Answer:
The molarity of the lead(II) ion in the original solution is 0.03M
Explanation:
<u>Step 1:</u> 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
<u>Step 2:</u> Calculate moles of PbI2
Moles of PbI2 = 0.862g / 461.01 g/mole
moles of PbI2 = 0.00187 moles
<u>Step 3:</u> 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+
<u>Step 4:</u> 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
