Question

Calculate the pH of \(0.01\ \text{M}\) strong dibasic acid.

• A. \(5.5\)
• B. \(2.5\)
• C. \(2.0\)
• D. \(1.7\)

Hint:

For strong acids:
• Monobasic acid: \[ [H^+] = C \]
• Dibasic acid: \[ [H^+] = 2C \]
• Tribasic acid: \[ [H^+] = 3C \] Always multiply concentration by the number of ionizable \(H^+\) ions before calculating pH.

Answer 1

The Correct Option is D

Concept: A strong dibasic acid ionizes completely in water and releases two hydrogen ions \((H^+)\) per molecule. General form: \[ H_2A \rightarrow 2H^+ + A^{2-} \] The pH of a solution is calculated using: \[ \text{pH} = -\log[H^+] \]

Step 1: Calculating hydrogen ion concentration.
Given concentration of strong dibasic acid: \[ 0.01\ \text{M} \] Since it is dibasic, each molecule gives: \[ 2H^+ \] Therefore, \[ [H^+] = 2 \times 0.01 \] \[ [H^+] = 0.02\ \text{M} \]

Step 2: Applying the pH formula.
\[ \text{pH} = -\log(0.02) \] Rewrite: \[ 0.02 = 2 \times 10^{-2} \] Thus, \[ \text{pH} = -\log(2 \times 10^{-2}) \] Using logarithm properties: \[ \text{pH} = -( \log 2 + \log 10^{-2}) \] \[ \text{pH} = -(0.3010 - 2) \] \[ \text{pH} = 2 - 0.3010 \] \[ \text{pH} \approx 1.7 \]

Step 3: Final conclusion.
Hence, the pH of the solution is: \[ \boxed{1.7} \] Therefore, the correct option is: \[ \boxed{(4)\ 1.7} \]