The hydrogen ion concentration of a sample of orange juice is 2.0 X 10$^{-11}$moldm$^{-3}$. What is its p$^{OH}$? [log102 = 0.3010]

1. Analyze the question

We are given the hydrogen ion concentration $[H^+]$ of a solution and asked to calculate the $pOH$.

• Given: $[H^+] = 2.0 \times 10^{-11} \, \text{mol dm}^{-3}$
• Given: $\log_{10} 2 = 0.3010$

2. Explain the concept

To solve this, we use two fundamental relationships in acid-base chemistry:

1. pH formula: $pH = -\log_{10}[H^+]$
2. Relationship between pH and pOH: $pH + pOH = 14.00$ (at 25°C)

3. Show the solution

Step 1: Calculate the pH $pH = -\log_{10}(2.0 \times 10^{-11})$

Using the laws of logarithms ($\log(a \times b) = \log a + \log b$): $pH = -(\log_{10} 2 + \log_{10} 10^{-11})$ $pH = -(0.3010 + (-11))$ $pH = -(0.3010 - 11)$ $pH = -(-10.699)$ $pH = 10.699 \approx 10.70$

Step 2: Calculate the pOH Using the relationship $pH + pOH = 14$: $10.70 + pOH = 14.00$ $pOH = 14.00 - 10.70$ $pOH = 3.30$

4. Identify the answer

Comparing our result to the given options, the correct value is 3.30. This matches option A.

5. Tip

Always check if the question asks for pH or pOH. A common mistake in JAMB exams is calculating the pH and stopping there, forgetting that the question specifically asked for the pOH. Remember: $pH + pOH$ is always 14.00!