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

Step 1: Concept:
The question asks for the hydroxide ion \([\text{OH}^-]\) concentration in an aqueous solution given the hydrogen ion \([\text{H}^+]\) concentration and the constant ion product of water (\(K_w\)).

Step 2: Key Formula or Approach:
At standard temperature (25\(^\circ\)C), the auto-ionization constant of water, \(K_w\), establishes an inverse relationship between \([\text{H}^+]\) and \([\text{OH}^-]\):
\[ K_w = [\text{H}^+] \times [\text{OH}^-] \]
To find \([\text{OH}^-]\), rearrange the formula:
\[ [\text{OH}^-] = \frac{K_w}{[\text{H}^+]} \]

Step 3: Step-by-step Explanation:

• Identify Given Values:
\( K_w = 1.0 \times 10^{-14} \)
\( [\text{H}^+] = 2.0 \times 10^{-4} \, \text{M} \)

• Perform the Calculation:
\[ [\text{OH}^-] = \frac{1.0 \times 10^{-14}}{2.0 \times 10^{-4}} \]
\[ [\text{OH}^-] = \left( \frac{1.0}{2.0} \right) \times \left( \frac{10^{-14}}{10^{-4}} \right) \]
\[ [\text{OH}^-] = 0.5 \times 10^{-14 - (-4)} \]
\[ [\text{OH}^-] = 0.5 \times 10^{-10} \, \text{M} \]

• Standardize into Scientific Notation:
To convert \( 0.5 \times 10^{-10} \) into standard scientific notation, move the decimal one place to the right, which decreases the exponent by 1:
\[ [\text{OH}^-] = 5.0 \times 10^{-11} \, \text{M} \]

Step 4: Final Answer:
The calculated concentration is exactly Option (B).