PART ONE: A solution of Na2CO3 is added dropwise to a solution that is 0.0402 M

Question

PART ONE:

A solution of Na2CO3 is added dropwise to a solution that is 0.0402 M in Y3+ and 0.000125 M in Cd2+.

The Ksp of Y2(CO3)3 is 1.03e-31.

The Ksp of CdCO3 is 1e-12.

(a) What concentration of CO32- is necessary to begin precipitation? (Neglect volume changes.) [CO32-] = _______M.

(b) Which cation precipitates first? Y3+ OR Cd2+

(c) What is the concentration of CO32- when the second cation begins to precipitate? [CO32-] =_______ M.

PART TWO:

(a) If the molar solubility of Sc(OH)3 at 25 oC is 9.52e-09 mol/L, what is the Ksp at this temperature?

Ksp = _____




(b) It is found that 1.42e-08 g of Al(OH)3 dissolves per 100 mL of aqueous solution at 25 oC. Calculate the solubility-product constant for Al(OH)3.

Ksp = ________




(c) The Ksp of YF3 at 25 oC is 8.62e-21. What is the molar solubility of YF3?

solubility = ______ mol/L

Explanation / Answer

PART-2:

a) Given that  molar solubility of Sc(OH)3 , s = 9.52 x 10-9 mol/L

Sc(OH)3 ---------> Sc3+ + 3 OH-

s 3s

Ksp= [Sc3+] [OH-]3

= s.(3s)3

= 27s4

= 27 x ( 9.52 x 10-9)4

= 221774 x 10-36

Ksp = 221774 x 10-36

b)

molar solubility of Al(OH)3 , s= (mass / molar mass) x (1/volume in Litres)

=  (1.42 x 10-8 g / 78 g/mol) x (1/0.1 L)

= 0.18x 10-8 mol/L

Al(OH)3 ---------> Al3+ + 3 OH-

s 3s

Ksp= [Al3+] [OH-]3

= s.(3s)3

= 27s4

= 27 x ( 0.18x 10-8)4

= 0.028 x 10-32

Ksp = 0.028 x 10-32

c) Given that

Ksp of YF3 = 8.62 x 10-21

Ksp = 27s4 , s= molar solubility

  8.62 x 10-21 =  27s4

s = ( 8.62 x 10-21/27)1/4

= 4.22 x 10-6 mol/L

Therefore,

molar solubility of YF3 =  4.22 x 10-6 mol/L

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