Question

Imagine that you are in chemistry lab and need to make 1.00 L of a solution with a pH of 2.40.

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 83.0 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

The desired solution will have P^H of 2.40. The equation for the P^H is as follows:

P^H = -log[H⁺]

log[H⁺] = -P^H

log[H⁺] = -2.40

[H⁺] = 10^-2.40 = 3.98 $\times$ 10^-3 M

The volume of HCl taken in a beaker is = (100-83) mL = 17 mL. The volume of NaOH added to the beaker unintentionally = (100-90) mL = 10 mL.

17 mL 6.00 $\times$ 10^-2 M HCl = {(17$\times$6.00$\times$10^-2) / (1000)} moles = 1.02$\times$10^-3 moles of HCl

10 mL 5.00 $\times$ 10^-2 M NaOH = {(10$\times$5.00$\times$10^-2) / (1000)} moles = 0.50$\times$10^-3 moles of NaOH

The reaction of HCl with NaOH is as following:

NaOH + HCl $\rightarrow$ NaCl + H₂O

Thus, 0.5$\times$10^-3 moles of NaOH neutralizes 0.5$\times$10^-3 moles of HCl. The remaining moles of HCl in the beaker is = (1.02$\times$10^-3 - 0.50$\times$10^-3) moles = 0.52$\times$10^-3 moles of HCl.

The HCl dissociates almost completely in solution. Thus [HCl] = [H⁺]

Since the final solution will be diluted to 1.00 L (1000 mL), therefore in order to achieve the [H⁺] value of 3.98$\times$10^-3 moles in 1 L of the solution, the moles of H+ needed is = (3.98$\times$10^-3 - 0.52$\times$10^-3) moles = 3.46 $\times$ 10^-3 moles of H⁺.

As mentioned above, 3.46 $\times$ 10^-3 moles of H⁺= 3.46 $\times$ 10^-3 moles of HCl

Lets assume still y mL of HCl is needed.

Thus,

{(y$\times$6.00$\times$10^-2) / (1000)} moles = 3.46$\times$10^-3 moles

y = {(3.46$\times$10^-3) / (6.00$\times$10^-5)} mL = 57.6 mL (answer)