Question

Calculate the pH of a \(0.001\,M\) \(NaOH\) solution at \(298\,K\).

• A. \(11\)
• B. \(10\)
• C. \(3\)
• D. \(7\)

Hint:

For strong bases, concentration of \(OH^-\) equals the base concentration. First compute pOH, then use \(pH + pOH = 14\).

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We need to determine the acidity/basicity (pH) of a basic solution.
The topic is Ionic Equilibrium.
Step 2: Key Formula or Approach:
For a strong base: \( [OH^-] = \text{Molarity} \).
Calculate \( pOH = -\log[OH^-] \).
Calculate \( pH = 14 - pOH \).
Step 3: Detailed Explanation:
\( NaOH \) is a strong base, so \( [OH^-] = 0.001\,M = 10^{-3}\,M \).
Calculate pOH:
\[ pOH = -\log(10^{-3}) = 3 \]
Now, calculate pH using the ionic product of water at \( 298\,K \):
\[ pH = 14 - pOH \]
\[ pH = 14 - 3 \]
Step 4: Final Answer:
\[ pH = 11 \]