Calculate the concentration of `H_3O^+` ions in 0.001 M solution of `Ba(OH)_2` assuming complete dissociation of `Ba(OH)_2`

Explanation

To calculate the concentration of \( H_3O^+ \) ions in a 0.001 M solution of \( Ba(OH)_2 \), we will follow these steps:

### Step 1: Understand the dissociation of \( Ba(OH)_2 \)
Barium hydroxide, \( Ba(OH)_2 \), is a strong base that completely dissociates in water. The dissociation can be represented as follows:

\[ Ba(OH)_2 \rightarrow Ba^{2+} + 2OH^- \]

This means that for every 1 mole of \( Ba(OH)_2 \) that dissociates, it produces 2 moles of hydroxide ions \( OH^- \).

### Step 2: Calculate the concentration of \( OH^- \) ions
Given that the concentration of \( Ba(OH)_2 \) is 0.001 M, we can calculate the concentration of \( OH^- \) ions produced:

\[ \text{Concentration of } OH^- = 2 \times \text{Concentration of } Ba(OH)_2 = 2 \times 0.001 \, M = 0.002 \, M \]

### Step 3: Use the ion product of water to find \( H_3O^+ \) concentration
The relationship between the concentrations of \( H_3O^+ \) and \( OH^- \) ions in water at 25°C is given by the ion product of water, \( K_w \):

\[ K_w = [H_3O^+][OH^-] = 1.0 \times 10^{-14} \]

We can rearrange this equation to solve for the concentration of \( H_3O^+ \):

\[ [H_3O^+] = \frac{K_w}{[OH^-]} \]

Substituting the known values:

\[ [H_3O^+] = \frac{1.0 \times 10^{-14}}{0.002} \]

### Step 4: Calculate \( [H_3O^+] \)
Now, we perform the calculation:

\[ [H_3O^+] = \frac{1.0 \times 10^{-14}}{0.002} = 5.0 \times 10^{-12} \, M \]

### Final Answer
The concentration of \( H_3O^+ \) ions in a 0.001 M solution of \( Ba(OH)_2 \) is \( 5.0 \times 10^{-12} \, M \).

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