Question

A solution containing potassium bromide is mixed with one containing lead acetate to form a solution that is 0.013 M in KBr and 0.0035 M in Pb(C2H3O2)2 . Part A Will a precipitate form in the mixed solution

Answer 1

Answer: The precipitate will not be formed in the above solution.

Step-by-step explanation:

The chemical equation for the reaction of potassium bromide and lead acetate follows:

`2KBr(aq.)+Pb(CH_3COO)_2(aq.)\rightarrow PbBr_2(s)+2CH_3COOK(aq.)`

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:

`PbBr_2(s)\rightleftharpoons Pb^(2+)(aq.)+2Br^-(aq.)`

The expression of
`Q_(sp)` for above equation follows:

`Q_(sp)=[Pb^(2+)]* [Br^-]^2`

Putting values of the concentrations in above expression, we get:

`Q_(sp)=(0.0035)* (0.013)^2\\\\Q_(sp)=5.92* 10^(-7)`

We know that:

`K_(sp)` for lead (II) bromide =
`4.67* 10^(-6)`

There are 3 conditions:

• When
  `K_(sp)>Q_(sp)`; the reaction is product favored. (No precipitation)
• When
  `K_(sp)<Q_(sp)`; the reaction is reactant favored. (Precipitation occurs)
• When
  `K_(sp)=Q_(sp)`; the reaction is in equilibrium. (sparingly soluble)

As, the
`Q_(sp)` is less than
`K_(sp)`. 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.