3) Solid Nal is slowly added to a solution that is 0.010 M in Cu\' and 0.010 M i

Question

3) Solid Nal is slowly added to a solution that is 0.010 M in Cu' and 0.010 M in Ag. (a) Which compound will begin to precipitate first? (b) Calculate [Ag'] when Cul just begins to precipitate. (c) What percent of Ag remains in solution at this point? (For Cul. Kn" 5.l × 10-12 : for AgI, K-,-8.3 × 10-17) 4) Find the approximate pH range sutable for the separation of Fe" and Zn2·ions by precipitation of Fe(OH)3 from a solution 'hat is Initially 0.0 10 Min both Fe" and Z Assume a 99 percent precipitation of Fe OH). (For Fe(OH), Ky.-1.1 x 10-36; For Zn(( )H)2. K. 1 .8 × 10-14)

Explanation / Answer

3. From the data

a) AgI having lower Ksp than CuI Ksp would precipitate out of solution first.

b) let us calculate [I-] needed for Cu+ precipitation,

[I-] = Ksp/[Cu+] = 5.1 x 10^-12/0.01 = 5.1 x 10^-10 M

the [Ag+] when CuI begins to precipitate = Ksp/[I-] = 8.3 x 10^-17/5.1 x 10^-10 = 1.63 x 10^-7 M

c) percent [Ag+] left in solution = 1.63 x 10^-7 x 100/0.01 = 0.00163%

4) Ksp of Fe(OH)3 is lower than Ksp of Zn(OH)2, therefore, Fe(OH)3 will precipitate out of solution first.

with 99% precipitation of Fe3+, amount left in solution = 0.01 - 0.01 x 0.99 = 0.0001 M

[OH-] needed for this Fe3+ when Zn(OH)2 begins to precipitate = sq.rt.(Ksp/[Zn2+])

= (1.8 x 10^-14/0.0001) = 1.8 x 10^-10 M

[H+] = 1 x 10^-14/1.8 x 10^-10 = 0.000055 M

pH needed for Zn(OH)2 precipitation = -log(0.000055) = 4.25

amount Fe3+ precipitated = 0.0099 M

[OH-] needed for this precipitation = cube.rt.(Ksp/[Fe3+])

= cube.rt(1.1 x 10^-36/0.0099) = 4.81 x 10^-12 M

[H+] = 1 x 10^-14/4.81 x 10^-12 = 0.0021 M

pH = -log[H+] = 2.68

So the pH needed for Fe(OH)3 precipitation = 2.68

Related Questions

Navigate

• Browse (All)
• Browse 3
• Subjects

---

---

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