Question

Imagine 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 7.00×10−2 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 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? Express your answer to three significant figures and include the appropriate units.

Answer 1

Volume of HCl = 100 mL – 8 mL = 16 mL

Volume of NaOH =100 mL – 86 mL = 14 mL

Moles of HCl = Volume x Molarity = 0.016 L x 7 x 10^-2 M = 1.12 x 10^-3 moles

Moles of NaOH = Volume x Molarity = 0.014 L x 5.00 x 10^-2 M = 7.0 x 10^-4 moles

Excess moles of HCl = 1.12 x 10^-3 – 7.0 x 10^-4 = 4.2 x 10^-4 moles

Total amount of HCl needed to reach pH of 2.8 can be calculated as

pH = 2.8

-log [H⁺] = 2.8

Therefore,

[H⁺] = 10^-2.8 =1.585 x 10^-3

1.585 x 10^-3 M of HCl is required to have a pH of 2.8

Moles of HCl required to reach pH 2.8 = 1.585 x 10^-3 M x 1 L = 1.585 x 10^-3 moles

Therefore,

Moles of HCl to be added =

HCl is needed to have a pH of 2.8. = Moles of HCl needed - Moles of HCl in solution

                                                     = (1.585 x 10^-3 moles) – (4.2 x 10^-4 moles) = 1.165 x 10^-3 moles

Liter of HCl needed = Moles/Molarity = (1.165 x 10^-3 moles) / (7 x 10^-2 M)

                                                                = 1.646 x 10^-2 L = 16. 64mL of HCl