Question

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

• A. \(3\)
• B. \(7\)
• C. \(11\)
• D. \(13\)

Hint:

For strong bases like \(NaOH\), assume complete dissociation. First calculate \(pOH = -\log[OH^-]\), then use \(pH + pOH = 14\) at \(25^\circ C\).

Answer 1

The Correct Option is C

Concept: \(NaOH\) is a strong base and completely dissociates in water: \[ NaOH \rightarrow Na^+ + OH^- \] Thus the hydroxide ion concentration equals the molarity of \(NaOH\). \[ [OH^-] = 0.001 = 10^{-3} \] Also, \[ pOH = -\log[OH^-] \] and \[ pH + pOH = 14 \quad \text{(at }25^\circ C\text{)} \]
Step 1: {Calculate \(pOH\).} \[ pOH = -\log(10^{-3}) \] \[ pOH = 3 \]
Step 2: {Find the pH.} \[ pH + pOH = 14 \] \[ pH = 14 - 3 \] \[ pH = 11 \]