If `K_(a_(1)) and K_(a_(2))` are the dissociation constants of two acids `HA_(1) and HA_(2)`, then the ratio of strengths of their solutions with equimolar concentration is ............. .

Answer

To solve the problem of finding the ratio of strengths of two acids \( HA_1 \) and \( HA_2 \) with equimolar concentrations given their dissociation constants \( K_{a_1} \) and \( K_{a_2} \), we can follow these steps:

### Step 1: Understand the dissociation of weak acids
When a weak acid \( HA \) dissociates in water, it can be represented as:
\[ HA \rightleftharpoons H^+ + A^- \]
For the acid \( HA_1 \), the dissociation constant \( K_{a_1} \) is given by:
\[ K_{a_1} = \frac{[H^+][A_1^-]}{[HA_1]} \]
Similarly, for the acid \( HA_2 \):
\[ K_{a_2} = \frac{[H^+][A_2^-]}{[HA_2]} \]

### Step 2: Define the strength of the acids
The strength of an acid is often defined in terms of its ability to dissociate in solution. For weak acids, the strength can be related to the concentration of the ions produced at equilibrium.

### Step 3: Set up the equilibrium concentrations
Assuming both acids are at the same initial concentration \( C \):
- Let \( x_1 \) be the amount dissociated for \( HA_1 \), then at equilibrium:
- \( [H^+] = x_1 \)
- \( [A_1^-] = x_1 \)
- \( [HA_1] = C - x_1 \)

Thus, we can express \( K_{a_1} \) as:
\[ K_{a_1} = \frac{x_1^2}{C - x_1} \]

For \( HA_2 \), let \( x_2 \) be the amount dissociated:
- At equilibrium:
- \( [H^+] = x_2 \)
- \( [A_2^-] = x_2 \)
- \( [HA_2] = C - x_2 \)

So, we can express \( K_{a_2} \) as:
\[ K_{a_2} = \frac{x_2^2}{C - x_2} \]

### Step 4: Relate the strengths of the acids
The strength of the acids can be defined as:
\[ \text{Strength of } HA_1 = [H^+] = x_1 \]
\[ \text{Strength of } HA_2 = [H^+] = x_2 \]

### Step 5: Derive the ratio of strengths
From the expressions for \( K_{a_1} \) and \( K_{a_2} \), we can derive:
\[ x_1 = \sqrt{K_{a_1} (C - x_1)} \]
\[ x_2 = \sqrt{K_{a_2} (C - x_2)} \]

Assuming \( C \) is large compared to \( x_1 \) and \( x_2 \) (which is valid for weak acids), we can approximate \( C - x \approx C \):
\[ x_1 \approx \sqrt{\frac{K_{a_1}}{C}} \]
\[ x_2 \approx \sqrt{\frac{K_{a_2}}{C}} \]

Thus, the ratio of strengths is:
\[ \frac{x_1}{x_2} = \frac{\sqrt{K_{a_1}}}{\sqrt{K_{a_2}}} = \sqrt{\frac{K_{a_1}}{K_{a_2}}} \]

### Final Answer
The ratio of strengths of their solutions with equimolar concentration is:
\[ \frac{\text{Strength of } HA_1}{\text{Strength of } HA_2} = \sqrt{\frac{K_{a_1}}{K_{a_2}}} \]