1) A general chemistry 2 student creates a solution which is prepared by adding

Question

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

A. Cadmium fluoride precipitates until the solution is saturated.

B.

The solution is unsaturated and no precipitate forms.

C.

The solubility of cadmium fluoride is increased by the presence of additional fluoride ions.

D.

One must know Ksp for cadmium nitrate to make meaningful predictions on this system.

E.

The presence of NaF will raise the solubility of Cd(NO3)2.

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

A. Cadmium fluoride precipitates until the solution is saturated.

B.

The solution is unsaturated and no precipitate forms.

C.

The solubility of cadmium fluoride is increased by the presence of additional fluoride ions.

D.

One must know Ksp for cadmium nitrate to make meaningful predictions on this system.

E.

The presence of NaF will raise the solubility of Cd(NO3)2.

Explanation / Answer

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)

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.