I NEED HELP ASAP

Question

which of the following Ksp is more likely to favor reactants?

1) Colbalt (II) hydroxide Co(OH)2; Ksp 5.92x10^-15

or

2) lead(II)chloride PbCI2 , Ksp 1.70x10^-5

Answer 1

Hello, Leya,

Let's look at the equations defined by the K_sp.

For a solubility of a compound that dissolves in water, we can write:

aA(s) = cC(aq)+dD(aq)

The solubility coefficient is defined as:

K_sp=[C]^c[D]^d

Now let's look at each of the questions:

1) Colbalt (II) hydroxide Co(OH)2; Ksp 5.92x10^-15

Co(OH)₂ = Co⁺⁺ + 2OH^-

K_sp = [Co⁺⁺]*[OH^-]²

If the concentration of Co⁺⁺ is x, then OH^- will be 2x. So we have:

Ksp = x³

5.92x10^-15 = x³

x = 1.8x10^-5

2) lead(II)chloride PbCI₂ , Ksp 1.70x10^-5

PbCL₂ = Pb⁺⁺ + 2Cl^-

K_sp = [Pb⁺⁺]*[Cl^-]²

We have the same situation with the product concentrations, so we can again write:

K_sp = x³

1.70x10^-5 = x³

x = 2.57x10^-2

When we compare the two, we can see that the larger K_sp favors the products, while the lower K_sp favors the reactants.(Remember that 10^-15 is smaller than 10^-5, despite the initial impression :) )

Bob