The ` [OH^ −] ` of 0.005M `H_2SO_4` is

Options

A. `2xx10^(-12)` M
B. `5xx10^(-3)` M
C. `10^(-2)` M
D. `10^(-12) `M

Answer

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

### Step 1: Determine the dissociation of sulfuric acid
Sulfuric acid is a strong acid and dissociates completely in water. The dissociation reaction is:
\[ H_2SO_4 \rightarrow 2H^+ + SO_4^{2-} \]
This means that for every 1 mole of \(H_2SO_4\), 2 moles of \(H^+\) ions are produced.

### 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:
\[ [H^+] = 2 \times [H_2SO_4] = 2 \times 0.005 \, M = 0.01 \, M \]

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

### Step 4: Substitute the values into the equation
Substituting the known values into the equation:
\[ [OH^-] = \frac{1.0 \times 10^{-14}}{0.01} \]
Calculating this gives:
\[ [OH^-] = 1.0 \times 10^{-12} \, M \]

### 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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