If 46.0 mL lead(II) nitrate solution reacts completely with excess sodium iodide
ID: 763536 • Letter: I
Question
If 46.0 mL lead(II) nitrate solution reacts completely with excess sodium iodide solution to yield 0.665 g precipitate, what is the molarity of lead(II) ion in the original solution?Explanation / Answer
If 38.0 mL lead(II) nitrate solution reacts completely with excess sodium iodide solution...more below? to yield 0.600 g precipitate, what is the molarity of lead(II) ion in the original solution? So lead(II) is going to be your limiter since the question says that it reacts COMPLETELY with EXCESS sodium iodide solution. The precipitate that is yielded has a mass of 0.600 grams. We need to go ahead and write out and balance the reaction: Pb(NO3)2 + 2NaI ----> 2NaNO3 + PbI2 Your precipitate will be PbI2. 0.600 grams of PbI2 will need to be multiplied by the reciprocal of molecular weight to get moles. Molecular weight of PbI2 is 461 g/mol. 0.600 g PbI2 (1 mol PbI2 / 461 g PbI2) = 0.0013015 mol PbI2 We can't stop there because we are trying to find the concentration of the lead(II) ion in the ORIGINAL solution. We will use our balanced reaction to obtain the number of moles of lead(II) ion. 0.0013015 mol PbI2 (1 mol Pb(NO3)2 / 1 mol PbI2) = 0.0013015 mol Pb(NO3)2 Molarity is M = (mols of solute / L of solution) 38.0 mL must be converted to L: 38.0 mL (1 L / 1000 mL) = 0.038 L solution So we will just put our moles of lead(II) over our liters of solution to obtain molarity: (0.0013015 mol Pb2+ / 0.038 L) = 0.0342504852 M With correct significant figures, the answer will be 0.0343 M or 3.43 x 10^-2 M