Week #2 Textbook Question 6B.3

[Carolina Gura Dis 1B]

Hi, for the textbook question 6B.3 the problem asks to a) calculate the pH of a laboratory technician's desired solution and b) calculate the pH of the actual solution prepared. The answer key shows that for part a), you simply take the negative log of the molarity for HCl(aq) but then for part b) the answer key multiplies the molarity by the desired volume of 200mL and then divides this answer by the actual volume of 250ml. I was just confused as to why you do this, are you supposed to cancel out both volumes in order to get your answer in molarity?

[Anita Sarransingh 2A]

Since the molarity of the actual solution is a value of moles per one liter of HCl, you would have to first find the amount of moles total in the solution, since that is the same amount that would have been added to the 200 mL instead of 250 mL. To find the value of m for a solution with given m / 1 L, you would multiply by the amount of liters in the desired solution, in this case 0.2 L. This would give the total amount of moles that were added. Next to get the new molarity, you would have to divide this number by 0.25 L, since that would give m / L of the actual solution.