A solution of Na 3 AsO 4 is added dropwise to a solution that is 0.0829 M in Bi

ID: 906198 • Letter: A

Question

A solution of Na3AsO4 is added dropwise to a solution that is 0.0829 M in Bi3+ and 4.26e-05 M in Cd2+.

The Ksp of BiAsO4 is 4.43e-10.
The Ksp of Cd3(AsO4)2 is 2.2e-33.

(a) What concentration of AsO43- is necessary to begin precipitation? (Neglect volume changes.)

[AsO43-] = M.

(b) Which cation precipitates first?

Bi3+

Cd2+

I am having trouble getting the answer for this one. An explanation would be helpful Thank You.

(a) What concentration of AsO43- is necessary to begin precipitation? (Neglect volume changes.)

Solution :- KSp of the Cd3(AsO4)2 is 2.2e-33 which very less

So using the ksp equation we can calculate the concentration of the AsO4^3-

Cd3(AsO4)2 ------> 3Cd^2+ + 2AsO4^3-

3x 2x

Ksp = [Cd^2+]^3 [AsO4^3-]^2

Ksp = [3x]^3 [2x]^2

2.2*10^-33 = [3x]^3 [2x]^2

2.2*10^-33 = 27x^3 * 4x^2

2.2*10^-33 = 108x^5

2.2*10^-33 / 108 =x^5

2.037*10^-35 =x^5

Taking the 5th root of both side we get

1.15*10^-7 =x

So the concentration of the AsO4^3- needed to begin the precipitation is 1.15*10^-7 M

(b) Which cation precipitates first?

Bi3+

Cd2+

Since the ksp of the Cd^2+ is very small therefore it will start precipitating first.

Solution

The Ksp of BiAsO4 is 4.43e-10.
so

BiAsO4 ------ > Bi^3+ + AsO4^3-

X x

Ksp = [x][x]

4.43*10^-10 = [x][x]

4.43*10^-10 = x^2

Taking square root of the both side we get

2.10*10^-5 =x

So the concentration of the AsO4^3- = 2.10*10^-5 M when the second cation start to precipitate.

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