Lead (II) nitrate (Pb(NO3)2) and sodium iodide (I) will break up into their ions in the solution and will react as shown:
Pb(NO3)2(aq) + 2 NaI(aq) → PbI2(s) + 2 NaNO3(aq)
If 38.5 mL of lead(II) nitrate solution reacts completely with excess sodium iodide solution to yield 0.628 g of precipitate, what is the molarity of lead(II) ion in the original solution?
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