Consider the following hydroxide salts: Zn(OH) 2 : Ksp = 4.5 x 10 -17 and La(OH)

Question

Consider the following hydroxide salts: Zn(OH)2 : Ksp = 4.5 x 10-17 and La(OH)3 : Ksp = 2.0 x 10-21

a) Will Zn(OH)2 precipitate when 100.0 mL each of 0.0010 M ZnCl2 and 0.00100M KOH are mixed? Attach your calculations.     a.) yes/ no =_______________

b ) Predict, without calculation, whether the molar solubility (in M) of Zn(OH)2 is greater than, less than, or about the same as La(OH)3.      b.) rel. sol. prediction=____________

c) Calculate the molar solubility (in M) of Zn(OH)2 and La(OH)3 in water. Is it possible to predict the relative solubility of two salts in water by direct comparison of their K sp values?

       c) solubility =_____________    Zn(OH)2 in M =_____________ La(OH)3 in M =_______________ Yes/No = ______________

d ) What is the solubility (in M) of zinc hydroxide at pH 9.36 (pH held at 9.36) ? Compare to your answer in part c (higher, lower, same). Hint: What is [OH-] at pH 9.36?

      d) solubility in M = ____________ Compare:__________________________________________

Explanation / Answer

a) ZnCl2 + 2 KOH ----------> Zn(OH)2(s) + 2HCl

100x 0.001 100x0.001

[Zn+2] = 100x0.001/200 = 0.0005

[OH-] = 100x0.001/200 =0.0005

Zn(OH)2 ------------> Zn+2 +2OH-

Ksp = [Zn+2] [OH-]2

ionic product = (0.0005)3

= 1.25 x 10-12

Thus ionic product > Ksp , henc precipitation occurs.

b) AS the Ksp of La(OH)3 is less than that of Zn(OH)2 , zinc hydroxide should be more solublethan lanthanum hydroxide.

c) Zn(OH)2 -------> Zn+2 + 2OH-

- s 2s

Ksp = sx (2s)2

= 4s3

=4.5x10-17

s3 = 4.5x10-17/ 4

= 2.24x10-6M

La(OH)3 ------> La+3 +3OH-

- s 3s

Ksp = s (3s)3

= 27s4 = 2.0x10-21

s4 = 2x10-21/27

=2.933x10-6 M

Thus la(OH)3 is slightly more soluble than Zn(oH02

d) When pH = 9.36 then pOh = 14-9.36 =4.64

and [OH-] = 2.29x 10-5

Zn(OH)2 -------> Zn+2 + 2OH-

s 2.29x10-5

Ksp =  sx(2.29x10-5)2 = 4.5x10-17

solubility = 8.581x10-8M

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