Question:medium

Calculate the pH of a \(0.001\, M\) \(NaOH\) solution at \(25^\circ C\).

Show Hint

For strong bases like \(NaOH\), assume complete dissociation. First calculate \(pOH = -\log[OH^-]\), then use \(pH + pOH = 14\) at \(25^\circ C\).
Updated On: Apr 22, 2026
  • \(3\)
  • \(7\)
  • \(11\)
  • \(13\)
Show Solution

The Correct Option is C

Solution and Explanation

Step 1: Understanding the Question:
We need to calculate the pH of a solution of sodium hydroxide (NaOH), which is a strong base, with a given molar concentration.
Step 2: Key Formula or Approach:
1. Since NaOH is a strong base, it dissociates completely in water: \( \text{NaOH} \rightarrow \text{Na}^+ + \text{OH}^- \). The concentration of hydroxide ions \( [OH^-] \) will be equal to the initial concentration of NaOH.
2. Calculate the pOH using the formula: \( \text{pOH} = -\log_{10}[\text{OH}^-] \).
3. Use the relationship between pH and pOH at \(25^\circ C\): \( \text{pH} + \text{pOH} = 14 \).
4. Solve for pH: \( \text{pH} = 14 - \text{pOH} \).
Step 3: Detailed Explanation:
1. Find \( [OH^-] \):
The concentration of the NaOH solution is \( 0.001 \, M \).
\( 0.001 \, M = 1 \times 10^{-3} \, M \).
Since NaOH dissociates completely, \( [OH^-] = 10^{-3} \, M \).
2. Calculate pOH:
\[ \text{pOH} = -\log_{10}[10^{-3}] \] \[ \text{pOH} = -(-3) = 3 \] 3. Calculate pH:
\[ \text{pH} = 14 - \text{pOH} \] \[ \text{pH} = 14 - 3 = 11 \] Step 4: Final Answer:
The pH of the \(0.001\, M\) NaOH solution is 11.
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