Calculate the `p^(H)` and `p^(OH)` of 0.0001MHCl Solution

Answer

To calculate the \( pH \) and \( pOH \) of a 0.0001 M HCl solution, follow these steps:

### Step 1: Determine the concentration of \( H^+ \) ions
Since HCl is a strong acid, it completely dissociates in solution. Therefore, the concentration of \( H^+ \) ions in the solution is equal to the concentration of the HCl solution.

\[ [H^+] = 0.0001 \, \text{M} = 10^{-4} \, \text{M} \]

### Step 2: Calculate the \( pH \)
The \( pH \) is calculated using the formula:

\[ pH = -\log[H^+] \]

Substituting the concentration of \( H^+ \):

\[ pH = -\log(10^{-4}) \]

Using the properties of logarithms:

\[ pH = -(-4) \cdot \log(10) = 4 \]

### Step 3: Calculate the \( pOH \)
We know that:

\[ pH + pOH = 14 \]

Now that we have calculated \( pH \):

\[ pOH = 14 - pH \]

Substituting the value of \( pH \):

\[ pOH = 14 - 4 = 10 \]

### Final Results
Thus, the values are:

\[ pH = 4 \]
\[ pOH = 10 \]

---