Question

Imagine that you are in chemistry lab and need to make 1.00 L of a solution with a pH of 2.70.

You have in front of you

* 100 mL of 7.00×10−2 M HCl
* 100 mL of 5.00×10−2 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 83.0 mL of HCl and 88.0 mL of NaOH left in their original containers.

Assuming the final solution will be diluted to 1.00 L, how much more HCl should you add to achieve the desired pH?

Answer 1

Start with 100 mL HCl, left with 83.0 means you have added 17.0 mL HCl to the beaker.
Start with 100 mL NaOH, left with 88 mL means you have added 12 mL NaOH.
Calculate moles HCl and moles NaOH added and subtract to determine the excess acid in the beaker. Then use pH 2.70 to determine (H^+) needed.
Finally, knowing how much H^+ is needed, calculate the amount of HCl needed to make the desired pH. Check my thinking. Post your work if you need further assistance.

Answer 2

what answer did you get??