If the concentration of two monobasic acids are same, their relative strength can be compared by

Options

A. `(alpha_(1))/(alpha_(2))`
B. `(K_(1))/(K_(2))`
C. `([H^(+)]_(1))/([H^(+)]_(2))`
D. `sqrt((K_(1))/(K_(2)))`

Answer

To determine the relative strength of two monobasic acids with the same concentration, we can analyze their dissociation constants (Ka) and the degree of dissociation (α). Here’s a step-by-step solution:

### Step 1: Define the Acids
Let’s consider two monobasic acids, HA and HB, both having the same concentration \( C \).

### Step 2: Write the Dissociation Equations
For acid HA:
\[ HA \rightleftharpoons H^+ + A^- \]
For acid HB:
\[ HB \rightleftharpoons H^+ + B^- \]

### Step 3: Express the Degree of Dissociation
Let \( \alpha_1 \) be the degree of dissociation of HA and \( \alpha_2 \) be the degree of dissociation of HB. The concentration of \( H^+ \) ions produced can be expressed as:
- For HA: \( [H^+] = C \alpha_1 \)
- For HB: \( [H^+] = C \alpha_2 \)

### Step 4: Write the Expression for the Acid Dissociation Constant
The dissociation constant \( K_a \) for each acid can be expressed as:
- For HA:
\[ K_{a1} = \frac{[H^+][A^-]}{[HA]} = \frac{(C \alpha_1)(C \alpha_1)}{C(1 - \alpha_1)} \approx \frac{C \alpha_1^2}{1} = C \alpha_1^2 \quad (\text{since } \alpha_1 \ll 1) \]
- For HB:
\[ K_{a2} = \frac{[H^+][B^-]}{[HB]} = \frac{(C \alpha_2)(C \alpha_2)}{C(1 - \alpha_2)} \approx \frac{C \alpha_2^2}{1} = C \alpha_2^2 \quad (\text{since } \alpha_2 \ll 1) \]

### Step 5: Relate the Degrees of Dissociation to the Dissociation Constants
From the expressions for \( K_{a1} \) and \( K_{a2} \):
\[ \alpha_1 = \sqrt{\frac{K_{a1}}{C}} \quad \text{and} \quad \alpha_2 = \sqrt{\frac{K_{a2}}{C}} \]

### Step 6: Compare the Relative Strengths
The relative strength of the two acids can be compared using their degrees of dissociation:
\[ \frac{\alpha_1}{\alpha_2} = \frac{\sqrt{\frac{K_{a1}}{C}}}{\sqrt{\frac{K_{a2}}{C}}} = \sqrt{\frac{K_{a1}}{K_{a2}}} \]
Thus, the relative strength can also be expressed as:
\[ \frac{[H^+]_1}{[H^+]_2} = \frac{C \alpha_1}{C \alpha_2} = \frac{\alpha_1}{\alpha_2} \]

### Conclusion
Therefore, the relative strength of the two monobasic acids can be compared by:
- \( \frac{[H^+]_1}{[H^+]_2} \)
- \( \sqrt{\frac{K_{a1}}{K_{a2}}} \)

### Final Answer
The correct options to compare the relative strength of the two acids are:
- \( \frac{[H^+]_1}{[H^+]_2} \)
- \( \sqrt{\frac{K_{a1}}{K_{a2}}} \)