Respuesta :
Answer: The precipitate will not be formed in the above solution.
Explanation:
The chemical equation for the reaction of potassium bromide and lead acetate follows:
[tex]2KBr(aq.)+Pb(CH_3COO)_2(aq.)\rightarrow PbBr_2(s)+2CH_3COOK(aq.)[/tex]
We are given:
Concentration of KBr = 0.013 M
Concentration of lead acetate = 0.0035 M
1 mole of KBr produces 1 mole of potassium ions and 1 mole of bromide ions
So, concentration of bromide ions = 0.013 M
1 mole of lead (II) acetate produces 1 mole of lead (II) ions and 2 moles of acetate ions
So, concentration of lead (II) ions = 0.0035 M
The salt produced is lead (II) bromide
The equation follows:
[tex]PbBr_2(s)\rightleftharpoons Pb^{2+}(aq.)+2Br^-(aq.)[/tex]
The expression of [tex]Q_{sp}[/tex] for above equation follows:
[tex]Q_{sp}=[Pb^{2+}]\times [Br^-]^2[/tex]
Putting values of the concentrations in above expression, we get:
[tex]Q_{sp}=(0.0035)\times (0.013)^2\\\\Q_{sp}=5.92\times 10^{-7}[/tex]
We know that:
[tex]K_{sp}[/tex] for lead (II) bromide = [tex]4.67\times 10^{-6}[/tex]
There are 3 conditions:
- When [tex]K_{sp}>Q_{sp}[/tex]; the reaction is product favored. (No precipitation)
- When [tex]K_{sp}<Q_{sp}[/tex]; the reaction is reactant favored. (Precipitation occurs)
- When [tex]K_{sp}=Q_{sp}[/tex]; the reaction is in equilibrium. (sparingly soluble)
As, the [tex]Q_{sp}[/tex] is less than [tex]K_{sp}[/tex]. The above reaction is product favored. This means that no salt or precipitate will be formed.
Hence, the precipitate will not be formed in the above solution.
