The acid ionization (hydrolysis) constant of `Zn^(2+)" is "1.0 xx 10^(-9)`
Calculate the pH of a 0.001 M solution of ZnCl2

Answer

To calculate the pH of a 0.001 M solution of ZnCl2, we will follow these steps:

### Step 1: Understand the Hydrolysis Reaction
When ZnCl2 is dissolved in water, the Zn²⁺ ions undergo hydrolysis, which can be represented as:
\[ \text{Zn}^{2+} + \text{H}_2\text{O} \rightleftharpoons \text{ZnOH}^+ + \text{H}^+ \]
This reaction indicates that Zn²⁺ acts as a weak acid.

### Step 2: Write the Expression for the Hydrolysis Constant (K_h)
The hydrolysis constant (K_h) for the reaction can be expressed as:
\[ K_h = \frac{[\text{H}^+][\text{ZnOH}^+]}{[\text{Zn}^{2+}]} \]
Given that the concentration of Zn²⁺ is \( C \) (0.001 M), we can simplify the expression.

### Step 3: Set Up the Equation
Assuming \( x \) is the concentration of \( \text{H}^+ \) produced from hydrolysis:
- The concentration of \( \text{Zn}^{2+} \) after hydrolysis will be \( C - x \approx C \) (since \( x \) will be very small compared to \( C \)).
- The concentration of \( \text{H}^+ \) will be \( x \).

Thus, we can write:
\[ K_h = \frac{x^2}{C} \]

### Step 4: Substitute Known Values
We are given:
- \( K_h = 1.0 \times 10^{-9} \)
- \( C = 0.001 \, \text{M} = 1.0 \times 10^{-3} \, \text{M} \)

Substituting these values into the equation:
\[ 1.0 \times 10^{-9} = \frac{x^2}{1.0 \times 10^{-3}} \]

### Step 5: Solve for \( x \)
Rearranging the equation gives:
\[ x^2 = 1.0 \times 10^{-9} \times 1.0 \times 10^{-3} \]
\[ x^2 = 1.0 \times 10^{-12} \]
Taking the square root:
\[ x = \sqrt{1.0 \times 10^{-12}} = 1.0 \times 10^{-6} \]

### Step 6: Calculate the pH
The concentration of \( \text{H}^+ \) ions is \( 1.0 \times 10^{-6} \, \text{M} \). The pH is calculated using the formula:
\[ \text{pH} = -\log[\text{H}^+] \]
Substituting the value:
\[ \text{pH} = -\log(1.0 \times 10^{-6}) = 6 \]

### Final Answer
The pH of the 0.001 M solution of ZnCl2 is **6**.

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