The pH of `M/(100) Ca(OH)_2` is

Options

A. 1.699
B. 12
C. 12.301
D. 12.699

Answer

To find the pH of a \( M/(100) \) solution of \( Ca(OH)_2 \), we can follow these steps:

### Step 1: Determine the concentration of \( Ca(OH)_2 \)
Given that the concentration is \( M/(100) \), we can express this as:
\[ \text{Concentration of } Ca(OH)_2 = \frac{1}{100} \, M = 10^{-2} \, M \]

### Step 2: Calculate the concentration of hydroxide ions \( OH^- \)
Calcium hydroxide dissociates in water as follows:
\[ Ca(OH)_2 \rightarrow Ca^{2+} + 2OH^- \]
From the dissociation, we see that 1 mole of \( Ca(OH)_2 \) produces 2 moles of \( OH^- \). Therefore, the concentration of hydroxide 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 hydroxide ions:
\[ pOH = -\log(2 \times 10^{-2}) \]

### Step 4: Calculate the logarithm
To calculate \( -\log(2 \times 10^{-2}) \):
\[ pOH = -(\log(2) + \log(10^{-2})) = -(\log(2) - 2) \]
Using \( \log(2) \approx 0.301 \):
\[ pOH = -(0.301 - 2) = 1.699 \]

### Step 5: Calculate the pH
Using the relationship between pH and pOH:
\[ pH + pOH = 14 \]
We can rearrange this to find pH:
\[ pH = 14 - pOH = 14 - 1.699 = 12.301 \]

### Final Answer
Therefore, the pH of the \( M/(100) \) solution of \( Ca(OH)_2 \) is approximately:
\[ \text{pH} \approx 12.301 \]

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