Suppose we have a solution of lead nitrate, Pb(NO3)2(aq). A solution of NaCl(aq) is added slowly until no further precipitation of PbCl2(s) occurs. The PbCl2(s) precipitate is collected by filtration, dried, and weighed. A total of 12.79 grams of PbCl2(s) is ob- tained from 200.0 milliliters of the original solution. Calculate the molarity of the Pb(NO3)2(aq) solution.

0.23 mol/L

Explanation:

Step 1:

The balanced equation for the reaction. This is illustrated below:

Pb(NO3)2(aq) + 2NaCl(aq) —> PbCl2(s) + 2NaNO3(aq)

Step 2:

Determination of the number of in 12.79g of PbCl2. This is illustrated below:

Mass of PbCl2 = 12.79g

Molar Mass of PbCl2 = 207 + (2x35.5) = 207 + 71 = 278g/mol

Number of mole of PbCl2 =?

Number of mole = Mass/Molar Mass

Number of mole of PbCl2 = 12.79/278

Number of mole of PbCl2 = 0.046 mole

Step 3:

Determination of the number of mole of Pb(NO3)2 that reacted.

This is illustrated below:

From the balanced equation above,

1 mole of Pb(NO3)2 reacted to produce 1 mole of PbCl2.

Therefore, it will also take 0.046 mole of Pb(NO3)2 to react to produce 0.046 mole of PbCl2.

Step 4:

Determination of the molarity of Pb(NO3)2. This is illustrated:

Mole of Pb(NO3)2 = 0.046 mole

Volume of the solution = 200 mL = 200/1000 = 0.2 L

Molarity =?

Molarity is defined as the mole of solute per unit litre of solution. It is given by:

Molarity = mole of solute /Volume

Molarity of Pb(NO3)2 = 0.046/0.2

Molarity of Pb(NO3)2 = 0.23 mol/L

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