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Determine the molar solubility for Cr(OH)3 Ksp = 6.3x10^-31 ....

a) Set up the ICE table Cr(OH)3 (s) <-> Cr^3+ (aq) + 3OH^- (aq)

b) Ksp expression

c) Determine molar solubility

Determine the molar solubility for CrOH3 Ksp 63x1031 a Set up the ICE table CrOH3 s ltgt Cr3 aq 3OH aq b Ksp expression c Determine molar solubility class=
Determine the molar solubility for CrOH3 Ksp 63x1031 a Set up the ICE table CrOH3 s ltgt Cr3 aq 3OH aq b Ksp expression c Determine molar solubility class=
Determine the molar solubility for CrOH3 Ksp 63x1031 a Set up the ICE table CrOH3 s ltgt Cr3 aq 3OH aq b Ksp expression c Determine molar solubility class=

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pstnonsonjoku

The equilibrium constant shows the extent of dissolution of a substance in solution. The molar solubility of the compound here is 1.2 * 10^-8 M.

What is solubility product?

The solubility product is the equilibrium constant that shows the extent to which a substnace is dissolved in soution.

Now let us set up the ICE table as shown;

               Cr(OH)3 (s) <-> Cr^3+ (aq) + 3OH^- (aq)

I                                           0                 0

C                                          +x               +3x

E                                          x                   3x

Ksp = [Cr^3+] [OH^- ]^3

6.3x10^-31 = [x] [3x]^3

6.3x10^-31 = 27x^4

x = 4√6.3x10^-31 /27

x = 1.2 * 10^-8 M

Lear n more about solubility product: https://brainly.com/question/857770

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