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 \( \frac{1}{100} \, \text{M} \, \text{Ca(OH)}_2 \) solution, we can follow these steps:

### Step 1: Determine the concentration of hydroxide ions
Calcium hydroxide, \( \text{Ca(OH)}_2 \), dissociates in water according to the following equation:

\[ \text{Ca(OH)}_2 \rightarrow \text{Ca}^{2+} + 2 \text{OH}^- \]

From the dissociation, we see that one mole of \( \text{Ca(OH)}_2 \) produces two moles of hydroxide ions (\( \text{OH}^- \)). Given the concentration of \( \text{Ca(OH)}_2 \) is \( \frac{1}{100} \, \text{M} \) or \( 0.01 \, \text{M} \):

\[ [\text{OH}^-] = 2 \times \left(\frac{1}{100}\right) = 0.02 \, \text{M} \]

### Step 2: Calculate the pOH
The pOH can be calculated using the formula:

\[ \text{pOH} = -\log[\text{OH}^-] \]

Substituting the concentration of hydroxide ions:

\[ \text{pOH} = -\log(0.02) \]

We can break this down further:

\[ \text{pOH} = -\log(2 \times 10^{-2}) = -(\log(2) + \log(10^{-2})) = -(\log(2) - 2) \]

Using \( \log(2) \approx 0.301 \):

\[ \text{pOH} = -0.301 + 2 = 1.699 \approx 1.70 \]

### Step 3: Calculate the pH
We know that:

\[ \text{pH} + \text{pOH} = 14 \]

Thus, we can find the pH:

\[ \text{pH} = 14 - \text{pOH} = 14 - 1.70 = 12.30 \]

### Final Answer
The pH of the \( \frac{1}{100} \, \text{M} \, \text{Ca(OH)}_2 \) solution is **12.30**.

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