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 a weak acid \(HA\) with a dissociation constant \(K_a = 4.9 \times 10^{-8}\), we can follow these steps:

### Step 1: Write the dissociation equation for the weak acid
The dissociation of the weak acid \(HA\) can be represented as:
\[ HA \rightleftharpoons H^+ + A^- \]

### Step 2: Set up the expression for the acid dissociation constant \(K_a\)
The expression for the dissociation constant \(K_a\) is given by:
\[ K_a = \frac{[H^+][A^-]}{[HA]} \]
Assuming that the initial concentration of \(HA\) is \(C = 0.1 \, \text{M}\) (decimolar), and let \(x\) be the concentration of \(H^+\) and \(A^-\) that dissociates:
- Initial concentration of \(HA\) = \(C = 0.1 \, \text{M}\)
- Change in concentration = \(-x\) for \(HA\) and \(+x\) for \(H^+\) and \(A^-\)

Thus, at equilibrium:
\[ [HA] = C - x = 0.1 - x \]
\[ [H^+] = x \]
\[ [A^-] = x \]

### Step 3: Substitute into the \(K_a\) expression
Substituting the equilibrium concentrations into the \(K_a\) expression gives:
\[ K_a = \frac{x \cdot x}{0.1 - x} = \frac{x^2}{0.1 - x} \]

### Step 4: Assume \(x\) is small compared to \(0.1\)
Since \(K_a\) is small, we can assume \(x\) is small compared to \(0.1\), so:
\[ 0.1 - x \approx 0.1 \]
Thus, the equation simplifies to:
\[ K_a \approx \frac{x^2}{0.1} \]

### Step 5: Solve for \(x\)
Substituting the value of \(K_a\):
\[ 4.9 \times 10^{-8} = \frac{x^2}{0.1} \]
Multiplying both sides by \(0.1\):
\[ 4.9 \times 10^{-9} = x^2 \]
Taking the square root:
\[ x = \sqrt{4.9 \times 10^{-9}} \approx 7 \times 10^{-5} \, \text{M} \]

### Step 6: Calculate \([OH^-]\) using the ion product of water
At 25°C, the ion product of water \(K_w\) is:
\[ K_w = [H^+][OH^-] = 1.0 \times 10^{-14} \]
Now substituting the value of \([H^+]\):
\[ 1.0 \times 10^{-14} = (7 \times 10^{-5})[OH^-] \]
Solving for \([OH^-]\):
\[ [OH^-] = \frac{1.0 \times 10^{-14}}{7 \times 10^{-5}} \approx 1.43 \times 10^{-10} \, \text{M} \]

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

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