If 62.3 ml of lead(ii) nitrate solution reacts completely with excess sodium iodide solution to yield 0.862 g of precipitate, what is the molarity of lead(ii) ion in the original solution?

The molarity of the lead(II) ion in the original solution is 0.03M

Explanation:

Step 1: The balanced reaction

Pb(NO3)2(aq) + 2 NaI aq) → PbI2(s) + 2 NaNO3 (aq)

Pb2+ + 2I- →PbI2

This means that for 1 mole Pb2+ consumed, there is needed 2 moles of I- to produce 1 mole of PbI2

Step 2: Calculate moles of PbI2

Moles of PbI2 = 0.862g / 461.01 g/mole

moles of PbI2 = 0.00187 moles

Step 3: calculate moles of Pb2+

For 1 mole of PbI2 produced we need 1 mole of Pb2+

This means for 0.00187 moles of PbI2 we need 0.00187 moles of Pb2+

Step 4: Calculate the molarity of the Pb2+ ion

Molarity of Pb2+ = moles of Pb2+ / volume =

Molarity of Pb2+ = 0.00187 moles / 62.3*10^-3

Molarity of Pb2+ ion = 0.03M

The molarity of the lead(II) ion in the original solution is 0.03M

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the answers would be

the first one and the second one