Imagine that you are in chemistry lab and need to make 1.00 L of a solution with a pH of 2.70.

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 85.0 mL of HCl and 89.0 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?

To calculate the required amount of HCl needed to achieve the desired pH, we need to consider the initial concentration of HCl, the volume of HCl already added, and the volume of the final solution.

1. Calculate the amount of HCl already added:
- initial volume of HCl: 100 mL
- unused volume of HCl: 85.0 mL
- total volume of HCl added: 100 mL - 85.0 mL = 15.0 mL

2. Calculate the molarity of the HCl already added:
- initial concentration of HCl: 6.00×10^-2 M
- moles of HCl added: concentration × volume = 6.00×10^-2 M × 0.015 L = 0.0009 moles

3. Calculate the amount of NaOH accidentally added:
- initial volume of NaOH: 100 mL
- unused volume of NaOH: 89.0 mL
- total volume of NaOH added: 100 mL - 89.0 mL = 11.0 mL

4. Calculate the molarity of the NaOH accidentally added:
- initial concentration of NaOH: 5.00×10^-2 M
- moles of NaOH added: concentration × volume = 5.00×10^-2 M × 0.011 L = 0.00055 moles

5. Determine the number of moles required to reach the desired pH:
- Since HCl is a strong acid and NaOH is a strong base, they react in a 1:1 ratio. So, the moles of HCl needed to neutralize the NaOH accidentally added could be considered as the moles of HCl needed to reach the desired pH.
- Moles of HCl needed: 0.00055 moles

6. Calculate the volume of the final solution:
- Desired volume of the final solution: 1.00 L
- Volume of water remaining: 1.00 L - (15.0 mL + 11.0 mL) = 0.974 L

7. Calculate the required concentration of HCl for the final solution:
- moles of HCl needed: 0.00055 moles
- required concentration of HCl: moles/volume = 0.00055 moles / 0.974 L ≈ 5.65×10^-4 M

Therefore, you should add enough HCl to the remaining volume of water (about 974 mL) to make a solution with a concentration of approximately 5.65×10^-4 M and achieve the desired pH of 2.70.

To determine how much more HCl you need to add to achieve the desired pH, you need to calculate the concentration of HCl and the resulting pH after adding the remaining HCl and NaOH.

First, let's determine the initial concentration of HCl before any dilution or addition. You have 100 mL of 6.00×10^(-2) M HCl, which means there are (6.00×10^(-2) mol/L) * (0.100 L) = 0.006 mol of HCl initially.

Next, let's calculate the concentration of NaOH added by considering its initial concentration and the amount mistakenly added. You have 100 mL of 5.00×10^(-2) M NaOH, which means there are (5.00×10^(-2) mol/L) * (0.100 L) = 0.005 mol of NaOH initially. However, you accidentally added 89.0 mL of NaOH, so the actual amount added is (5.00×10^(-2) mol/L) * (0.0890 L) = 0.00445 mol of NaOH.

Now, let's calculate the total moles of HCl and NaOH in the mixture:

Total moles of HCl = initial HCl - HCl used + HCl added
= 0.006 mol - (0.100 L/L00 mL) * (0.0850 L) + HCl added

Total moles of NaOH = initial NaOH - NaOH used + NaOH added
= 0.005 mol - (0.100 L/100 mL) * (0.0890 L) + 0.00445 mol

Since HCl and NaOH react in a 1:1 stoichiometric ratio, the total moles of water (H2O) formed from their reaction will be the same for both. Thus, the moles of H2O formed can be calculated as follows:

Moles of H2O formed = initial HCl - HCl used + HCl added
= 0.006 mol - (0.100 L/100 mL) * (0.0850 L) + HCl added

To achieve a pH of 2.70, we need to calculate the concentration of H3O+ (hydronium ion) in the solution. The pH=-log[H3O+], so rearranging this equation gives [H3O+] = 10^(-pH).

[H3O+] = 10^(-2.70) = 0.001995 mol/L

The concentration of H3O+ is directly related to the concentration of HCl. Since HCl is a strong acid that dissociates completely in water, the concentration of H3O+ is equal to the concentration of HCl.

0.001995 mol/L = 0.006 mol - (0.100 L/100 mL) * (0.0850 L) + HCl added

Now, isolate HCl added:

HCl added = 0.001995 mol/L - (0.006 mol - (0.100 L/100 mL) * (0.0850 L))

Finally, calculate the volume of HCl required to achieve the desired pH:

Volume of HCl required = HCl added / (0.006 mol/L)

Remember to convert the volume to the proper unit (L or mL) before adding it to the solution.

You added 85ml of acid and 89ml base. So an excess of 4 ml of base you started with.

So, add 4 ml of acid at the end to bring the original system to neutral.

2.70=-logH= log x

x= 1.99E-3

That is the H concentration desired.
So, 1.99E-3moles=6E-2*volumeinliters
voluem of acid= 33.3ml
and add the 4ml we determined above, so 37.3ml of the HCl solution.

Check my thinking and calcs.

I have interpreted this problem differently than Bob Pursley.

I think the problem states that 85 mL HCl and 89 mL NaOH WERE LEFT when the accident was discovered; therefore, we have added 100-85 = 15 mL HCl and 100-89 = 11 mL NaOH.
Therefore, mmoles HCl added = 15 mL x 0.06 M = 0.9 mmoles HCl.
mmoles NaOH added = 11 mL x 0.05 M = 0.55 mmoles NaOH.
0.9 mmoles HCl - 0.55 mmoles NaOH = 0.35 mmoles HCl in excess. How much more do we need to add?
For pH 2.70, (H^+) = 1.995 x 10^-3 which I have rounded to 2.00 x 10^-3 moles/ L or 2.00 mmoles/1000 mL.
2.00 mmoles needed - 0.35 mmoles in solution = 1.65 mmoles additional HCl to be added.
mL x M = mmoles or
mL = mmoles/M = 1.65/0.06 = 27.5 mL additional HCl to be added.
Check my work. Check my thinking. Note that I have used a different concentration for HCl and NaOH (one 0.06 and the other 0.05).