Q: Calculate the pH of a solution prepared by diluting 3.0mL of 2.5M HCl to a final volume of 100mL with H2O.
A: pH = 1.1
I first found the pH of 2.5M HCl by using: pH = -log [H+]. I calculated a pH of -0.398. I then substituted this into the Henderson-Hasselbalch equation as the pKa: pH = pKa + log ([base]/[acid]). For the ratio within the log, I used 97% for the base (water) and 3.0% for the acid (HCl). I used these numbers because the ratio was out of 100mL = 3.0mL(acid) + 97ml(base). After calculating I got the correct answer of pH = 1.1
Did I do this problem correctly or did I simply get lucky? I understand the steps that I performed...I didn't just plug in numbers, but I feel as if I could have done it differently. Is there a different way to calculate this problem?
Thank you