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 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 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?

Answer 1

pH = 2.7, therefore amount of H+ needed is 10^-2.7 M = 0.002 M
i have 82.0 ml of hcl and 86.0 ml of naoh left, therefore 18 ml of hcl used, and 14.0 ml of naoh used.
mols of H+ = 0.018 x 6 x 10^-2 = 1.08 x 10^-3
mols of OH- = 0.014 x 5 x 10^-2 = 7.0 x 10^-4
H+ and OH- neutralise each other, so remaining moles of H+ = 3.8 x 10^-4
i need 10^-2.7 mols of H+, so

10^-2.7 - (3.8 x 10^-4) = 0.00162 mol

volume of HCl needed = 0.00162 / 6.00 x 10^-2 = 0.027 L = 27 mL