Question

Will a precipitate form if 100.0 mL of 2.5 x 10^-3 M Cd(NO₃)₂ and 75.0 mL of 0.0500 M NaOH are mixed at 25^o C? (Yes or No) Calculate the concentration of cadmiun ion in solution at equilibrium after the two solutions are mixed. (K_sp is 7.2 x 10^-15 for Cd(OH)₂ at 25^o C) Please show all work.

Answer 1

we know that

moles = molarity x volume (L)

so

moles of Cd(N03)2 = 2.5 x 10-3 x 0.1 = 2.5 x 10-4

moles of NaOH = 0.05 x 75 x 10-3 = 3.75 x 10-3

now

final volume = 100 + 75

final volume = 175 ml

now

conc = moles / volume

so

[NaOH ] = 3.75 x 10-3 / 175 x 10-3

[NaOH ] = 0.02143

[Cd(N03)2] = 2.5 x 10-4 / 175 x 10-3

[Cd(N03)2] = 1.4286 x 10-3

now

ionic product of Cd(OH)2 = [Cd+2] [OH-]^2

= [ 1.4286 x 10-3 ] [ 0.02143]^2

= 6.56 x 10-7

now

given Ksp = 7.2 x 10-15

as

ionic product > Ksp

Yes , a precipitate will form

Cd+2 + 2OH- ---> Cd(OH)2

now

[Cd+2] required = 0.5 x [OH-]

[Cd+2] required= 0.5 x 0.02143

[Cd+2] required = 0.010715

now

but

only 0.0014826 [Cd+2] is present

so

all the [Cd+2] is reacted and [OH-] in excess

so

the answers are

Yes , a precipitate will form

the [Cd+2]eq = 0 M