The conductivity ofa solution may be taken to be directly proportional to the total concentration of the charge carriers (ions) present in it in many cases. Using the above, find the percent decrease in conductivity (k) of a solution of a week monoacidic base BOH when its 0.1 M solution is diluted to double its original volume. `(K_(b) = 10^(-5) for BOH) (take `sqrt(50) = 7.07`) (mark the answer to nearest integer)

Explanation

To find the percent decrease in conductivity (k) of a solution of a weak monoacidic base BOH when its 0.1 M solution is diluted to double its original volume, we can follow these steps:

### Step 1: Understand the dissociation of the weak base
The weak base BOH dissociates in water as follows:
\[ \text{BOH} \rightleftharpoons \text{B}^+ + \text{OH}^- \]

### Step 2: Determine the initial concentration of ions
Given:
- Initial concentration of BOH, \( C = 0.1 \, \text{M} \)
- The dissociation constant \( K_b = 10^{-5} \)

Using the formula for the concentration of ions at equilibrium:
\[ [B^+] = [OH^-] = \sqrt{K_b \cdot C} \]

Substituting the values:
\[ [B^+] = [OH^-] = \sqrt{10^{-5} \cdot 0.1} = \sqrt{10^{-6}} = 10^{-3} \, \text{M} \]

### Step 3: Calculate the initial conductivity
The total concentration of charge carriers (ions) is:
\[ [B^+] + [OH^-] = 10^{-3} + 10^{-3} = 2 \times 10^{-3} \, \text{M} \]

Thus, the initial conductivity \( K_i \) is proportional to:
\[ K_i \propto 2 \times 10^{-3} \]

### Step 4: Determine the new concentration after dilution
When the solution is diluted to double its volume, the new concentration \( C' \) becomes:
\[ C' = \frac{0.1}{2} = 0.05 \, \text{M} \]

### Step 5: Calculate the new concentration of ions
Using the same formula for the new concentration of ions:
\[ [B^+] = [OH^-] = \sqrt{K_b \cdot C'} = \sqrt{10^{-5} \cdot 0.05} = \sqrt{5 \times 10^{-7}} = \frac{\sqrt{5}}{10^{3.5}} \]

Using \( \sqrt{50} = 7.07 \):
\[ [B^+] = [OH^-] = \frac{\sqrt{5}}{10^{3.5}} = \frac{7.07}{1000} = 7.07 \times 10^{-4} \, \text{M} \]

### Step 6: Calculate the new conductivity
The total concentration of charge carriers after dilution is:
\[ [B^+] + [OH^-] = 7.07 \times 10^{-4} + 7.07 \times 10^{-4} = 2 \times 7.07 \times 10^{-4} = 1.414 \times 10^{-3} \]

Thus, the final conductivity \( K_f \) is proportional to:
\[ K_f \propto 1.414 \times 10^{-3} \]

### Step 7: Calculate the percent decrease in conductivity
The percent decrease in conductivity can be calculated using the formula:
\[ \text{Percent Decrease} = \frac{K_f - K_i}{K_i} \times 100 \]

Substituting the values:
\[ \text{Percent Decrease} = \frac{1.414 \times 10^{-3} - 2 \times 10^{-3}}{2 \times 10^{-3}} \times 100 \]

Calculating the numerator:
\[ 1.414 \times 10^{-3} - 2 \times 10^{-3} = -0.586 \times 10^{-3} \]

Now substituting back:
\[ \text{Percent Decrease} = \frac{-0.586 \times 10^{-3}}{2 \times 10^{-3}} \times 100 = -29.3\% \]

Taking the absolute value for the decrease:
\[ \text{Percent Decrease} \approx 29.3\% \]

### Step 8: Round to the nearest integer
Rounding 29.3% to the nearest integer gives:
\[ \text{Final Answer} = 29\% \]