What is the hydronium ion concentration of a `0.25` M HA solution? `(K_(a)=4xx10^(-8))`

Options

A. `10^(-4)`
B. `10^(-5)`
C. `10^(-7)`
D. `10^(-10)`

Answer

To find the hydronium ion concentration of a 0.25 M HA solution with a dissociation constant \( K_a = 4 \times 10^{-8} \), we can follow these steps:

### Step 1: Write the dissociation equation
The weak acid HA dissociates in water according to the following equation:
\[ HA \rightleftharpoons H^+ + A^- \]

### Step 2: Set up the initial concentrations
Let the initial concentration of HA be 0.25 M. At the start (before dissociation), the concentrations are:
- \([HA] = 0.25 \, \text{M}\)
- \([H^+] = 0 \, \text{M}\)
- \([A^-] = 0 \, \text{M}\)

### Step 3: Define the change in concentration
Let \( x \) be the amount of HA that dissociates. At equilibrium, the concentrations will be:
- \([HA] = 0.25 - x \, \text{M}\)
- \([H^+] = x \, \text{M}\)
- \([A^-] = x \, \text{M}\)

### Step 4: Write the expression for \( K_a \)
The expression for the acid dissociation constant \( K_a \) is given by:
\[ K_a = \frac{[H^+][A^-]}{[HA]} \]
Substituting the equilibrium concentrations into this expression gives:
\[ K_a = \frac{x \cdot x}{0.25 - x} = \frac{x^2}{0.25 - x} \]

### Step 5: Make an approximation
Since \( K_a \) is very small, we can assume that \( x \) is much smaller than 0.25 M. Therefore, we can approximate:
\[ 0.25 - x \approx 0.25 \]
This simplifies our equation to:
\[ K_a \approx \frac{x^2}{0.25} \]

### Step 6: Substitute the value of \( K_a \)
Substituting the given value of \( K_a \):
\[ 4 \times 10^{-8} = \frac{x^2}{0.25} \]

### Step 7: Solve for \( x^2 \)
Rearranging the equation gives:
\[ x^2 = 4 \times 10^{-8} \times 0.25 \]
Calculating the right side:
\[ x^2 = 4 \times 10^{-8} \times 0.25 = 1 \times 10^{-8} \]

### Step 8: Take the square root to find \( x \)
Now, take the square root of both sides:
\[ x = \sqrt{1 \times 10^{-8}} = 10^{-4} \]

### Step 9: Conclusion
The hydronium ion concentration \([H^+]\) is equal to \( x \):
\[ [H^+] = 10^{-4} \, \text{M} \]

### Final Answer
The hydronium ion concentration of the 0.25 M HA solution is \( 10^{-4} \, \text{M} \).

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