If 25.0 mL of sodium chloride solution reacted with excess lead (I) nitrate solution to produce 0.117 g of precipitate, what was the molarity of the sodium chloride solution?

First, write the balanced chemical equation. This is a double replacement reaction, as we have two aqueous solutions of salts reacting with each other. Our equation is:

NaCl (aq) + PbNO3 (aq) --> PbCl (s) + NaNO3 (aq)

This means our precipitate is PbCl.

Recall that molarity is the moles/L.

Solving

moles of NaCl

Since we made 0.117g of PbCl, and that we have used excess lead (I) nitrate, this means that the sodium chloride is our limiting reactant in this reaction. However much PbCl was formed will indicate how much NaCl we had used in this equation.

First, we need to find the molar mass of PbCl, which is the molar mass of Pb + molar mass of Cl, which is:

207.20 g/mol + 35.45 g/mol = 242.65 g/mol

In that case, we can set up the following stoich equation:

[tex]0.117 g* \frac{1 mol PbCl}{242.65g} * \frac{1 mol NaCl}{1 mol PbCl} = 0.000482[/tex] moles NaCl (rounded to 3 sig figs)

Molarity of NaCl solution

Now, we have the amount of moles of NaCl, but now we need to know the volume of the NaCl solution.

We were given that we had 25.0 mL of this solution, but remember that molarity is calculated in moles/L. So, convert 25.0 mL to liters.

[tex]25.0 mL * \frac{1 L }{1000 mL } = 0.0250 L[/tex]

Now, divide the moles by the liters.

[tex]\frac{0.000482 moles}{0.0250 L } = 0.0193 M[/tex]

The molarity of the NaCl solution is 0.0193M