Question

What is the pH of 0.02 M NaOH solution?

• A. 10.3
• B. 11.3
• C. 11.7
• D. 12.3

Hint:

To quickly estimate logarithms of the form $a \times 10^{-b}$, the negative log is simply $b - \log(a)$. Memorizing $\log(2) \approx 0.3$ and $\log(3) \approx 0.48$ will save you lots of calculation time!

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We need to find the pH of a strong base solution given its molarity ($0.02\text{ M NaOH}$).

Step 2: Key Formula or Approach:
Since NaOH is a strong base, it dissociates completely in water: $\text{NaOH} \rightarrow \text{Na}^+ + \text{OH}^-$.
The concentration of hydroxide ions $[\text{OH}^-]$ will be equal to the concentration of the NaOH solution.
Calculate the pOH first using $pOH = -\log_{10}[\text{OH}^-]$.
Then, use the relationship at standard temperature: $pH + pOH = 14$ to find the pH.

Step 3: Detailed Explanation:
The concentration $[\text{OH}^-] = 0.02\text{ M} = 2 \times 10^{-2}\text{ M}$.
Calculate the pOH:
$$pOH = -\log_{10}(2 \times 10^{-2})$$
$$pOH = -(\log_{10}2 + \log_{10}10^{-2})$$
$$pOH = -(0.3010 - 2)$$
$$pOH = 2 - 0.3010 = 1.699$$
Now, calculate the pH:
$$pH = 14 - pOH$$
$$pH = 14 - 1.699 = 12.301$$

Step 4: Final Answer:
The pH of the solution is approximately 12.3, which corresponds to option (D).