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 7.00×10⁻²M HCl,

100 mL of 5.00×10⁻²M 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 87.0 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

The final concentration of H+ that will give pH=2.70 can be calculated as below:-
pH = -log[H+]
So [H+] = 10^-pH
[H+] = 0.001995 mol/L

Therefore,

0.001995 moles of H+ need to be in the flask for pH=2.70

We have already added 17.0 ml of 0.0700 M HCl
moles H+ = M x L = 0.0700 M x 0.0170 L = 0.00119 moles HCl

We have also added 13.0 ml of 0.0500 M NaOH
moles OH- = 0.0500 M x 0.0130 L = 0.000650 moles NaOH

1 mol OH- neutralises 1 mole H+

So number of H+ moles = 0.00119 moles - 0.000650 moles = 0.00054 moles

After the distraction we have 0.00054 moles of HCl in the flask

We need to add a further
0.001995 mol - 0.00054 = 0.001455 moles

Volume of the HCl solution that has this many moles:

Molarity = moles / litres
Therefore litres = moles / molarity
= 0.001455 mol / 0.0700 M
= 0.02078 L
= 20.78 ml

Therefore, we need to add further 20.78 ml of the HCl solution to the flask.