Equal volume of 2 solution having pH=2, pH=10 are mixed together at `90^@`C. The pH of resulting solution is : (Take `K_(w)=10^(-12)` at `90^@` C)

Explanation

To find the pH of the resulting solution when equal volumes of two solutions with pH 2 and pH 10 are mixed together at 90°C, we can follow these steps:

### Step 1: Calculate the concentration of H⁺ ions in each solution.
1. For the solution with pH = 2:
\[ \text{[H⁺]} = 10^{-pH} = 10^{-2} \, \text{M} = 0.01 \, \text{M} \]

2. For the solution with pH = 10:
\[ \text{[H⁺]} = 10^{-pH} = 10^{-10} \, \text{M} = 0.0000000001 \, \text{M} \]

### Step 2: Calculate the concentration of OH⁻ ions in the basic solution.
Since we are given \( K_w = 10^{-12} \) at 90°C, we can calculate the concentration of OH⁻ ions in the solution with pH = 10:
\[ K_w = [H⁺][OH⁻] \]
From this, we can find [OH⁻]:
\[ [OH⁻] = \frac{K_w}{[H⁺]} = \frac{10^{-12}}{10^{-10}} = 10^{-2} \, \text{M} \]

### Step 3: Mix the two solutions.
When equal volumes of both solutions are mixed, the concentrations of H⁺ and OH⁻ will be halved:
- For the acidic solution:
\[ \text{[H⁺]} = \frac{0.01}{2} = 0.005 \, \text{M} \]
- For the basic solution:
\[ \text{[OH⁻]} = \frac{0.01}{2} = 0.005 \, \text{M} \]

### Step 4: Determine the resulting concentrations.
After mixing, we have:
- [H⁺] = 0.005 M
- [OH⁻] = 0.005 M

### Step 5: Calculate the resulting pH.
Since the concentrations of H⁺ and OH⁻ are equal, they will neutralize each other:
\[ \text{[H⁺]} = \text{[OH⁻]} = 0.005 \, \text{M} \]
Now, we can use the \( K_w \) to find the new concentration of H⁺:
\[ K_w = [H⁺][OH⁻] = 10^{-12} \]
Since we have equal concentrations:
\[ [H⁺]^2 = 10^{-12} \implies [H⁺] = 10^{-6} \, \text{M} \]

### Step 6: Calculate the pH of the resulting solution.
Now we can find the pH:
\[ \text{pH} = -\log[H⁺] = -\log(10^{-6}) = 6 \]

### Final Answer:
The pH of the resulting solution is **6**.

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