Question

Imagine that you are in chemistry lab and need to make 1.00 Lof a solution with a pH of 2.80.

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 HCland 90.0 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

we know that

pH = -log [H+]

so

-log [H+] = 2.8

[H+] = 1.5849 x 10-3 M

we know that

moles = molarity x volume (L)

so

moles of H+ = 1.5849 x 10-3 x 1

moles of H+ = 1.5849 x 10-3

volume of HCl used = 100 - 81 = 19 ml

volume of NaOH used = 100 - 90 = 10 ml

now

we know that

moles = molarity x volume (L)

moles of HCl added = 0.07 x 19 x 10-3 = 1.33 x 10-3

moles of NaOH added = 0.05 x 10 x 10-3 = 0.5 x 10-3

now

the reaction is

HCl + NaOH ---> NaCl + H20

moles of HCl reacted = moles of NaOH added = 0.5 x 10-3

so

moles of HCl left = 1.33 x 10-3 - 0.5 x 10-3

moles of HCl left = 0.83 x 10-3

now

HCl---> H+ + Cl-

we can see that

moles of H+ present = moles of HCl = 0.83 x 10-3

now

moles of H+ needed = final moles of H+ - moles of H+ present

moles of H+ needed = ( 1.5849 x 10-3 ) - ( 0.83 x 10-3)

moles of H+ needed = 0.7549 x 10-3

so

moles of HCl need = 0.7549 x 10-3

now

volume(L) = moles / molarity

so

volume of HCl needed (L) = 0.7549 x 10-3 / 0.07

volume (L) = 0.010784

volume (ml) = 0.010784 x 1000

volume (ml) = 10.784

so

further 10.784 ml of HCl is to be added to acheive the desired pH