right)}^{-} \rightarrow P b B {r}_{2 \left(s\right)}$

If we have a saturated solution of lead(II) bromide then the following equilibrium exists:

$P b B {r}_{2 \left(s\right)} r i g h t \le f t h a r p \infty n s P {b}_{\left(a q\right)}^{2 +} + 2 B {r}_{\left(a q\right)}^{-}$

For which:

${K}_{s p} = \left[P {b}_{\left(a q\right)}^{2 +}\right] {\left[B {r}_{\left(a q\right)}^{-}\right]}^{2} = 4 \times {10}^{- 5} {\text{mol"^2."l}}^{- 2}$ at ${25}^{\circ} \text{C}$

This is known as the solubility product.

The solid will be in equilibrium with its constituent ions and the solution will be saturated.

If the concentration of the ions increase to the point where ${K}_{s p}$ is exceeded then solid lead(II) bromide will form.

We can test this by putting in the concentration values given:

$\left[P {b}_{\left(a q\right)}^{2 +}\right] {\left[B {r}_{\left(a q\right)}\right]}^{2} = 0.015 \times {0.035}^{2} = 1.8 \times {10}^{- 5} {\text{mol"^2."l}}^{- 2}$

So you can see that this is less than the ${K}_{s p}$ value at ${25}^{\circ} \text{C}$ so I would suggest that a precipitate will not form under these conditions.