Assuming the final solution will be diluted to 1.00 , how much more should you add to achieve the desired pH?
Information:
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 85.0 mL of HCl and 89.0 ml of NaOH left in their original containers.
You added 15 mL x 0.07M HCl = 1.05 millimoles initially.
Then you added 11 mL x 0.05M NaOH = 0.55 mmoles NaOH. Those react in 1:1 ratio to form 0.55 mmoles NaCl and this leaves 1.05-0.55 = 0.50 mmoles HCl remaining.
A pH of 2.80 = -log(H^+) and solve for (H^+); I obtained 0.00158M is what you want to produce.
0.00158 moles desired for the 1 L.
-0.0005 on hand with the 26 ml.
-------------
0.00108 moles extra needed.
M = moles/L and
L = moles/M = 0.00108/0.07 = ? L HCl needed, the make to 1L final volume.
Check my work.
Thank you!
To calculate how much more solution you should add to achieve the desired pH, you need to consider the concentrations and volumes of the HCl and NaOH solutions you currently have.
First, you need to determine the moles of HCl and NaOH you have remaining:
Moles of HCl = concentration of HCl × volume of HCl remaining
= (7.00×10^(-2) M) × (85.0 mL) / 1000 mL/L
Moles of NaOH = concentration of NaOH × volume of NaOH remaining
= (5.00×10^(-2) M) × (89.0 mL) / 1000 mL/L
Next, you need to determine the moles of HCl and NaOH that were accidentally added:
Moles of HCl added accidentally = concentration of HCl × volume of NaOH added accidentally
= (7.00×10^(-2) M) × (100 mL) / 1000 mL/L
Moles of NaOH added accidentally = concentration of NaOH × volume of HCl added accidentally
= (5.00×10^(-2) M) × (100 mL) / 1000 mL/L
Now, you can calculate the total moles of acid (HCl) and base (NaOH) in the solution:
Total moles of acid = moles of HCl remaining - moles of NaOH added accidentally
= Moles of HCl - Moles of NaOH added accidentally
Total moles of base = moles of NaOH remaining - moles of HCl added accidentally
= Moles of NaOH - Moles of HCl added accidentally
Since the solution is diluted to 1.00 L, you can calculate the final concentration of the acid and base:
Final concentration of acid = Total moles of acid / 1.00 L
Final concentration of base = Total moles of base / 1.00 L
Finally, you can determine the pH of the final solution based on the Henderson-Hasselbalch equation:
pH = -log10 ([acid] / [base])
Substitute the final concentrations of acid and base into the equation to find the pH. If the calculated pH is not 2.80, you will need to add more solution to adjust it to the desired pH. The amount to be added will depend on the difference between the calculated pH and the desired pH.
To solve this problem, we need to calculate how much more HCl or NaOH we should add to achieve the desired pH of 2.80 in the final solution.
First, let's determine the initial pH of the solution before adding any HCl or NaOH. We can do this by comparing the concentration of H+ ions (acidic) and OH- ions (basic) in the solution.
Given that the initial solution contains 100 ml of 7.00×10−2 M HCl, we can calculate the concentration of H+ ions using the equation:
[H+] = concentration of HCl = 7.00×10−2 M
Since HCl is a strong acid and completely dissociates in water, the concentration of H+ ions is equal to the concentration of HCl.
Now, let's calculate the initial concentration of OH- ions. Since NaOH is a strong base, it also completely dissociates in water.
Given that the initial solution contains 100 ml of 5.00×10−2 M NaOH, we can calculate the concentration of OH- ions using the equation:
[OH-] = concentration of NaOH = 5.00×10−2 M
Since the initial solution contains equal volumes of HCl and NaOH, the concentration of OH- ions is also equal to the concentration of NaOH.
Since the solution is neutral when the concentration of H+ ions is equal to the concentration of OH- ions, we can assume that the initial pH of the solution is 7.
Now, let's calculate the amount of excess H+ or OH- ions in the solution after the accidental addition of NaOH.
Given that 85.0 ml of HCl and 89.0 ml of NaOH are left in their original containers, we can calculate the concentrations of H+ and OH- ions in the remaining solutions.
[H+] = (85.0 ml / 100 ml) * 7.00×10−2 M
[OH-] = (89.0 ml / 100 ml) * 5.00×10−2 M
Now, let's determine which ion is in excess by comparing their concentrations.
If [H+] > [OH-], then there is an excess of H+ ions.
If [H+] < [OH-], then there is an excess of OH- ions.
If [H+] = [OH-], then the solution is neutral.
Compare the concentrations of H+ and OH- ions calculated above to determine which ion is in excess in the remaining solution.
Once we know which ion is in excess, we can calculate the additional amount of the other ion needed to achieve the desired pH of 2.80.
The pH scale is logarithmic, which means that each unit represents a tenfold difference in acidity or basicity. So, to increase the pH from 7 to 2.80, we need to decrease the concentration of H+ ions or increase the concentration of OH- ions by a factor of 10^(7 - 2.80).
If H+ ions are in excess, we need to add more NaOH to increase the concentration of OH- ions. The additional volume of NaOH needed can be calculated using the equation:
Additional volume of NaOH = (Desired increase factor) * (Volume of remaining HCl) / (Initial concentration of NaOH)
If OH- ions are in excess, we need to add more HCl to decrease the concentration of OH- ions. The additional volume of HCl needed can be calculated using the equation:
Additional volume of HCl = (Desired decrease factor) * (Volume of remaining NaOH) / (Initial concentration of HCl)
Plug in the values into the respective equations to calculate the additional volume of NaOH or HCl needed.
Finally, add the calculated additional volume of NaOH or HCl to the remaining solutions and mix thoroughly to achieve the desired pH of 2.80 in the final solution.