1) A general chemistry 2 student creates a solution which is prepared by adding 0.029 mol of NaCl to 1.00 L of 0.0140 M Pb(NO3)2. Which of the following statements is correct? (Ksp = 1.7 x 10^-5 for PbCL2)
A. Both sodium nitrate and lead (II) chloride precipitate.
B. The data is not sufficient to predict precipitation.
C. Sodium nitrate precipitates until the solution is saturated
D. Lead(II) chloride precipitates until the solution is saturated.
E. The solution is unsaturated and no precipitate forms.
14) A student in General Chemistry 2 lab adds 0.20 mol of NaF to 1.00 L of 0.35 M cadmium nitrate, Cd(NO3)2. Which of the following statements is correct? Ksp = 6.44×10-3 for CdF2
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1) Ksp of Pb(NO3)2 = [Pb2+]*[NO3-]2 = s*(2s)2 = 4s3 ; where s = molar solubility
Thus, s = (Ksp/4)1/3 = 0.0162 M
Pb(NO3)2(aq) + 2NaCl(aq) ---------> 2NaNO3(aq) + PbCl2(s)
Moles of Pb(NO3)2 reacting = molarity*volume of solution in litres = 0.014
moles of NaCl reacting = 0.029
Clearly, moles of NaCl is in excess
Now, moles of PbCl2 formed = moles of Pb(NO3)2 reacting = 0.014 which is less than 's'
Thus, no precipitation will occur
Hence the correct option is :- (E)
2) Ksp of CdF2 = [Cd2+]*[F-]2 = s*(2s)2 = 4s3 ; where s = molar solubility
Thus, s = (Ksp/4)1/3 = 0.1172 M
Cd(NO3)2(aq) + 2NaF(aq) ---------> 2NaNO3(aq) + CdF2(s)
Moles of Cd(NO3)2 present = molarity*volume of solution in litres = 0.35
moles of NaF present = 0.2
Clearly, moles of NaF is limiting
Now, moles of CdF2 formed = (1/2)*moles of NaF reacting = 0.01 which is less than 's'
Thus, no precipitation will occur
Hence the correct option is :- (B)
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