The pH of `10^(-11)` M HCI at `25^@C` is

Options

A. 11
B. 3
C. Slightly greater than 7
D. Slightly less than 7

Answer

To find the pH of a `10^(-11)` M HCl solution at `25°C`, we will follow these steps:

### Step 1: Understand the dissociation of HCl
HCl is a strong acid, which means it completely dissociates in water:
\[ \text{HCl} \rightarrow \text{H}^+ + \text{Cl}^- \]
Thus, the concentration of H⁺ ions from HCl will be `10^(-11)` M.

### Step 2: Consider the autoionization of water
At `25°C`, water also undergoes autoionization:
\[ \text{H}_2\text{O} \rightleftharpoons \text{H}^+ + \text{OH}^- \]
The ion product of water (\(K_w\)) is given by:
\[ K_w = [\text{H}^+][\text{OH}^-] = 10^{-14} \]

### Step 3: Calculate the concentration of H⁺ from water
Let the concentration of H⁺ ions contributed by the autoionization of water be \(x\). Thus, we have:
\[ [\text{H}^+] = 10^{-11} + x \]
\[ [\text{OH}^-] = x \]
Substituting these into the expression for \(K_w\):
\[ 10^{-14} = (10^{-11} + x)(x) \]

### Step 4: Assume \(x\) is small
Since `10^(-11)` is very small compared to `10^(-7)`, we can assume that \(x\) is negligible compared to `10^(-11)`:
\[ 10^{-14} \approx 10^{-11} \cdot x \]
From this, we can solve for \(x\):
\[ x = \frac{10^{-14}}{10^{-11}} = 10^{-3} \]
Thus, the concentration of H⁺ ions from water is approximately \(10^{-3}\) M.

### Step 5: Calculate the total concentration of H⁺
Now, we can find the total concentration of H⁺ ions:
\[ [\text{H}^+] = 10^{-11} + 10^{-7} \approx 10^{-7} \text{ M} \]
This is because \(10^{-11}\) is negligible compared to \(10^{-7}\).

### Step 6: Calculate the pH
Now we can calculate the pH using the formula:
\[ \text{pH} = -\log[\text{H}^+] \]
Substituting the total concentration:
\[ \text{pH} = -\log(1.0001 \times 10^{-7}) \]
Calculating this gives:
\[ \text{pH} \approx 6.99 \]

### Final Answer
Thus, the pH of `10^(-11)` M HCl at `25°C` is approximately **6.99**.

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