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 82.0 mL of HCl and 86.0 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

HCl is in the solution:

100 mL(Amount started with) – 82 mL(amount left) = 18 mL of HCl.

NaOH is in the solution:

100 mL(amount started with) – 86 mL(amount left) = 14 mL of NaOH.

Converting volume into moles

(18 / 1000) L x 7.00 x 10^-2 M = 1.26 x 10^-3 mol of HCl

(14 / 1000) L x 5.00 x 10^-2 M = 7.0 x 10^-4 mol of NaOH

There is excess HCl = 1.26 x 10^-3 – 7.0 x 10^-4 = 5.6 x 10^-4 moles of HCl

Now, to determine the total amount of HCl needed to reach a pH of 2.8.

pH = 2.8

-log [H⁺] = 2.8

[H⁺] = 10^-2.8 =1.585 x 10^-3

Hence, 1.585 x 10^-3 M HCl is the concentration needed to have a pH of 2.8.

Moles = Molarity x Liters

1.585 x 10^-3 M x 1.00 L = 1.585 x 10^-3 moles HCl is needed to have a pH of 2.8.

So, (Total moles of HCl needed) - (Moles of HCl in solution) = Moles of HCl needed to be added

1.585 x 10^-3 moles - 5.6 x 10^-4 moles = 10.25 x 10^-4 moles HCl

Now, converting moles of HCl to Liter (or mL) of HCl.

Molarity = Moles/Liter

Liter = Moles/Molarity

(10.25 x 10^-4 moles) / (7.00 x 10^-2 M) = 1.46 x 10^-2 L = 14.6 mL of HCl

Hence, 14.6 mL of HCl is needed to be added to the solution to have a pH of 2.8. The Total solution should be 1.00 L which is achieved by adding water to your solution. The dilution with water will not affect your pH level.