The balanced equation that describes the reaction between lead (II) nitrate and sodium iodide to produce lead (II) iodide and sodium nitrate is expressed Pb(NO3)2 + 2NaI → PbI2 + 2NaNO3. All nitrates are soluble so lead (II) iodide is the precipitate. given 0.0628 g PbI2 is equal to 1.3623x10^-4 moles PbI2. This is equal to 1.3623x10^-4 moles also of Pb(NO3)2. Hence the molarity of the solution is 1.3623x10^-4 moles per 0.0385 L or 0.003538 M.
