Photo AI
10^{-2} mole of HCl is added to exactly 1.00 L of pure water at 25°C - VCE - SSCE Chemistry - Question 5 - 2004 - Paper 1
Question 5
10^{-2} mole of HCl is added to exactly 1.00 L of pure water at 25°C. The change in pH of the water is closest to A. 10^{-2} B. 2 C. 5 D. 7
Worked Solution & Example Answer:10^{-2} mole of HCl is added to exactly 1.00 L of pure water at 25°C - VCE - SSCE Chemistry - Question 5 - 2004 - Paper 1
Step 1
Step 2
Determine the pH of the solution
Answer
Hydrochloric acid (HCl) is a strong acid, which means it fully dissociates in solution. Thus, the concentration of hydrogen ions, [ [H^+] ], is equal to the concentration of HCl:
[ [H^+] = 10^{-2} \text{ M} ]
Now, compute the pH:
[ pH = -\log[H^+] = -\log(10^{-2}) = 2 ]
Step 3
Determine the change in pH
Answer
Since the initial pH of pure water at 25°C is 7, the change in pH can be calculated as follows:
[ \Delta pH = pH_{ ext{final}} - pH_{ ext{initial}} = 2 - 7 = -5 ]
The absolute value of the change is 5, so the closest answer choice reflecting the change in pH is 2 (since it indicates a shift towards acidity).