Which relation is wrong

Options

A. `10^(-pH)+10^(-pOH)=10^(-14)`
B. `pH alpha(1)/([H^(+)])`
C. `K_(w) prop T`
D. dissociation constant of water `K=1.8 xx 10^(-16)`

Explanation

To determine which relation is wrong, we need to analyze the given relations step by step. Let's break it down:

### Step 1: Understand the given relations
We are given several relations involving pH, pOH, and the ion product of water (Kw). The key relations we need to consider are:
1. \( pH + pOH = 14 \) (at 25°C)
2. \( Kw = [H^+][OH^-] \)
3. The relationship between pH and pOH.

### Step 2: Analyze the first relation
The first relation mentioned is \( 10^{-pH} + 10^{-pOH} = 10^{-14} \). This is derived from the definition of pH and pOH:
- \( pH = -\log[H^+] \)
- \( pOH = -\log[OH^-] \)

From the ion product of water, we know:
\[ Kw = [H^+][OH^-] = 10^{-14} \text{ at 25°C} \]

This means:
\[ 10^{-pH} \times 10^{-pOH} = 10^{-14} \]

### Step 3: Verify the correctness of the relation
Now, if we take the logarithm of both sides, we get:
\[ -pH - pOH = -14 \]
or
\[ pH + pOH = 14 \]

This relation holds true only at 25°C. At different temperatures, the value of Kw changes, and thus the relation \( pH + pOH = 14 \) does not hold.

### Step 4: Identify the wrong relation
The statement that \( pOH = Kw \) is incorrect because Kw changes with temperature and cannot be a constant relation like \( pOH \). Therefore, the relation \( pOH = Kw \) is the wrong relation.

### Conclusion
The wrong relation is \( pOH = Kw \).

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