Calculate the pH value in 0.001 M solution of `Ba(OH)_2`

Answer

To calculate the pH of a 0.001 M solution of barium hydroxide (Ba(OH)₂), follow these steps:

### Step 1: Write the dissociation equation
Barium hydroxide dissociates in water as follows:
\[ \text{Ba(OH)}_2 (s) \rightarrow \text{Ba}^{2+} (aq) + 2 \text{OH}^- (aq) \]

### Step 2: Determine the concentration of hydroxide ions
From the dissociation, we see that one mole of Ba(OH)₂ produces two moles of hydroxide ions (OH⁻). Therefore, if the concentration of Ba(OH)₂ is 0.001 M, the concentration of hydroxide ions will be:
\[ [\text{OH}^-] = 2 \times [\text{Ba(OH)}_2] = 2 \times 0.001 \, \text{M} = 0.002 \, \text{M} \]

### Step 3: Calculate the pOH
The pOH is calculated using the formula:
\[ \text{pOH} = -\log[\text{OH}^-] \]
Substituting the concentration of OH⁻:
\[ \text{pOH} = -\log(0.002) \]

Calculating this gives:
\[ \text{pOH} \approx 2.7 \]

### Step 4: Calculate the pH
Using the relationship between pH and pOH:
\[ \text{pH} + \text{pOH} = 14 \]
We can rearrange this to find the pH:
\[ \text{pH} = 14 - \text{pOH} \]
Substituting the value of pOH:
\[ \text{pH} = 14 - 2.7 = 11.3 \]

### Final Answer
The pH of the 0.001 M solution of Ba(OH)₂ is **11.3**.

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