magine that you are in chemistry lab and need to make 1.00 L of a solution with

Question

magine that you are in chemistry lab and need to make 1.00 L of a solution with a pH of 2.80. You have in front of you 100 mL of 7.00×102 M HCl, 100 mL of 5.00×102 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 80.0 mL of HCl and 90.0 mL of NaOH left in their original containers. Part A Assuming the final solution will be diluted to 1.00 L , how much more HCl should you add to achieve the desired pH?

Explanation / Answer

Dilution

Part A)

Final pH of solution = 2.80

pH = -log[H+]

so [H+] needed = 1.6 x 10^-3 M

moles [H+] needed = 1.6 x 10^-3 M x 1000 ml = 1.6 mmol

we have till now,

ml of HCl already added = 100 - 80 = 20 ml

moles HCl added = 7 x 10^-2 M x 20 ml = 1.4 mmol

ml of NaOH already added = 100 - 90 = 10 ml

moles NaOH added = 5 x 10^-2 M x 10 ml = 0.5 mmol

some HCl neutraizes NaOH in solution

excess moles HCl left = moles [H+] in solution = 1.4 - 0.5 = 0.9 mmol

So,

we need = 1.6 - 0.9 = 0.7 mmol of more H+ in solution

volume of additional HCl to be added = 0.7 mmol/7 x 10^-2 M = 10 ml

So we need to added another 10 ml of 7 x 10^-2 M HCl to make a solution with pH 2.80

Related Questions

Navigate

• Browse (All)
• Browse M
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.