Calculate The `[H^(+)]` ions of `0.008 M Ca(OH)_(2)` is

Options

A. `2 xx 10^(-7) M`
B. `6.25 xx 10^(-13) M`
C. `1.4 xx 10^(-12) M`
D. `1.25 xx 10^(-11) M`

Explanation

To calculate the concentration of \([H^+]\) ions in a \(0.008 \, M\) solution of \(Ca(OH)_2\), we can follow these steps:

### Step 1: Determine the dissociation of \(Ca(OH)_2\)
Calcium hydroxide, \(Ca(OH)_2\), is a strong base and dissociates completely in water:
\[ Ca(OH)_2 \rightarrow Ca^{2+} + 2OH^- \]
From the dissociation, we can see that for every mole of \(Ca(OH)_2\), we get 2 moles of \(OH^-\).

### Step 2: Calculate the concentration of \(OH^-\) ions
Given the concentration of \(Ca(OH)_2\) is \(0.008 \, M\), the concentration of \(OH^-\) ions will be:
\[ [OH^-] = 2 \times [Ca(OH)_2] = 2 \times 0.008 \, M = 0.016 \, M \]

### Step 3: Use the ion product of water to find \([H^+]\)
The ion product of water (\(K_w\)) at \(25^\circ C\) is:
\[ K_w = [H^+][OH^-] = 1.0 \times 10^{-14} \]
We can rearrange this equation to find \([H^+]\):
\[ [H^+] = \frac{K_w}{[OH^-]} \]

### Step 4: Substitute the values
Now, substituting the values we have:
\[ [H^+] = \frac{1.0 \times 10^{-14}}{0.016} \]

### Step 5: Calculate \([H^+]\)
Now, performing the calculation:
\[ [H^+] = \frac{1.0 \times 10^{-14}}{0.016} = 6.25 \times 10^{-13} \, M \]

### Final Answer
The concentration of \([H^+]\) ions in the \(0.008 \, M\) \(Ca(OH)_2\) solution is:
\[ [H^+] = 6.25 \times 10^{-13} \, M \]

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