The hydrolysis constant for `ZnCl_(2)` will be
where `K_(b)` is effective dissociation constant of base `Zn^(++)`

Options

A. `K_(h)=(K_(w))/(K_(b))`
B. `K_(h)=(K_(w)^(2))/(K_(b))`
C. `K_(h)=(K_(w)^(2))/(K_(b)^(2))`
D. `K_(h)=(K_(b))/(K_(w)^(2))`

Explanation

To find the hydrolysis constant \( K_h \) for \( ZnCl_2 \), we can follow these steps:

### Step 1: Understand the Hydrolysis Reaction
When \( ZnCl_2 \) dissolves in water, it dissociates into \( Zn^{2+} \) and \( Cl^- \) ions. The \( Zn^{2+} \) ion can undergo hydrolysis, reacting with water to form \( Zn(OH)_2 \) and releasing \( H^+ \) ions:

\[ Zn^{2+} + 2H_2O \rightleftharpoons Zn(OH)_2 + 2H^+ \]

### Step 2: Write the Expression for Hydrolysis Constant \( K_h \)
The hydrolysis constant \( K_h \) can be expressed using the concentrations of the products and reactants at equilibrium:

\[ K_h = \frac{[Zn(OH)_2][H^+]^2}{[Zn^{2+}]} \]

### Step 3: Relate \( K_h \) to \( K_b \)
The base dissociation constant \( K_b \) for the reaction of \( Zn(OH)_2 \) can be expressed as:

\[ K_b = \frac{[OH^-]^2}{[Zn(OH)_2]} \]

### Step 4: Use the Ion Product of Water \( K_w \)
We know that the ion product of water \( K_w \) is given by:

\[ K_w = [H^+][OH^-] \]

From this, we can express \( [OH^-] \) in terms of \( [H^+] \):

\[ [OH^-] = \frac{K_w}{[H^+]} \]

### Step 5: Substitute \( [OH^-] \) into the \( K_b \) Expression
Substituting \( [OH^-] \) into the \( K_b \) expression gives:

\[ K_b = \frac{\left(\frac{K_w}{[H^+]}\right)^2}{[Zn(OH)_2]} \]

### Step 6: Rearranging the Equations
Rearranging the equation for \( K_h \) and substituting \( K_b \):

\[ K_h = \frac{[Zn(OH)_2][H^+]^2}{[Zn^{2+}]} \]

### Step 7: Final Expression for Hydrolysis Constant
Combining the expressions for \( K_h \) and \( K_b \) and using \( K_w \):

\[ K_h = \frac{K_w^2}{K_b} \]

### Conclusion
Thus, the hydrolysis constant \( K_h \) for \( ZnCl_2 \) is given by:

\[ K_h = \frac{K_w^2}{K_b} \]