Imagine that you are in chemistry lab and need to make 1.00 L of a solution with
ID: 1015383 • Letter: I
Question
Imagine that you are in chemistry lab and need to make 1.00 L of a solution with a pH of 2.60. You have in front of you 100 mL of 6.00×102 M HCl, 100 mL of 5.00×102 M NaOH, and plenty of distilled water. You start to add HCl to a beaker of water when someone asks you a question. When you return to your dilution, you accidentally grab the wrong cylinder and add some NaOH. Once you realize your error, you assess the situation. You have 81.0 mL of HCl and 85.0 mL of NaOH left in their original containers. Part A Assuming the final solution will be diluted to 1.00 L , how much more HCl should you add to achieve the desired pH? Express your answer to three significant figures and include the appropriate units.
Explanation / Answer
You have 81.0 mL of HCl and 85.0 mL of NaOH left in their original containers.
SO you have added 19 mL of HCl and 15 mL of NaOH
So moles of HCl remaining unreacted is
0.019 x 0.06 -0.015 x 0.05 = 0.00039 moles of HCl is present
pH 2.6 = 10-2.6 = 0.002511 M H+
You need to have 0.002511 moles of HCl in 1 L
You already have 0.00039 moles so you need 0.002511 - 0.00039 = 0.00212 moles
which will be V x 0.06 = 0.00212 moles
V = 0.0354 L or 35.4 mL
So you must add 35.4 mL more of HCl and then dilute to 1 L to get a 1L solution of 2.6 pH