What is the `P^(H)` of a solution obtained by mixing equal volumes of two solution of HCl and KOH with PH values 5 and 9 at `25^(@)`C ?

Answer

To find the pH of the solution obtained by mixing equal volumes of two solutions of HCl and KOH with pH values of 5 and 9 at 25°C, we can follow these steps:

### Step 1: Determine the concentration of H⁺ ions in HCl
Given that the pH of the HCl solution is 5, we can calculate the concentration of H⁺ ions using the formula:
\[ [H^+] = 10^{-pH} \]
\[ [H^+] = 10^{-5} \, \text{M} \]

### Step 2: Determine the concentration of OH⁻ ions in KOH
The pH of the KOH solution is given as 9. First, we calculate the pOH:
\[ pOH = 14 - pH \]
\[ pOH = 14 - 9 = 5 \]
Now, we can calculate the concentration of OH⁻ ions:
\[ [OH^-] = 10^{-pOH} \]
\[ [OH^-] = 10^{-5} \, \text{M} \]

### Step 3: Calculate the number of moles of H⁺ and OH⁻ ions
Assuming equal volumes of both solutions (let's say volume = V for each solution), we can calculate the number of moles of H⁺ and OH⁻ ions:
- Moles of H⁺ from HCl:
\[ \text{Moles of } H^+ = [H^+] \times V = 10^{-5} \times V \]
- Moles of OH⁻ from KOH:
\[ \text{Moles of } OH^- = [OH^-] \times V = 10^{-5} \times V \]

### Step 4: Neutralization reaction
The neutralization reaction is:
\[ H^+ + OH^- \rightarrow H_2O \]
Since the moles of H⁺ and OH⁻ are equal, they will completely neutralize each other:
- Initial moles of H⁺ = \( 10^{-5} V \)
- Initial moles of OH⁻ = \( 10^{-5} V \)

After the reaction:
- Moles of H⁺ remaining = \( 10^{-5} V - 10^{-5} V = 0 \)
- Moles of OH⁻ remaining = \( 10^{-5} V - 10^{-5} V = 0 \)

### Step 5: Determine the pH of the resulting solution
Since there are no excess H⁺ or OH⁻ ions left after the neutralization, the resulting solution is neutral. The pH of a neutral solution at 25°C is:
\[ \text{pH} = 7 \]

### Final Answer
The pH of the solution obtained by mixing equal volumes of the two solutions is:
\[ \text{pH} = 7 \]

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