The pH of `10^(-2)` M NaOH solution is .............. Times the pH of `10^(-2)` M HCl solution

Explanation

To solve the problem, we need to find the pH of both the `10^(-2)` M NaOH solution and the `10^(-2)` M HCl solution, and then compare them.

### Step 1: Calculate the pH of `10^(-2)` M HCl solution
HCl is a strong acid and dissociates completely in water:
\[ \text{HCl} \rightarrow \text{H}^+ + \text{Cl}^- \]

For a `10^(-2)` M HCl solution:
- The concentration of H⁺ ions is also `10^(-2)` M.

Now, we can calculate the pH using the formula:
\[ \text{pH} = -\log[\text{H}^+] \]

Substituting the concentration:
\[ \text{pH} = -\log(10^{-2}) \]
\[ \text{pH} = 2 \]

### Step 2: Calculate the pH of `10^(-2)` M NaOH solution
NaOH is a strong base and also dissociates completely in water:
\[ \text{NaOH} \rightarrow \text{Na}^+ + \text{OH}^- \]

For a `10^(-2)` M NaOH solution:
- The concentration of OH⁻ ions is `10^(-2)` M.

To find the pH, we first need to calculate the pOH:
\[ \text{pOH} = -\log[\text{OH}^-] \]

Substituting the concentration:
\[ \text{pOH} = -\log(10^{-2}) \]
\[ \text{pOH} = 2 \]

Now, we can find the pH using the relation:
\[ \text{pH} + \text{pOH} = 14 \]
Thus,
\[ \text{pH} = 14 - \text{pOH} \]
\[ \text{pH} = 14 - 2 \]
\[ \text{pH} = 12 \]

### Step 3: Compare the pH values
Now we have:
- pH of `10^(-2)` M HCl = 2
- pH of `10^(-2)` M NaOH = 12

To find how many times the pH of NaOH is compared to HCl:
\[ \text{Ratio} = \frac{\text{pH of NaOH}}{\text{pH of HCl}} \]
\[ \text{Ratio} = \frac{12}{2} = 6 \]

### Final Answer
The pH of `10^(-2)` M NaOH solution is **6 times** the pH of `10^(-2)` M HCl solution.

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