Question

Imagine that you are in chemistry lab and need to make 1.00 L of a solution with a pH of 2.50.

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 81.0 mL of HCl and 85.0 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 = 2.5 L required

a)

V = 81 mL of HCl, V = 85 mL of NaOH

if V final = 1 L

[H+] required = 10 ^-pH = 10^-2.5 = 0.0031622 M

for 1L --> mol of H+ = 0.0031622*! = 0.0031622mol

then...

mol of H+ --> 0.0031622

we have already:

mol of HCl = MV = (100-81)*(7*10^-2) = 1.33 mmol of H+ added

mol of NaOH= MV = (100-85)*(5*10^-2) = 0.75 mmol of OH- added

mmol of H+ left = 1.33 -0.75 = 0.58 mmol =0.00058 mol

we need 0.0031622mol

so

mol of H+ = 0.0031622 - 0.00058 = 0.0025822 mol of H+ extra

V of Hcl solution required extra = (mol/M) = (0.0025822)/(7*10^-2) = 0.03688 Liter = 0.03688*10^3 mL = 36.8 mL of HCl is required