Question

What is the concentration of OH\(^-\) in a solution with an H\(^+\) concentration of \( 2 \times 10^{-4} \, \text{M} \) ? The ion product of water at 25\(^\circ\)C is \( 1.0 \times 10^{-14} \, \text{M}^2 \).

• A. \( 2 \times 10^{10} \, \text{M} \)
• B. \( 5 \times 10^{-11} \, \text{M} \)
• C. \( 0.5 \times 10^{-12} \, \text{M} \)
• D. \( 0.2 \times 10^{10} \, \text{M} \)

Hint:

An alternative check: \( \text{pH} + \text{pOH} = 14 \).
\( \text{pH} = -\log(2 \times 10^{-4}) \approx 4 - 0.3 = 3.7 \).
\( \text{pOH} = 14 - 3.7 = 10.3 \).
\( [\text{OH}^-] = 10^{-10.3} \approx 5 \times 10^{-11} \). Both methods yield the exact same result!

Answer 1

The Correct Option is B