To solve the problem step by step, we will follow the reasoning and calculations as discussed in the video transcript.
### Step 1: Determine the initial concentration of H⁺ ions
Given that the pH of the solution is 2, we can calculate the concentration of H⁺ ions using the formula:
\[
\text{pH} = -\log[H^+]
\]
So,
\[
[H^+] = 10^{-\text{pH}} = 10^{-2} \, \text{M}
\]
### Step 2: Determine the new pH after dilution
The problem states that the new pH becomes double the initial pH. Thus:
\[
\text{New pH} = 2 \times 2 = 4
\]
Now, we can calculate the new concentration of H⁺ ions:
\[
[H^+] = 10^{-\text{New pH}} = 10^{-4} \, \text{M}
\]
### Step 3: Set up the equilibrium expression for the acid dissociation
The acid HA dissociates as follows:
\[
HA \rightleftharpoons H^+ + A^-
\]
Let the initial concentration of HA be \(C\) (which is 1 M). At equilibrium, if \(\alpha\) is the degree of ionization, we have:
- Concentration of \(H^+\) = \(C \alpha\)
- Concentration of \(A^-\) = \(C \alpha\)
- Concentration of undissociated \(HA\) = \(C(1 - \alpha)\)
The equilibrium expression for the dissociation constant \(K_a\) is given by:
\[
K_a = \frac{[H^+][A^-]}{[HA]} = \frac{C\alpha \cdot C\alpha}{C(1 - \alpha)} = \frac{C^2 \alpha^2}{C(1 - \alpha)}
\]
Given \(K_a = 10^{-4}\), we can write:
\[
10^{-4} = \frac{C^2 \alpha^2}{C(1 - \alpha)}
\]
### Step 4: Substitute known values
Substituting \(C = 1\) M into the equation:
\[
10^{-4} = \frac{1^2 \alpha^2}{1(1 - \alpha)} \implies 10^{-4} = \frac{\alpha^2}{1 - \alpha}
\]
### Step 5: Solve for \(\alpha\)
Rearranging gives:
\[
10^{-4}(1 - \alpha) = \alpha^2
\]
This simplifies to:
\[
10^{-4} - 10^{-4}\alpha = \alpha^2
\]
Rearranging further:
\[
\alpha^2 + 10^{-4}\alpha - 10^{-4} = 0
\]
Using the quadratic formula:
\[
\alpha = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-10^{-4} \pm \sqrt{(10^{-4})^2 + 4 \cdot 1 \cdot 10^{-4}}}{2}
\]
Calculating the discriminant:
\[
(10^{-4})^2 + 4 \cdot 10^{-4} = 10^{-8} + 4 \cdot 10^{-4} = 4.0001 \cdot 10^{-4}
\]
Thus:
\[
\alpha \approx 0.5
\]
### Step 6: Calculate the initial concentration \(C\)
From \(C \alpha = 10^{-4}\):
\[
C \cdot 0.5 = 10^{-4} \implies C = \frac{10^{-4}}{0.5} = 2 \times 10^{-4} \, \text{M}
\]
### Step 7: Use dilution formula
Using the dilution formula \(C_1 V_1 = C_2 V_2\):
\[
1 \, \text{M} \cdot 1 \, \text{L} = (2 \times 10^{-4} \, \text{M}) \cdot V_2
\]
Solving for \(V_2\):
\[
V_2 = \frac{1}{2 \times 10^{-4}} = 5000 \, \text{L}
\]
### Step 8: Convert to mL
Converting liters to mL:
\[
5000 \, \text{L} = 5000 \times 1000 \, \text{mL} = 5 \times 10^6 \, \text{mL}
\]
### Step 9: Identify \(y\) and \(z\)
From \(5 \times 10^6 \, \text{mL}\), we can identify:
- \(y = 5\)
- \(z = 6\)
### Step 10: Calculate \((y + z) / 2\)
\[
\frac{y + z}{2} = \frac{5 + 6}{2} = \frac{11}{2} = 5.5
\]
### Final Answer
The value of \((y + z) / 2\) is **5.5**.
