Imagine that you are in chemistry lab and need to make 1.00 L of a solution with
ID: 506702 • Letter: I
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 6.00×102M HCl,
- 100 mL of 5.00×102M 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 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?
Explanation / Answer
Ans. #1. Given,
[HCl] stock = 6.00 x 10-2 M ; Volume = 100.0 mL
[NaOH] stock = 5.0 x 10-2 M ; Volume = 100.0 mL
#2. C1V1 = C2V2 - equation 1
Total content of reaction mixture-
(100.0 mL – 84.0 mL) HCl + (100.0 mL – 90.0 mL) NaOH
= 16.0 mL HCl + 10.0 mL NaOH
= 26.0 mL
Moles of HCl = Volume in liters x Molarity
= 0.016 L x (6.00 x 10-2 M)
= 0.00096 mol = 0.096 x 10-2 mol
Moles of NaOH = Volume in liters x Molarity
= 0.010 L x (5.00 x 10-2 M)
= 0.0005 mol = 0.05 x 10-2 mol
1 mol NaOH neutralizes 1 mol HCl.
So,
Remaining moles of HCl in reaction mixture = Moles of NaOH – Moles of HCl
= 0.00096 mol - 0.0005 mol
= 0.00046 moles
#3. pH = -log [H+]
Or, 2.40 = -log [H+]
Or, [H+] = antilog (- 2.40) = 0.0039810717055
Therefore, required concentration of [H+] to get pH 2.40 = 0.003981 M
Moles of H+ required for 1.0 L solution = Volume in liters x Molarity
= 1.0 L x 0.003981 M
= 0.003981 mol
#4. 1 mol HCl donates 1 mol H+.
So, moles of HCl required to get pH 2.40 is equal to moles of H+ required for the same.
So,
Required moles of HCl in 1.0 L solution = 0.003981 moles
Moles of HCl remaining in the reaction mixture = 0.00046 moles ; [see #2]
Moles of HCl required to be added =
Required moles for 1.0 L – Moles of HCl remaining in reaction mixture
= 0.003981 moles - 0.00046 moles
= 0.003521 mol
Now,
Volume of HCl required to give 0.003521 mol HCl is given by-
Moles of HCl = Volume in liters x Molarity
Or, 0.003521 mol = V x (6.00 x 10-2 M) = V x (0.06 mol/ L)
Or, V = 0.003521 mol / (0.06 mol/ L) = 0.05868 L = 56.68 mL
Thus, required volume of additional HCl needed to be added to reaction mixture = 56.68 mL
#6. Final solution 1.0 L = 1000.0 mL
Volume of water required =
1000.0 mL – (Vol/ of reaction mixture + Vol. of additional HCl )
= 1000.0 mL – (26.0 mL + 56.68 mL)
= 917.32 mL
Solution preparation: To the 26.0 mL reaction mixture, add 56.68 mL additional HCl from stock. Make the final volume to 1000.0 mL with distilled water by adding 917.32 mL distilled water. The solution has volume of 1.0 L, and pH of 2.40.