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)

the maximum mass of nh3 is nh3 in grams

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