The `[H_(3)O^(+)]` in a solution having `[OH^(-)]=5xx10^(-3)N` is :-

Options

A. `2xx10^(-3)N`
B. `2xx10^(-12)N`
C. `5xx10^(-3)N`
D. none

Answer

To find the concentration of \([H_3O^+]\) in a solution where \([OH^-] = 5 \times 10^{-3} N\), we can follow these steps:

### Step 1: Use the Ion Product of Water
The ion product of water (\(K_w\)) at 25°C is given by the equation:
\[ K_w = [H_3O^+][OH^-] \]
The value of \(K_w\) is \(1 \times 10^{-14}\).

### Step 2: Substitute the Known Value
We know the concentration of hydroxide ions \([OH^-]\):
\[ [OH^-] = 5 \times 10^{-3} N \]
Now, we can substitute this value into the equation:
\[ K_w = [H_3O^+](5 \times 10^{-3}) \]

### Step 3: Rearrange the Equation
To find \([H_3O^+]\), rearrange the equation:
\[ [H_3O^+] = \frac{K_w}{[OH^-]} \]

### Step 4: Substitute \(K_w\) and Calculate
Now substitute \(K_w\) and \([OH^-]\) into the equation:
\[ [H_3O^+] = \frac{1 \times 10^{-14}}{5 \times 10^{-3}} \]

### Step 5: Perform the Calculation
Calculating the right-hand side:
\[ [H_3O^+] = \frac{1 \times 10^{-14}}{5 \times 10^{-3}} = 1 \times 10^{-14} \times \frac{1}{5} \times 10^{3} = \frac{1}{5} \times 10^{-11} \]
\[ [H_3O^+] = 0.2 \times 10^{-11} = 2 \times 10^{-12} N \]

### Conclusion
Thus, the concentration of \([H_3O^+]\) in the solution is:
\[ [H_3O^+] = 2 \times 10^{-12} N \]