A 250.0 mL sample of aqueous solution contains an unknown amount of dissolved NaBr. Excess aqueous Pb (NO3) 2is then added to this solution and a precipitate forms.

A. If the mass of precipitate recovered was 3.006 grams, what was the molar concentration of Br-ions in the original solution?

B. What is the chemical formula of the precipitate?

C. Write the net ionic equation for this precipitation reaction.

A. 0.0655 mol/L.

B. PbBr2.

C. Pb2 + (aq) + Br - - -> PbBr2 (s).

Explanation:

Balanced equation of the reaction:

Pb (NO3) 2 (aq) + 2NaBr (aq) - -> PbBr2 (s) + 2NaNO3 (aq)

A.

Number of moles

PbBr2

Molar mass = 207 + (80*2)

= 367 g/mol.

Moles = mass/molar mass

= 3.006/367

= 0.00819 mol.

Since 2 moles of NaBr reacted to form 1 mole of PbBr2. Therefore, moles of NaBr = 2*0.00819

= 0.01638 moles of NaBr.

Since, the ionic equation is

NaBr (aq) - -> Na + (aq) + Br - (aq)

Since 1 moles of NaBr dissociation in solution to give 1 mole of Br-

Therefore, molar concentration of Br-

= 0.0164/0.25 L

= 0.0655 mol/L.

B.

PbBr2

C.

Pb (NO3) 2 (aq) - -> Pb2 + (aq) + 2No3^2 - (aq)

2NaBr (aq) - -> 2Na + (aq) + 2Br - (aq)

Net ionic equation:

Pb2 + (aq) + 2Br - - -> PbBr2 (s)