Calculate the pH of a `10^(-3)`M solution of` Ba(OH)_(2)` if it undergose complete ionisation
`(Kw=1xx10^(-10))` :-

Options

A. 12. 30
B. 11 .30
C. 10. 00
D. 9. 00

Explanation

To calculate the pH of a `10^(-3)` M solution of `Ba(OH)₂` that undergoes complete ionization, follow these steps:

### Step 1: Determine the concentration of hydroxide ions (OH⁻)
`Ba(OH)₂` dissociates in water as follows:
\[ \text{Ba(OH)}_2 \rightarrow \text{Ba}^{2+} + 2 \text{OH}^- \]

Since the concentration of `Ba(OH)₂` is `10^(-3)` M, and it produces 2 moles of OH⁻ for every mole of `Ba(OH)₂`, the concentration of OH⁻ will be:
\[ [\text{OH}^-] = 2 \times [\text{Ba(OH)}_2] = 2 \times 10^{-3} \, \text{M} = 2 \times 10^{-3} \, \text{M} \]

### Step 2: Calculate the pOH
To find the pOH, use the formula:
\[ \text{pOH} = -\log[\text{OH}^-] \]

Substituting the value of OH⁻ concentration:
\[ \text{pOH} = -\log(2 \times 10^{-3}) \]

Using logarithmic properties:
\[ \text{pOH} = -(\log(2) + \log(10^{-3})) \]
\[ \text{pOH} = -(\log(2) - 3) \]
\[ \text{pOH} = 3 - \log(2) \]

Using the approximate value of \(\log(2) \approx 0.301\):
\[ \text{pOH} = 3 - 0.301 = 2.699 \]

### Step 3: Calculate the pH
Using the relationship between pH and pOH:
\[ \text{pH} + \text{pOH} = 14 \]

Substituting the value of pOH:
\[ \text{pH} = 14 - \text{pOH} \]
\[ \text{pH} = 14 - 2.699 \]
\[ \text{pH} \approx 11.301 \]

### Final Answer
The pH of the `10^(-3)` M solution of `Ba(OH)₂` is approximately **11.3**.

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