I am completely lost with this question. If 50 ml of 0.5M HCl is added to 800 ml of pure water, what will be the approximate pH of the resulting solution?

Answer 1

First you must assume that the volumes are additive; i.e., the total volume will be 850
(HCl) = 0.5M x (50/850) = ?
Then pH = -log(HCl) = ?

Answer 2

Thank you so much!