The dissociation constant for an acid HA is `1.6 xx 10^(-5)`. Calculate its `H_(3)O^(+)` in concentration in 0.01 M solution .

Explanation

To calculate the concentration of \( H_3O^+ \) in a 0.01 M solution of the weak acid HA with a dissociation constant \( K_a = 1.6 \times 10^{-5} \), we can follow these steps:

### Step 1: Write the dissociation equation
The dissociation of the weak acid HA in water can be represented as:
\[ HA \rightleftharpoons H^+ + A^- \]
In aqueous solution, \( H^+ \) is often represented as \( H_3O^+ \).

### Step 2: Set up the expression for the dissociation constant
The expression for the dissociation constant \( K_a \) is given by:
\[ K_a = \frac{[H_3O^+][A^-]}{[HA]} \]
Assuming that the initial concentration of HA is \( C = 0.01 \, M \), and at equilibrium, let \( \alpha \) be the degree of dissociation. Thus, we can express the concentrations at equilibrium as follows:
- \( [H_3O^+] = C \alpha \)
- \( [A^-] = C \alpha \)
- \( [HA] = C(1 - \alpha) \)

### Step 3: Substitute into the \( K_a \) expression
Substituting these values into the \( K_a \) expression gives:
\[ K_a = \frac{(C \alpha)(C \alpha)}{C(1 - \alpha)} = \frac{C^2 \alpha^2}{C(1 - \alpha)} = \frac{C \alpha^2}{1 - \alpha} \]

### Step 4: Simplify the equation
Assuming \( \alpha \) is small (which is typical for weak acids), we can approximate \( 1 - \alpha \approx 1 \). Thus, the equation simplifies to:
\[ K_a \approx C \alpha^2 \]

### Step 5: Solve for \( \alpha \)
Rearranging the equation gives:
\[ \alpha^2 = \frac{K_a}{C} \]
Substituting the values:
\[ \alpha^2 = \frac{1.6 \times 10^{-5}}{0.01} = 1.6 \times 10^{-3} \]
Taking the square root:
\[ \alpha = \sqrt{1.6 \times 10^{-3}} \approx 0.04 \]

### Step 6: Calculate the concentration of \( H_3O^+ \)
Now, we can find the concentration of \( H_3O^+ \):
\[ [H_3O^+] = C \alpha = 0.01 \times 0.04 = 0.0004 \, M = 4 \times 10^{-4} \, M \]

### Final Answer
The concentration of \( H_3O^+ \) in a 0.01 M solution of the acid HA is:
\[ [H_3O^+] = 4 \times 10^{-4} \, M \]

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