Calculate the pH of a `10^(-2)`M solution of` Ba(OH)_(2)` if it undergoes complete ionisation
`(Kw=10^(-14))` :-

Options

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

Explanation

To calculate the pH of a \(10^{-2}\) M solution of \(Ba(OH)_2\) that undergoes complete ionization, we can follow these steps:

### Step 1: Determine the ionization of \(Ba(OH)_2\)
Barium hydroxide, \(Ba(OH)_2\), dissociates completely in water to give:
\[ Ba(OH)_2 \rightarrow Ba^{2+} + 2OH^{-} \]

### Step 2: Calculate the concentration of hydroxide ions
Since the concentration of the \(Ba(OH)_2\) solution is \(10^{-2}\) M, the concentration of hydroxide ions (\(OH^{-}\)) produced will be:
\[ [OH^{-}] = 2 \times [Ba(OH)_2] = 2 \times 10^{-2} \, M = 2 \times 10^{-2} \, M \]

### Step 3: Calculate the pOH
The pOH can be calculated using the formula:
\[ pOH = -\log[OH^{-}] \]
Substituting the concentration of \(OH^{-}\):
\[ pOH = -\log(2 \times 10^{-2}) \]
Using the logarithmic property:
\[ pOH = -\log(2) - \log(10^{-2}) = -\log(2) + 2 \]
Using the approximate value \(\log(2) \approx 0.301\):
\[ pOH \approx 2 - 0.301 = 1.699 \approx 1.7 \]

### Step 4: Calculate the pH
Using the relationship between pH and pOH:
\[ pH + pOH = pK_w \]
Where \(pK_w = -\log(K_w)\) and \(K_w = 10^{-14}\):
\[ pK_w = 14 \]
Now substituting the pOH value:
\[ pH + 1.7 = 14 \]
Solving for pH:
\[ pH = 14 - 1.7 = 12.3 \]

### Final Answer
The pH of the \(10^{-2}\) M solution of \(Ba(OH)_2\) is approximately:
\[ \boxed{12.3} \]