The `[OH^(-)]` of `0.005 M` is `H_(2)SO_(4)`

Options

A. `2 xx 10^(-12) M`
B. `5 xx 10^(-3) M`
C. `10^(-12) M`
D. `10^(-14) M`

Explanation

To find the concentration of hydroxide ions \([OH^-]\) in a \(0.005 M\) solution of \(H_2SO_4\), we can follow these steps:

### Step 1: Determine the dissociation of \(H_2SO_4\)
Sulfuric acid (\(H_2SO_4\)) is a strong acid that dissociates completely in water. The dissociation can be represented as:
\[ H_2SO_4 \rightarrow 2H^+ + SO_4^{2-} \]
From this equation, we can see that one mole of \(H_2SO_4\) produces two moles of \(H^+\) ions.

### Step 2: Calculate the concentration of \(H^+\) ions
Given that the concentration of \(H_2SO_4\) is \(0.005 M\), we can calculate the concentration of \(H^+\) ions produced:
\[ \text{Concentration of } H^+ = 2 \times [H_2SO_4] = 2 \times 0.005 = 0.01 M \]

### Step 3: Use the ionic product of water
The ionic product of water (\(K_w\)) at \(25^\circ C\) is:
\[ K_w = [H^+][OH^-] = 1.0 \times 10^{-14} \]
Now we can use this to find the concentration of \(OH^-\) ions.

### Step 4: Rearrange the equation to find \([OH^-]\)
We can rearrange the equation to solve for \([OH^-]\):
\[ [OH^-] = \frac{K_w}{[H^+]} \]
Substituting the known values:
\[ [OH^-] = \frac{1.0 \times 10^{-14}}{0.01} = 1.0 \times 10^{-12} M \]

### Final Answer
The concentration of hydroxide ions \([OH^-]\) in a \(0.005 M\) solution of \(H_2SO_4\) is:
\[ [OH^-] = 1.0 \times 10^{-12} M \]

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