Question

The pH of \(10^{-8}\) M HCl solution will be

• A. 8.00
• B. 6.00
• C. 6.96
• D. 1.00

Hint:

For very dilute strong acids (\(<10^{-6}\) M), water autoionization must be considered.
Total [H+] = [H+] from acid + [H+] from water.
pH of \(10^{-8}\) M HCl = 6.96 (not 8.00).

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
pH is defined as the negative logarithm of hydrogen ion concentration in a solution.
For strong acids like HCl, complete dissociation occurs in aqueous solution.
However, at very low concentrations (\(<10^{-6}\) M), the contribution of H+ ions from water autoionization cannot be neglected.

Step 2: Key Formula or Approach:
pH = -log[H+].
For very dilute strong acids, total [H+] = [H+] from acid + [H+] from water.
Water contributes \(10^{-7}\) M H+ ions (from \(K_w = 10^{-14}\)).

Step 3: Detailed Explanation:
For \(10^{-8}\) M HCl:
- The acid contributes \(10^{-8}\) M H+ ions.
- Water contributes \(10^{-7}\) M H+ ions (from autoionization).
Total [H+] = \(10^{-8} + 10^{-7} = 1.1 \times 10^{-7}\) M.
pH = -log(1.1 \(\times\) 10\(^{-7}\)) = -[log(1.1) + log(10\(^{-7}\))] = -(0.0414 - 7) = 6.96.
Thus, pH is 6.96, not 8.00 (alkaline) because the solution remains acidic due to water's H+ contribution.

Step 4: Final Answer:
Thus, the pH of \(10^{-8}\) M HCl is 6.96, which corresponds to option (C).
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