Weak acids and bases are not completely ionised when dissolved in polar medium like water
`HA rarr H^(+) +A^(-)`
`{:(t_(0),C,O,O),(t_(eq),C-Calpha,Calpha,Calpha):}`
`K_(a)=(Calpha^(2))/(1-alpha)=Calpha^(2),alphasqrt((K_(a))/(C))`
`:. (alpha_(1))/(alpha_(2))=sqrt((Ka_(1))/(Ka_(2))),(alpha_(1))/(alpha_(2))=sqrt((C_(2))/(C_(1)))`
( For two acids at same conc.) (for same acid at diff conc. )
Relative strength of two weak monoprotic acids may be given as

Options

A. `([H^(+)]_(1))/([H^(+)]_(2))`
B. `(alpha_(1))/(alpha_(2))`
C. `sqrt((Ka_(1))/(Ka_(2)))`
D. All of the above

Answer

To solve the problem regarding the relative strength of two weak monoprotic acids, we will follow these steps:

### Step-by-Step Solution:

1. **Understanding the Dissociation of Weak Acids:**
A weak acid (HA) dissociates in water according to the equation:
\[ HA \rightleftharpoons H^+ + A^- \]
Here, \( HA \) is the weak acid, \( H^+ \) is the hydrogen ion, and \( A^- \) is the conjugate base.

2. **Setting Up the Equilibrium Expression:**
For the dissociation of the weak acid, the acid dissociation constant (\( K_a \)) is defined as:
\[ K_a = \frac{[H^+][A^-]}{[HA]} \]

3. **Defining Concentrations at Equilibrium:**
Let \( C \) be the initial concentration of the acid. At equilibrium, if \( \alpha \) is the degree of dissociation, the concentrations will be:
- \([H^+] = C \alpha\)
- \([A^-] = C \alpha\)
- \([HA] = C(1 - \alpha) \approx C\) (since \( \alpha \) is small for weak acids)

Thus, the expression for \( K_a \) becomes:
\[ K_a = \frac{(C \alpha)(C \alpha)}{C(1 - \alpha)} \approx C \alpha^2 \]

4. **Relating the Degrees of Dissociation:**
For two weak acids, \( HA_1 \) and \( HA_2 \), with \( K_{a1} \) and \( K_{a2} \) respectively, we can write:
\[ K_{a1} = C \alpha_1^2 \quad \text{and} \quad K_{a2} = C \alpha_2^2 \]

5. **Finding the Ratio of Degrees of Dissociation:**
From the expressions for \( K_a \):
\[ \frac{\alpha_1^2}{\alpha_2^2} = \frac{K_{a1}}{K_{a2}} \]
Taking the square root gives:
\[ \frac{\alpha_1}{\alpha_2} = \sqrt{\frac{K_{a1}}{K_{a2}}} \]

6. **Conclusion on Relative Strength:**
The relative strength of the two weak acids can be expressed in terms of their dissociation constants or their degrees of dissociation:
- \(\frac{[H^+]_1}{[H^+]_2}\)
- \(\frac{\alpha_1}{\alpha_2}\)
- \(\sqrt{\frac{K_{a1}}{K_{a2}}}\)

Therefore, the relative strength of two weak monoprotic acids can be given as:
\[ \text{Relative strength} = \frac{[H^+]_1}{[H^+]_2} = \frac{\alpha_1}{\alpha_2} = \sqrt{\frac{K_{a1}}{K_{a2}}} \]