# Assuming the final solution will be diluted to 1.00 L , how much more HCl should you add to achieve the desired pH?  ?

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 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 84.0 mL of HCl and 88.0 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?

Relevance

(100 mL HCl initially) - (84.0 mL HCl left) = 16.0 mL HCl used

(16.0 mL HCl) x (6.00×10^−2 M HCl) = 0.960 mmol HCl used

(100 mL NaOH initially) - (88.0 mL NaOH left) = 12.0 mL NaOH used

(12.0 mL NaOH) x (5.00×10^−2 M NaOH) = 0.600 mmol NaOH used

10^-2.50 = 0.003162 M (the desire concentration of H+ and HCl)

(1.00 L) x (0.003162 M) = 0.003162 mol =

3.162 mmol HCl needed in the final solution

NaOH and HCl react in equimolar amounts.  So 0.600 millimoles of NaOH will neutralize 0.600 millimoles of HCl.

(3.162 mmol HCl for the final solution) +

(0.600 mmol HCl to correct the NaOH mistake) = 3.762 mmol HCl total needed

(3.762 mmol HCl total needed) - (0.960 mmol HCl already added) =

2.802 mmol HCl additional required after the NaOH mistake

(2.802 x 10^-3 mol HCl) / (6.00×10^−2 mol HCl/L) = 0.0467 L =

46.7 mL HCl to be added