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 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?

Express your answer to three significant figures and include the appropriate units.

Answer 1

Volume of HCl to be added = 33.3 mL

Explanation

pH = 2.50

[H⁺] = 10^-pH

[H⁺] = 10^-2.50

[H⁺] = 3.16 x 10^-3 M

moles H⁺ to be added = (molarity H⁺) * (volume of solution)

moles H⁺ to be added = (3.16 x 10^-3 M) * (1.00 L)

moles H⁺ to be added = 3.16 x 10^-3 mol

moles HCl to be added = moles H⁺ to be added

moles HCl to be added = 3.16 x 10^-3 mol

moles HCl to be added = 3.16 x 10^-3 mol * (1000 mmol / 1 mol)

moles HCl to be added = 3.16 mmol

moles of HCl already added = (molarity HCl) * (volume HCl)

moles of HCl already added = (7.00 x 10^-2 M) * (100 mL - 81.0 mL)

moles of HCl already added = 1.33 mmol

moles of NaOH added = (molarity NaOH) * (volume NaOH)

moles of NaOH added = (5.00 x 10^-2 M) * (100 mL - 90.0 mL)

moles of NaOH added = 0.5 mmol

NaOH will neutralize HCl

effective moles of HCl in solution = (moles of HCl already added) - (moles of NaOH added)

effective moles of HCl in solution = (1.33 mmol) - (0.5 mmol)

effective moles of HCl in solution = 0.83 mmol

remaining moles of HCl to be added = (moles HCl to be added) - (effective moles of HCl in solution)

remaining moles of HCl to be added = (3.16 mmol) - (0.83 mmol)

remaining moles of HCl to be added = 2.33 mmol

volume of HCl to be added = (remaining moles of HCl to be added) / (molarity HCl)

volume of HCl to be added = (2.33 mmol) / (7.00 x 10^-2 M)

volume of HCl to be added = 33.3 mL