At `25^(@)`C dissociation constants of acid HA and base BOH in aqueous solution are same. The pH of 0.01 M solution of HA is 5. The pOH of `10^(-4)` M solution of BOH at the same temperature is :

Options

A. `3.5`
B. 4
C. 6
D. None of these

Answer

To solve the problem step by step, we will follow the reasoning provided in the video transcript.

### Step 1: Understanding the Given Information
We know that:
- The pH of a 0.01 M solution of the acid HA is 5.
- The dissociation constants (Ka for HA and Kb for BOH) are the same at 25°C.

### Step 2: Calculate the Ka Value for HA
Using the pH value, we can find the concentration of hydrogen ions \([H^+]\):
\[ \text{pH} = 5 \implies [H^+] = 10^{-5} \, \text{M} \]

For a weak acid HA, the dissociation can be represented as:
\[ HA \rightleftharpoons H^+ + A^- \]
The expression for the dissociation constant \(K_a\) is:
\[ K_a = \frac{[H^+][A^-]}{[HA]} \]

Assuming that the initial concentration of HA is 0.01 M and that \(x\) is the amount that dissociates, we have:
\[ [H^+] = x = 10^{-5} \, \text{M} \]
\[ [A^-] = x = 10^{-5} \, \text{M} \]
\[ [HA] \approx 0.01 - x \approx 0.01 \, \text{M} \, (\text{since } x \text{ is very small}) \]

Thus, we can write:
\[ K_a = \frac{(10^{-5})(10^{-5})}{0.01} = \frac{10^{-10}}{0.01} = 10^{-8} \]

### Step 3: Determine the Kb Value for BOH
Since the dissociation constants are the same:
\[ K_b = K_a = 10^{-8} \]

### Step 4: Calculate the pOH of the BOH Solution
For the base BOH, we have a concentration of 10^-4 M. The relationship between \(K_b\) and the concentration of the base is given by:
\[ K_b = \frac{[OH^-]^2}{[BOH]} \]

Let \(y\) be the concentration of hydroxide ions produced:
\[ K_b = \frac{y^2}{10^{-4} - y} \approx \frac{y^2}{10^{-4}} \, (\text{since } y \text{ is small}) \]

Substituting \(K_b\):
\[ 10^{-8} = \frac{y^2}{10^{-4}} \implies y^2 = 10^{-8} \times 10^{-4} = 10^{-12} \]
\[ y = 10^{-6} \, \text{M} \implies [OH^-] = 10^{-6} \, \text{M} \]

### Step 5: Calculate the pOH
The pOH can be calculated using:
\[ \text{pOH} = -\log[OH^-] = -\log(10^{-6}) = 6 \]

### Final Answer
The pOH of the \(10^{-4} M\) solution of BOH is:
\[ \text{pOH} = 6 \]

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