Question

If 38.5 mL of lead (II) nitrate solution reacts completely with excess sodium iodide to yield 0.0628 g of precipitate, what is the molarity of lead (II) in the original solution?

Answer 1

In this question, we have the following reaction occurring:

Pb(NO3)2 (aq) + 2 NaI (aq) -> PbI2 (s) + 2 NaNO3 (aq)

The question says that we have:

38.5 mL of Pb(NO3)2

and 0.0628 grams of the precipitate, and the precipitate will be the solid that is formed on the products side, therefore we have 0.0628 grams of PbI2

To find the molarity of Pb(NO3)2, we have the volume in liters, which we already have, and the number of moles, which we will find now, but first we need to find the number of moles of PbI2, we will use the molar mass, 461.01g/mol, and the given mass, 0.0628 grams

461.01g = 1 mol

0.0628g = x moles

x = 1.36*10^-4 moles of PbI2

Now we have the number of moles of PbI2, and according to the molar ratio in the reaction, 1:1 of molar ratio, if we have 1.36*10^-4 moles of PbI2, we will also have 1.36*10^-4 moles of Pb(NO3)2, now using the molarity formula:

M = n/V

M = 1.36*10^-4/0.0385

M = 0.0035

The Molarity of Pb(NO3)2 is 0.0035 M