Question

The pH of 0.005 M KOH is 9.95. Calculate the \([OH^-]\)?

• A. \( 6.71 \times 10^{-4} \, \text{M} \)
• B. \( 1.12 \times 10^{-4} \, \text{M} \)
• C. \( 4.45 \times 10^{-5} \, \text{M} \)
• D. \( 8.91 \times 10^{-5} \, \text{M} \)

Hint:

If a strong base like KOH is given with a pH that does not match its concentration, always use the given pH to find [OH⁻] via pOH = 14 – pH.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
pH of the solution is given as 9.95. We need to find hydroxide ion concentration \([OH^-]\).

Step 2: Key Formula or Approach:
\( \text{pOH} = 14 - \text{pH} \) and \([OH^-] = 10^{-\text{pOH}}\).

Step 3: Detailed Explanation:
Given pH = 9.95, so pOH = 14 – 9.95 = 4.05. \([OH^-] = 10^{-4.05}\). \(10^{-4.05} = 10^{-4} \times 10^{-0.05}\). \(10^{-0.05} = 10^{-1/20} \approx 0.891\). Thus \([OH^-] \approx 8.91 \times 10^{-5} \, \text{M}\).

Step 4: Final Answer:
Option (D) is correct.