The hydrogen ion concentration in weak acid of dissociation constant `K_(a)` and concentration `c` is nearly equal to

Options

A. `sqrt(K_(a)//c)`
B. `c//K_(a)`
C. `K_(a)c`
D. `sqrt(K_(a)c)`

Explanation

To find the hydrogen ion concentration in a weak acid with a dissociation constant \( K_a \) and an initial concentration \( c \), we can follow these steps:

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

### Step 2: Set up the initial concentrations
Let the initial concentration of the weak acid \( HA \) be \( c \). At equilibrium, if \( x \) is the amount that dissociates, the concentrations will be:
- \([HA] = c - x\)
- \([H^+] = x\)
- \([A^-] = x\)

### Step 3: Write the expression for the dissociation constant \( K_a \)
The expression for the dissociation constant \( K_a \) is given by:
\[ K_a = \frac{[H^+][A^-]}{[HA]} = \frac{x \cdot x}{c - x} = \frac{x^2}{c - x} \]

### Step 4: Make an assumption for weak acids
Since \( HA \) is a weak acid, we can assume that \( x \) is much smaller than \( c \) (i.e., \( x \ll c \)). Therefore, we can approximate:
\[ c - x \approx c \]

### Step 5: Simplify the equation
Substituting this approximation into the \( K_a \) expression gives:
\[ K_a \approx \frac{x^2}{c} \]

### Step 6: Solve for \( x \)
Rearranging the equation to solve for \( x \) (which represents \([H^+]\)):
\[ x^2 = K_a \cdot c \]
\[ x = \sqrt{K_a \cdot c} \]

### Conclusion
Thus, the hydrogen ion concentration \([H^+]\) in the weak acid is nearly equal to:
\[ [H^+] \approx \sqrt{K_a \cdot c} \]