Question

The pH of a solution is \( 4 \) then its \( \text{OH}^- \) ion concentration (in mol dm\(^{-3}\)) is

• A. \( 10^{-4} \)
• B. \( 10^{-10} \)
• C. \( 10^{-2} \)
• D. \( 10^{-12} \)
• E. \( 10^{-6} \)

Hint:

Always remember: \[ \text{pH} + \text{pOH} = 14 \] So if pH = 4, then pOH = 10 and \( [\text{OH}^-] = 10^{-10} \).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This question relates the pH of a solution to its hydroxide ion concentration ([OH\(^-\)]). This involves using the relationship between pH, pOH, and the ion product constant of water (\(K_w\)). Note that dm\(^3\) is equivalent to a litre (L).
Step 2: Key Formula or Approach:
There are two key relationships for aqueous solutions at 25°C:
1. pH + pOH = 14
2. pOH = -log\(_{10}\) [OH\(^-\)] or [OH\(^-\)] = 10\(^{-pOH}\)
Alternatively, we can use the ion product of water:
[H\(^+\)][OH\(^-\)] = K\(_w\) = 1.0 \(\times\) 10\(^{-14}\)
Step 3: Detailed Explanation:
Method 1: Using pOH
First, find the pOH of the solution.
Given: pH = 4
Using the formula pH + pOH = 14:
\[ 4 + \text{pOH} = 14 \] \[ \text{pOH} = 14 - 4 = 10 \] Now, calculate the hydroxide ion concentration from the pOH.
\[ [\text{OH}^-] = 10^{-\text{pOH}} \] \[ [\text{OH}^-] = 10^{-10} \text{ mol dm}^{-3} \] Method 2: Using K\(_w\)
First, find the hydrogen ion concentration [H\(^+\)] from the pH.
Given: pH = 4
\[ [\text{H}^+] = 10^{-\text{pH}} = 10^{-4} \text{ M} \] Now, use the ion product constant of water, \(K_w\).
\[ [\text{H}^+][\text{OH}^-] = 1.0 \times 10^{-14} \] Rearrange to solve for [OH\(^-\)]:
\[ [\text{OH}^-] = \frac{1.0 \times 10^{-14}}{[\text{H}^+]} \] \[ [\text{OH}^-] = \frac{1.0 \times 10^{-14}}{10^{-4}} = 10^{-14 - (-4)} = 10^{-10} \text{ mol dm}^{-3} \] Both methods yield the same result.
Step 4: Final Answer:
The OH\(^-\) ion concentration is 10\(^{-10}\) mol dm\(^{-3}\). This corresponds to option (B).