Question

In treating an industrial wastewater, we add NaOH to remove Cr3+ ions (atomic weight) = 52). The chemical equation for the dissolution of Cr(OH)3 is
Cr(OH)3 (s) ↔ Cr3+ + 3OH-    (Ksp = 6.7*10-31)
What is the final equilibrium concentration (in mg/L) of Cr3+ ions in a solution of Cr(OH)3 when the water has a pH of 5?

Answer 1

Given :

Ksp = 6.7 E -31

Lets calculate OH- concentration in the solution

pH = 5

pOH = 14-pH = 14 – 5 = 9

[OH-] = Antilog ( -pOH)

= Antilog ( -9)

=1.0 E-9 M

ICE chart

              Cr(OH)3 (s)    ↔    Cr3+      +       3OH-

I                                            o                   1.0E-9

C                                         +x                    +3x

E                                            x                  (1.0E-9+3x)

          Ksp = [Cr3+] [ OH-] ³

       6.7 E-31 = x (1.0E-9+3x)³

Value of ksp is very low so we neglect 3x in bracket.

       6.7 E-31 = x (1.0E-9)³

x = 0.00067 M

[Cr3+] = 0.00067 M

Lets assume volume = 1.0 L

Mol of [Cr3+] = 0.00067 mol

Mass of Cr3+= 0.00067 mol * molar mass of Cr3+

= 0.00067 mol * 51.996 g /mol

= 0.03483 g

Concentration = 0.03483 g/L

Now concentration in mg/L
= 0.03483 * 1000 mg / L

= 34.8 mg /L