Answer:
0.34 M
Explanation:
From the equation of reaction:
[tex]Pb(NO_3)_2 +2NaI --> 2NaNO_3 + PbI_2[/tex]
1 mole of lead (II) nitrate yields 1 mole of the PbI2 precipitate
mole of the precipitate = mass/molar mass
= 0.733/461.01 = 0.00159 mole
If 1 mole of lead (II) nitrate gives 1 mole of PbI2, then 0.00159 mole of PbI2 will require:
0.00159 x 1 = 0.00159 mole of lead (II) nitrate.
1 mole of lead (II) nitrate contains 2 moles of lead (II) ion
0.00159 mole of lead (II) nitrate will contain:
0.00159 x 2 = 0.00318 mole of lead (II) ion
volume of lead (II) nitrate = 93.3 mL = 0.0933 Liter
Hence, molarity of lead (II) ion = mole/volume
= 0.00318/0.0933
= 0.034 M
The molarity of lead(II) ion in the original solution is 0.034 M.
