Which of the following will be observed after 10.6 g of lead(II) nitrate and 1.4
ID: 831321 • Letter: W
Question
Which of the following will be observed after 10.6 g of lead(II) nitrate and 1.47 g of NaCl are added to 2.0 L of water without changing the volume significantly? (Ksp = 1.6 x 10-5.)
- Both lead(II) nitrate and NaCl will dissolve, but no precipitate will form. - All of the NaCl will dissolve, but some lead(II) nitrate will be left as a solid. - None of the lead(II) nitrate will dissolve, but all of the NaCl will dissolve. - Lead(II) chloride will precipitate. - All of the lead(II) nitrate will dissolve, but some NaCl will be left as a solid.Explanation / Answer
The answer is:
Both lead(II) nitrate and NaCl will dissolve, but no precipitate will form.
Pb(NO3)2 + 2 NaCl => PbCl2 + 2 NaNO3
PbCl2 <=> Pb2+ + 2 Cl-
Moles of Pb2+ = moles of Pb(NO3)2 = mass/molar mass of Pb(NO3)2
= 10.6/331.22 = 0.03200 mol
Moles of Cl- = moles of NaCl = mass/molar mass of NaCl
= 1.47/58.443 = 0.02515 mol
[Pb2+] = moles of Pb2+/total volume
= 0.03200/2.0 = 0.01600 M
[Cl-] = moles of Cl-/total volume
= 0.02515/2.0 = 0.01258 M
Ionic product Qsp = [Pb2+][Cl-]^2
= 0.01600 x 0.01258^2 = 2.5 x 10^(-6)
Since Qsp < Ksp (= 1.6 x 10^(-5)) => no precipitate will form
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