The dissociation constant of a weak monobasic acid in `0.01 M` solution is `10^(-8)`. What is its `[OH^(-)]` concentration?

Options

A. `10^(-6)`
B. `10^(-8)`
C. `10^(-9)`
D. `10^(-10)`

Explanation

To find the concentration of hydroxide ions \([OH^-]\) in a \(0.01 M\) solution of a weak monobasic acid with a dissociation constant \(K_a = 10^{-8}\), we can follow these steps:

### Step 1: Write the expression for the dissociation of the weak acid
For a weak monobasic acid \(HA\), the dissociation can be represented as:
\[ HA \rightleftharpoons H^+ + A^- \]
The dissociation constant \(K_a\) for this reaction is given by:
\[ K_a = \frac{[H^+][A^-]}{[HA]} \]

### Step 2: Set up the equation using the given values
Let the initial concentration of the acid \(C = 0.01 M\). Since the acid is weak, we can assume that the change in concentration due to dissociation is small. Therefore, we can approximate:
\[ [H^+] = [A^-] = x \quad \text{and} \quad [HA] \approx C - x \approx C \]
Thus, the expression for \(K_a\) becomes:
\[ K_a = \frac{x^2}{C} \]
Substituting the known values:
\[ 10^{-8} = \frac{x^2}{0.01} \]

### Step 3: Solve for \(x\) (which represents \([H^+]\))
Rearranging the equation gives:
\[ x^2 = 10^{-8} \times 0.01 = 10^{-10} \]
Taking the square root of both sides:
\[ x = \sqrt{10^{-10}} = 10^{-5} \]
Thus, \([H^+] = 10^{-5} M\).

### Step 4: Use the ion product of water to find \([OH^-]\)
The ion product of water \(K_w\) at \(25^\circ C\) is:
\[ K_w = [H^+][OH^-] = 10^{-14} \]
Now, substituting the value of \([H^+]\):
\[ 10^{-14} = (10^{-5})[OH^-] \]
Solving for \([OH^-]\):
\[ [OH^-] = \frac{10^{-14}}{10^{-5}} = 10^{-9} M \]

### Final Answer
The concentration of hydroxide ions \([OH^-]\) in the solution is:
\[ [OH^-] = 10^{-9} M \]

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