Question
100 ml of 0.01N HCl is added to 0.9L of distilled water. What would the final pH of the solution be?
Answers
DrBob222
0.01N HCl x (100 mL/1000 mL) = ?
Then pH = -log(HCl).
Note: This assumes the volumes are additive. Strictly speaking they are not; however, I think the intent of the question is to assume they are additive.
Then pH = -log(HCl).
Note: This assumes the volumes are additive. Strictly speaking they are not; however, I think the intent of the question is to assume they are additive.