Let `K_(w)` at `100^(@)C` be `5.5xx10^(-13) M^(2)`. If an aqueous solution at this temperature has `pH=6.2`. Its nature will be

Options

A. acidic
B. alkaline
C. neutral
D. can't say

Explanation

To determine the nature of the solution with a given pH of 6.2 at 100°C, we will follow these steps:

### Step 1: Calculate \( K_w \) at 100°C
We are given:
\[ K_w = 5.5 \times 10^{-13} \, M^2 \]

### Step 2: Calculate \( pK_w \)
The relationship between \( K_w \) and \( pK_w \) is given by:
\[ pK_w = -\log(K_w) \]
Substituting the value of \( K_w \):
\[ pK_w = -\log(5.5 \times 10^{-13}) \]
Using the logarithmic property \( \log(ab) = \log(a) + \log(b) \):
\[ pK_w = -(\log(5.5) + \log(10^{-13})) \]
Calculating \( \log(10^{-13}) = -13 \):
\[ pK_w = -\log(5.5) + 13 \]
Using \( \log(5.5) \approx 0.74 \):
\[ pK_w \approx -0.74 + 13 = 12.26 \]

### Step 3: Calculate the pH of neutral water at 100°C
For a neutral solution at any temperature:
\[ pK_w = pH + pOH \]
At neutrality, \( pH = pOH \), thus:
\[ pK_w = 2 \times pH \]
From the previous calculation:
\[ 12.26 = 2 \times pH \]
Solving for \( pH \):
\[ pH = \frac{12.26}{2} = 6.13 \]

### Step 4: Compare the given pH with the neutral pH
Given:
\[ \text{pH of the solution} = 6.2 \]
Neutral pH at 100°C is:
\[ pH = 6.13 \]
Since \( 6.2 > 6.13 \), the solution is basic.

### Conclusion
The nature of the solution is basic (alkaline).

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