A student followed the procedure of this experiment to determine the K_sp of zin

Question

A student followed the procedure of this experiment to determine the K_sp of zinc (II) iodate, Zn(IO_3)_2 Solutions of Z_n (NO_3)_2 of known concentrations were titrated with 0.200 MKIO_3 solutions to the first appearance of a white precipitate. For each of the zinc nitrate solution concentrations below, calculate the expected concentration of iodate that would be required to initiate precipitation of zinc (II) iodate. Show all calculations. (Assume that K_sp = 3.9 times 10^-6 at. 25^degree for zinc(II) iodate). 0.100 M Zn(NO_3)_2 0.0100 M Zn (NO_3) _2

Explanation / Answer

a)

Zn(NO3)2(aq) --> Zn+2 + 2NO3-(aq)

[Zn+2] = 0.1 M

[NO3-] = 0.1*2 = 0.2 M

now, for solution:

Zn(IO3)2(s) <--> Zn+2(aq) + 2IO3-(aq)

Ksp = [Zn+2][IO3-]^2

3.9*10^-6 = 0.1 * (2S)^2

S = sqrt((3.9*10^-6)/(0.1))

S = 0.006244 mol of Zn(IO3)2 per liter

b)

similar:

Zn(NO3)2(aq) --> Zn+2 + 2NO3-(aq)

[Zn+2] = 0.01 M

[NO3-] = 0.01*2 = 0.02 M

now, for solution:

Zn(IO3)2(s) <--> Zn+2(aq) + 2IO3-(aq)

Ksp = [Zn+2][IO3-]^2

3.9*10^-6 = 0.1 * (2S)^2

S = sqrt((3.9*10^-6)/(0.01))

S = 0.0197484 mol of Zn(IO3)2 per liter

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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