What is the pH value of `N/10^(7) KOH` solution?

Options

A. 7
B. between 7 and 8
C. 8
D. between 6 and 7

Explanation

To find the pH value of a `N/10^(7)` KOH solution, we can follow these steps:

### Step 1: Understand the concentration of KOH
KOH is a strong base that completely dissociates in solution:
\[ \text{KOH} \rightarrow \text{K}^+ + \text{OH}^- \]

Given that the concentration of KOH is \( N/10^7 \) or \( 10^{-7} \) N, this means the concentration of hydroxide ions \([OH^-]\) produced is also \( 10^{-7} \) M.

### Step 2: Consider the contribution of water
Pure water also contributes to the \([H^+]\) and \([OH^-]\) concentrations. At 25°C, the concentration of \([H^+]\) in pure water is \( 10^{-7} \) M, which means:
\[ [H^+] = 10^{-7} \, \text{M} \]
\[ [OH^-] = 10^{-7} \, \text{M} \]

### Step 3: Calculate the total concentration of hydroxide ions
Since both the KOH and water contribute to the \([OH^-]\), we can add them together:
\[ [OH^-]_{total} = [OH^-]_{KOH} + [OH^-]_{H_2O} = 10^{-7} + 10^{-7} = 2 \times 10^{-7} \, \text{M} \]

### Step 4: Calculate the pOH
To find the pOH, we use the formula:
\[ \text{pOH} = -\log[OH^-] \]
Substituting the total concentration of hydroxide ions:
\[ \text{pOH} = -\log(2 \times 10^{-7}) \]

Using the logarithmic property:
\[ \text{pOH} = -\log(2) - \log(10^{-7}) \]
\[ \text{pOH} = -\log(2) + 7 \]

The value of \(\log(2) \approx 0.301\):
\[ \text{pOH} = 7 - 0.301 = 6.699 \]

### Step 5: Calculate the pH
Using the relationship between pH and pOH:
\[ \text{pH} + \text{pOH} = 14 \]
We can find the pH:
\[ \text{pH} = 14 - \text{pOH} = 14 - 6.699 = 7.301 \]

### Final Answer
The pH value of the \( N/10^7 \) KOH solution is approximately **7.301**.

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