Step 1: Understanding the Concept:
The pH of a solution is defined as the negative logarithm (base 10) of the molar concentration of hydrogen ions (\( [H^+] \)).
\[ \text{pH} = -\log[H^+] \]
A dibasic acid (or diprotic acid) is an acid that can release two hydrogen ions (\( H^+ \)) per molecule upon complete ionization.
Examples include sulfuric acid (\( H_2SO_4 \)).
Step 2: Key Formula or Approach:
For a strong dibasic acid \( H_2A \) of concentration \( C \):
\[ H_2A \rightarrow 2H^+ + A^{2-} \]
Since the acid is strong, it ionizes 100%.
The concentration of hydrogen ions will be twice the concentration of the acid:
\[ [H^+] = 2 \times C \]
Step 3: Detailed Explanation:
Given molarity of acid \( C = 0.01 \text{ M} \).
First, calculate the hydrogen ion concentration:
\[ [H^+] = 2 \times 0.01 = 0.02 \text{ M} \]
\[ [H^+] = 2 \times 10^{-2} \text{ M} \]
Now, apply the pH formula:
\[ \text{pH} = -\log[2 \times 10^{-2}] \]
Using the property of logarithms \( \log(m \times n) = \log m + \log n \):
\[ \text{pH} = -(\log 2 + \log 10^{-2}) \]
\[ \text{pH} = -(0.3010 + (-2)) \]
\[ \text{pH} = -(0.3010 - 2) \]
\[ \text{pH} = -(-1.699) \]
\[ \text{pH} \approx 1.7 \]
Step 4: Final Answer:
The pH of the 0.01 M strong dibasic acid is 1.7.