The concentration of NaOH solution is `10^(-8)`M . Find out the `(OH^(-))` concentration

Options

A. `10^(-8)`
B. Greater than `10^(-6)`
C. `10^(-6)`
D. Lies between `10^(-6)` and `10^(-7)`

Explanation

To find the concentration of hydroxide ions \((OH^-)\) in a \(10^{-8} M\) NaOH solution, we can follow these steps:

### Step 1: Understand the dissociation of NaOH
NaOH is a strong base, which means it completely dissociates in water. The dissociation can be represented as:

\[ NaOH \rightarrow Na^+ + OH^- \]

This means that for every mole of NaOH that dissolves, one mole of \(OH^-\) ions is produced.

### Step 2: Calculate the concentration of \(OH^-\) from NaOH
Given that the concentration of NaOH is \(10^{-8} M\), the concentration of \(OH^-\) ions produced from the dissociation of NaOH will also be \(10^{-8} M\):

\[ [OH^-] = [NaOH] = 10^{-8} M \]

### Step 3: Consider the contribution of water
However, we must also consider that pure water itself contributes to the \(OH^-\) concentration. The \(pH\) of pure water at 25°C is 7, which corresponds to:

\[ [H^+] = [OH^-] = 10^{-7} M \]

### Step 4: Calculate the total \(OH^-\) concentration
Since the \(OH^-\) concentration from NaOH is \(10^{-8} M\) and the \(OH^-\) concentration from water is \(10^{-7} M\), we need to add these two contributions together to find the total \(OH^-\) concentration:

\[ [OH^-]_{total} = [OH^-]_{NaOH} + [OH^-]_{water} = 10^{-8} M + 10^{-7} M \]

### Step 5: Convert to a common exponent
To add these concentrations, we can express \(10^{-7} M\) as \(10^{-7} M = 10^{-7} M + 0.1 \times 10^{-7} M\):

\[ [OH^-]_{total} = 10^{-7} M + 0.1 \times 10^{-7} M = 1.1 \times 10^{-7} M \]

### Final Answer
Thus, the total concentration of \(OH^-\) in the solution is:

\[ [OH^-]_{total} = 1.1 \times 10^{-7} M \]

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