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 85.0 mL of HCl and 90.0 mL of NaOH left in their original containers.

Q: 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

if we want fina pH = 2.7

[H+] = 10^-pH = 10^-2.7 = 0.001995 M

so..

V final = 1 L

so final mol of H+ = MV = 0.001995 mol of H+

mmol of HCl initialy = 85*(7*10^-2) = 5.95 mmol of H+

mmol of NaOH initially = 90*(5*10^-2) = 4.5 mmol of OH-

so, there was reaction

5.95-4.5 = 1.45 mmol of H+ left

1.45*10^-3 mol of H+

so we need = 0.001995 - (0.00145) = 0.000545 mol of H+

we need this form HCl solution

so

V = mol/M

V = (0.000545)/(7*10^-2) = 0.00778571 L = 7.78 mL of HCl solution will do