The `pH` of `0.01 (M) KOH` is `12`, if the temperature of the given `KOH` solution is increased which of the following would occur?

Options

A. Both `pH` and `pOH` would remain constant
B. `pH` would be decreased but `pOH` remains constant
C. `pH` would be increased while `pOH` remains constant
D. `pH` would be increased while `pOH` would be decreased

Explanation

To solve the question regarding the effect of temperature increase on the pH of a 0.01 M KOH solution, we will follow these steps:

### Step 1: Understand the initial conditions
The pH of the 0.01 M KOH solution is given as 12. Since KOH is a strong base, it completely dissociates in water to produce hydroxide ions (OH⁻).

### Step 2: Calculate the pOH
Using the relationship between pH and pOH:
\[ \text{pH} + \text{pOH} = 14 \]
Given that the pH is 12:
\[ \text{pOH} = 14 - 12 = 2 \]

### Step 3: Calculate the concentration of OH⁻ ions
Using the pOH to find the concentration of hydroxide ions:
\[ \text{pOH} = -\log[\text{OH}^-] \]
Thus:
\[ 2 = -\log[\text{OH}^-] \]
This implies:
\[ [\text{OH}^-] = 10^{-2} \, \text{M} = 0.01 \, \text{M} \]

### Step 4: Understand the effect of temperature on Kw
As the temperature increases, the ion product of water (Kw) also increases. At 25°C, Kw is \(1.0 \times 10^{-14}\), but it increases with temperature. For example, at higher temperatures, Kw might be \(1.0 \times 10^{-12}\) or higher.

### Step 5: Determine the new pKw
The new pKw can be calculated as:
\[ \text{pKw} = -\log(K_w) \]
If Kw increases, pKw will decrease. For instance, if Kw becomes \(1.0 \times 10^{-12}\):
\[ \text{pKw} = 12 \]

### Step 6: Analyze the relationship between pH, pOH, and pKw
Since:
\[ \text{pH} + \text{pOH} = \text{pKw} \]
If pKw decreases (let's say to 12), and since the pOH remains constant at 2 (because KOH is a strong base and fully dissociates), we can find the new pH:
\[ \text{pH} = \text{pKw} - \text{pOH} \]
Thus:
\[ \text{pH} = 12 - 2 = 10 \]

### Step 7: Conclusion
As the temperature increases, the pKw decreases, leading to a decrease in pH. Therefore, the pH of the solution will decrease.

### Final Answer
The correct conclusion is that the pH would decrease while the pOH remains constant.

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