Answer:
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
<em>Ksp = x²</em>
<em>This is the Ksp in terms of molar solubility for AB(s)</em>
<em>b) AB2(s)</em>
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
<em>Ksp = 4x³</em>
<em>c) AB3(s)</em>
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
<em>Ksp = 27x⁴</em>
<em>d) A3B2</em>
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
<em>Ksp = 108x⁵</em>
<em>And that's all the ways of expressing the Ksp</em>
