Imagine that you are in chemistry lab and need to make 1.00 L of a solution with
ID: 502379 • Letter: I
Question
Imagine that you are in chemistry lab and need to make 1.00 L of a solution with a pH of 2.80.
You have in front of you
100 mL of 6.00×102M HCl,
100 mL of 5.00×102M 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 82.0 mL of HCl and 86.0 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?
Explanation / Answer
The solution now contains (100-86)= 14 mL of NaOH and (100-82) = 18 mL of Hcl
HCl + NaOH -------------> NaCl + H2O
18x6 x10-2 14x 5 x10-2 0 0 initial mmoles
38x10-2 0 70x10-2 - after reaction.
The solution now contains 0.38 mmoles of HCl in it .
The pH of the required solution = 2.8 ,
The [H+] to be present in it = antilog -2.8
= 1.585x10-3 M
Since the volume of solution is 1 L the solution must contain [H+] = [HCl] = 1.585x10-3 M
or the mmoles of HCl that has to be in 1L of solution = 1.585 mmoles
The solution already has 0.38 mmoles
TThe mmoles of HCl that has to be added = 1.585 -0.38
= 1.205 mmoles
We know this comes form the acid we have and
mmoles = molarity x volume in mL
1.205 mmoles = 6.0x10-2 xV
or v = 20.08 mL
So to the solution , we need to add 20.08 mL Hcl and dilute it to 1L.