The dissociation constant of a weak acid HA is `4.9 xx 10^(-8)`. Calculate for a decimolar solution of acid:.
`OH^(-)` concentration

Answer

To solve the problem of finding the concentration of hydroxide ions \([OH^-]\) in a decimolar solution of the weak acid \(HA\) with a dissociation constant \(K_a = 4.9 \times 10^{-8}\), we can follow these steps:

### Step 1: Define the initial concentration of the acid
Given that the solution is decimolar, the initial concentration of the acid \(HA\) is:
\[ C = 0.1 \, \text{M} \]

### Step 2: Set up the dissociation equation
The dissociation of the weak acid \(HA\) can be represented as follows:
\[ HA \rightleftharpoons H^+ + A^- \]
Let \( \alpha \) be the degree of dissociation. At equilibrium, the concentrations will be:
- \([HA] = C(1 - \alpha)\)
- \([H^+] = C\alpha\)
- \([A^-] = C\alpha\)

### Step 3: Write the expression for the dissociation constant \(K_a\)
The dissociation constant \(K_a\) is given by:
\[ K_a = \frac{[H^+][A^-]}{[HA]} = \frac{(C\alpha)(C\alpha)}{C(1 - \alpha)} = \frac{C^2\alpha^2}{C(1 - \alpha)} \]
Simplifying this gives:
\[ K_a = \frac{C\alpha^2}{1 - \alpha} \]

### Step 4: Substitute known values into the equation
Substituting \(K_a = 4.9 \times 10^{-8}\) and \(C = 0.1\):
\[ 4.9 \times 10^{-8} = \frac{0.1\alpha^2}{1 - \alpha} \]

### Step 5: Assume \(\alpha\) is small
Since \(K_a\) is small, we can assume that \(\alpha\) is small compared to 1, thus \(1 - \alpha \approx 1\):
\[ 4.9 \times 10^{-8} \approx 0.1\alpha^2 \]
Rearranging gives:
\[ \alpha^2 = \frac{4.9 \times 10^{-8}}{0.1} = 4.9 \times 10^{-7} \]

### Step 6: Calculate \(\alpha\)
Taking the square root:
\[ \alpha = \sqrt{4.9 \times 10^{-7}} \approx 7 \times 10^{-4} \]

### Step 7: Calculate \([H^+]\)
Now, we can find the concentration of hydrogen ions:
\[ [H^+] = C\alpha = 0.1 \times 7 \times 10^{-4} = 7 \times 10^{-5} \, \text{M} \]

### Step 8: Use the ion product of water to find \([OH^-]\)
At 25°C, the ion product of water \(K_w\) is:
\[ K_w = [H^+][OH^-] = 1 \times 10^{-14} \]
We can rearrange this to find \([OH^-]\):
\[ [OH^-] = \frac{K_w}{[H^+]} = \frac{1 \times 10^{-14}}{7 \times 10^{-5}} \approx 1.43 \times 10^{-10} \, \text{M} \]

### Final Answer
The concentration of hydroxide ions \([OH^-]\) in the solution is:
\[ [OH^-] \approx 1.43 \times 10^{-10} \, \text{M} \]