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

Answer

To solve the problem of calculating the percentage of ionization of the weak acid HA with a dissociation constant \( K_a = 4.9 \times 10^{-8} \) in a decimolar (0.1 M) solution, we can follow these steps:

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

### Step 2: Set up the initial and equilibrium concentrations
Let the initial concentration of HA be \( C = 0.1 \, \text{M} \). At equilibrium, if \( \alpha \) is the degree of ionization, the concentrations will be:
- \([HA] = C(1 - \alpha) \approx C\) (since \(\alpha\) is small)
- \([H^+] = C\alpha\)
- \([A^-] = C\alpha\)

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

### Step 4: Solve for \( \alpha \)
Rearranging the equation gives:
\[ \alpha^2 = \frac{K_a}{C} \]
Substituting the values of \( K_a \) and \( C \):
\[ \alpha^2 = \frac{4.9 \times 10^{-8}}{0.1} = 4.9 \times 10^{-7} \]
Taking the square root:
\[ \alpha = \sqrt{4.9 \times 10^{-7}} = 7 \times 10^{-4} \]

### Step 5: Calculate the percentage of ionization
The percentage of ionization is given by:
\[ \text{Percentage of ionization} = \alpha \times 100\% \]
Substituting the value of \( \alpha \):
\[ \text{Percentage of ionization} = 7 \times 10^{-4} \times 100 = 0.07\% \]

### Final Answer
The percentage of ionization of the weak acid HA in a decimolar solution is:
\[ \boxed{0.07\%} \]