The pH of `10^(-2)` M `Ca(OH)_(2)` is

Options

A. 12.3010
B. 6.96
C. 7.2
D. 6

Explanation

To find the pH of a `10^(-2)` M solution of `Ca(OH)₂`, we can follow these steps:

### Step 1: Understand the dissociation of `Ca(OH)₂`
Calcium hydroxide, `Ca(OH)₂`, is a strong base that completely dissociates in water. The dissociation can be represented as:
\[ Ca(OH)₂ \rightarrow Ca^{2+} + 2OH^{-} \]
From this equation, we can see that for every 1 mole of `Ca(OH)₂`, we get 2 moles of hydroxide ions (`OH^{-}`).

### Step 2: Calculate the concentration of `OH^{-}`
Given the concentration of `Ca(OH)₂` is `10^(-2)` M, the concentration of `OH^{-}` ions will be:
\[ [OH^{-}] = 2 \times [Ca(OH)₂] = 2 \times 10^{-2} \, M = 2 \times 10^{-2} \, M \]

### 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:
\[ pOH = -\log(2) - \log(10^{-2}) \]
\[ pOH = -\log(2) + 2 \]
Using the approximate value of \(\log(2) \approx 0.301\):
\[ pOH \approx -0.301 + 2 = 1.699 \]

### Step 4: Calculate the pH
We can find the pH using the relationship:
\[ pH + pOH = 14 \]
Thus,
\[ pH = 14 - pOH \]
Substituting the value of pOH:
\[ pH = 14 - 1.699 \]
\[ pH \approx 12.301 \]

### Final Answer
The pH of the `10^(-2)` M `Ca(OH)₂` solution is approximately **12.301**.

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