posted by Saira .
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 6.00×10−2 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 85.0 mL of HCl and 89.0 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?
You added 85ml of acid and 89ml base. So an excess of 4 ml of base you started with.
So, add 4 ml of acid at the end to bring the original system to neutral.
2.70=-logH= log x
That is the H concentration desired.
voluem of acid= 33.3ml
and add the 4ml we determined above, so 37.3ml of the HCl solution.
Check my thinking and calcs.
I have interpreted this problem differently than Bob Pursley.
I think the problem states that 85 mL HCl and 89 mL NaOH WERE LEFT when the accident was discovered; therefore, we have added 100-85 = 15 mL HCl and 100-89 = 11 mL NaOH.
Therefore, mmoles HCl added = 15 mL x 0.06 M = 0.9 mmoles HCl.
mmoles NaOH added = 11 mL x 0.05 M = 0.55 mmoles NaOH.
0.9 mmoles HCl - 0.55 mmoles NaOH = 0.35 mmoles HCl in excess. How much more do we need to add?
For pH 2.70, (H^+) = 1.995 x 10^-3 which I have rounded to 2.00 x 10^-3 moles/ L or 2.00 mmoles/1000 mL.
2.00 mmoles needed - 0.35 mmoles in solution = 1.65 mmoles additional HCl to be added.
mL x M = mmoles or
mL = mmoles/M = 1.65/0.06 = 27.5 mL additional HCl to be added.
Check my work. Check my thinking. Note that I have used a different concentration for HCl and NaOH (one 0.06 and the other 0.05).