Question

An aqueous solution of HCl with pH 1.0 is diluted by adding equal volume of water (ignoring dissociation of water). The pH of HCl solution would be: $ \text{(Given log 2 = 0.30)} $

• A. reduce to 0.5
• B. increase to 1.3
• C. remain same
• D. increase to 2

Hint:

When diluting an acidic solution, the concentration of hydrogen ions is halved, leading to an increase in pH. Always use the logarithmic formula to calculate the change in pH.

Answer 1

The Correct Option is B

Step 1: Understanding pH and dilution
The pH of a solution is related to the concentration of hydrogen ions (\([H^+]\)) in the solution by the equation: \[ \text{pH} = -\log [H^+] \] For an aqueous HCl solution with pH 1.0, the concentration of hydrogen ions \([H^+]\) is: \[ \text{pH} = 1.0 \quad \Rightarrow \quad [H^+] = 10^{-1} = 0.1 \, \text{M} \] 
Step 2: Diluting the solution
When an equal volume of water is added to the solution, the concentration of hydrogen ions is halved (since the volume doubles). 
Therefore, the new concentration of \([H^+]\) will be: \[ [H^+]_{\text{new}} = \frac{0.1}{2} = 0.05 \, \text{M} \] 
Step 3: Calculating the new pH
The pH of the diluted solution is given by: \[ \text{pH}_{\text{new}} = -\log (0.05) \] Using the logarithm property \(\log 0.05 = \log (5 \times 10^{-2}) = \log 5 + \log 10^{-2}\), we get: \[ \log 0.05 = \log 5 - 2 = 0.69897 - 2 = -1.30103 \] Thus: \[ \text{pH}_{\text{new}} = -(-1.30103) = 1.30103 \approx 1.3 \] Therefore, the pH increases to 1.3 after dilution. 
Thus, the correct answer is option (2).

Answer 2

The given data: \[ \text{HCl}_{\text{aq}} \quad pH = 1 \quad ; \quad [H^+] = 10^{-1} \] If equal volume of water is added, the concentration will become half: \[ [H^+]_{\text{sol}} = \frac{10^{-1}}{2} \] Therefore, the new pH is: \[ pH = 1.3 \]