Certainly! Let's solve this problem step by step.
We start with the dissociation reaction:
[tex]\[X \leftrightarrow Y + 2Z.\][/tex]
Let the initial molar solubility of [tex]\(X\)[/tex] be [tex]\(s\)[/tex].
At equilibrium:
- The concentration of [tex]\(X\)[/tex] is [tex]\([X] = s\)[/tex].
- The concentration of [tex]\(Y\)[/tex] is [tex]\([Y] = s\)[/tex].
- The concentration of [tex]\(Z\)[/tex] is [tex]\([Z] = 2s\)[/tex].
The equilibrium constant [tex]\(K_s\)[/tex] is expressed as:
[tex]\[K_s = [Y][Z]^2.\][/tex]
Substituting the equilibrium concentrations, we get:
[tex]\[K_s = (s) \cdot (2s)^2.\][/tex]
Simplifying, we have:
[tex]\[K_s = s \cdot 4s^2 = 4s^3.\][/tex]
Now, let's consider the condition where the molar solubility of [tex]\(X\)[/tex] is multiplied by 2. Therefore, the new molar solubility is:
[tex]\[s_{\text{new}} = 2s.\][/tex]
We now need to find the new equilibrium constant [tex]\(K_s\)[/tex] when the solubility is [tex]\(s_{\text{new}}\)[/tex].
At this new solubility:
- [tex]\([X] = s_{\text{new}} = 2s\)[/tex].
- [tex]\([Y] = s_{\text{new}} = 2s\)[/tex].
- [tex]\([Z] = 2s_{\text{new}} = 2(2s) = 4s\)[/tex].
Substituting these new values into the expression for [tex]\(K_s\)[/tex], we get:
[tex]\[K_s' = [Y][Z]^2 = (2s) \cdot (4s)^2.\][/tex]
Simplifying this, we have:
[tex]\[K_s' = 2s \cdot 16s^2 = 32s^3.\][/tex]
Thus, the new equilibrium constant [tex]\(K_s\)[/tex] is:
[tex]\[K_s' = 32s^3.\][/tex]
We need to determine the factor by which [tex]\(K_s\)[/tex] is multiplied. Initially, we had:
[tex]\[K_s = 4s^3.\][/tex]
Now, with the increased solubility:
[tex]\[K_s' = 32s^3.\][/tex]
The multiplication factor is:
[tex]\[\frac{K_s'}{K_s} = \frac{32s^3}{4s^3} = 8.\][/tex]
Therefore, the value of [tex]\(K_s\)[/tex] is multiplied by [tex]\(8\)[/tex].
The correct answer is [tex]\(8\)[/tex].
