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