Express ksp in terms of molar solubility, x, for a salt with a formula of ab(s) ab2(s) ab3(s) a3b2(s)

See explanation below

Explanation:

The previous answer is wrong. To do this, we need to write the equilibrium equation for each species there, and then, write the Ksp in terms of molar solubility.

Molar solubility will be called as "x". So let's write the chemical equation for each species:

a) AB(s)

In this case, the equation would be:

AB(s) <> A(aq) + B(aq)

Doing an ICE chart we have:

        AB(s) <> A + B

I:                             0    0

C:                           +x   +x

E:                            x     x

In this case Ksp would be:

Ksp = [A] [B]

However, A and B are X so:

Ksp = x * x

Ksp = x²

This is the Ksp in terms of molar solubility for AB(s)

b) AB2(s)

In this case, we have two atoms of B so equation would be like this:

AB2(s) <> A(aq) + 2B(aq)

ICE chart:

          AB2(s) <> A(aq) + 2B(aq)

I:                                   0           0

C:                                 +x         +2x

E:                                  x           2x

Therefore Ksp:

Ksp = [A] [B]² ---> replacing with x:

Ksp = x * (2x)²

Ksp = x * 4x²

Ksp = 4x³

c) AB3(s)

We do the same as we did in part b, but changing concentration of B:

        AB3(s) <> A(aq) + 3B(aq)

I:                                   0           0

C:                                 +x         +3x

E:                                  x           3x

Therefore Ksp:

Ksp = [A] [B]³ ---> replacing with x:

Ksp = x * (3x)³

Ksp = x * 27x³

Ksp = 27x⁴

d) A3B2

We do the same, but changing concentration of A and B:

        A3B2(s) <> 3A(aq) + 2B(aq)

I:                                   0           0

C:                                +3x         +2x

E:                                  3x           2x

Therefore Ksp:

Ksp = [A]³ [B]² ---> replacing with x:

Ksp = (3x)³ * (2x)²

Ksp = 27x³ * 4x²

Ksp = 108x⁵

And that's all the ways of expressing the Ksp

you need to take a look at the periodic system in order to sort these elements into pairs that would most likely exhibit similar chemical properties. first, you need to find elements that are found in the same group (groups are vertical in the periodic system), and then to see whether they have the same valence number. if they meet these criteria, they will have similar properties.

the first pair: ne (neon) and kr (krypton), are noble gases found in group 8

the second pair: li (lithium) and k (potassium), found in group 1

the third pair: s (sulfur) and te (tellurium), found in group 6