The relation between pH, `pK_(a)` and concentration c of the solution of a weak acid is .......... .

Answer

To find the relation between pH, pK_a, and the concentration (c) of a weak acid, we can follow these steps:

### Step 1: Understanding Weak Acid Dissociation
A weak acid (HA) partially dissociates in water:
\[ HA \rightleftharpoons H^+ + A^- \]

### Step 2: Establishing Initial Concentrations
Let the initial concentration of the weak acid be \( c \). At equilibrium, if \( \alpha \) is the degree of dissociation, the concentrations will be:
- \([HA] = c(1 - \alpha)\)
- \([H^+] = c\alpha\)
- \([A^-] = c\alpha\)

### Step 3: Writing the Equilibrium Constant Expression
The equilibrium constant \( K_a \) for the dissociation of the weak acid is given by:
\[ K_a = \frac{[H^+][A^-]}{[HA]} \]

Substituting the equilibrium concentrations, we get:
\[ K_a = \frac{(c\alpha)(c\alpha)}{c(1 - \alpha)} \]

### Step 4: Simplifying the Expression
This simplifies to:
\[ K_a = \frac{c\alpha^2}{1 - \alpha} \]

### Step 5: Assuming \( \alpha \) is Small
For weak acids, \( \alpha \) is usually small, so we can assume \( 1 - \alpha \approx 1 \):
\[ K_a \approx c\alpha^2 \]

### Step 6: Solving for \( \alpha \)
Rearranging gives:
\[ \alpha \approx \sqrt{\frac{K_a}{c}} \]

### Step 7: Finding \([H^+]\)
The concentration of \( H^+ \) ions is:
\[ [H^+] = c\alpha \]
Substituting for \( \alpha \):
\[ [H^+] \approx c\sqrt{\frac{K_a}{c}} = \sqrt{K_a \cdot c} \]

### Step 8: Relating pH to \([H^+]\)
The pH is defined as:
\[ pH = -\log[H^+] \]
Substituting for \([H^+]\):
\[ pH = -\log\left(\sqrt{K_a \cdot c}\right) \]
This can be rewritten using logarithmic properties:
\[ pH = -\frac{1}{2}\log(K_a \cdot c) \]
\[ pH = -\frac{1}{2}(\log K_a + \log c) \]
This can be expressed as:
\[ pH = -\frac{1}{2} \log K_a - \frac{1}{2} \log c \]

### Step 9: Relating pH to pK_a
Since \( pK_a = -\log K_a \), we can substitute:
\[ pH = \frac{1}{2} pK_a - \frac{1}{2} \log c \]

### Final Relation
Thus, the final relation between pH, pK_a, and concentration \( c \) is:
\[ pH = \frac{1}{2} pK_a - \frac{1}{2} \log c \]

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