The pH of a solution of HCL in water is found to be 2.50.
What volume of water would you add to 1.00L of this solution to raise the pH to 3.10?
pH = -log (H^+)
2.50 = -log (H^+)
(H^+) = 0.00316 M is current.
What you want is
3.10 = -log (H^+)
(H^+) = 0.00125 M
Plug in to the dilution formula.
mL1 x M1 = mL2 x M2
1000 mL x 0.00316 = mL2 x 0.00125
Solve for mL2 = total volume
Then assuming volumes are additive
total volume - 1000 mL = mL to be added.
To find the volume of water needed to raise the pH from 2.50 to 3.10, we need to use the formula for calculating pH:
pH = -log[H+]
where [H+] represents the concentration of the hydrogen ions in the solution.
Step 1: Calculate the concentration of H+ ions in the initial solution:
pH = 2.50
[H+] = 10^(-pH)
[H+] = 10^(-2.50) = 3.162 * 10^(-3) mol/L
Step 2: Calculate the concentration of H+ ions in the desired solution:
pH = 3.10
[H+] = 10^(-3.10) = 9.548 * 10^(-4) mol/L
Step 3: Calculate the change in concentration:
Δ[H+] = [H+]_desired - [H+]_initial
Δ[H+] = 9.548 * 10^(-4) - 3.162 * 10^(-3) = -2.208 * 10^(-3) mol/L
Step 4: Calculate the moles of H+ ions in 1.00L of the initial solution:
moles_initial = [H+]_initial * volume_initial
moles_initial = 3.162 * 10^(-3) * 1.00L = 3.162 * 10^(-3) mol
Step 5: Calculate the volume of water needed to achieve the desired pH:
volume_water = Δ[H+] / [H+]_initial * volume_initial
volume_water = (-2.208 * 10^(-3) mol/L) / (3.162 * 10^(-3) mol/L) * 1.00L
volume_water ≈ -0.70L
Since we cannot have a negative volume, we can conclude that no water should be added to the solution to raise the pH from 2.50 to 3.10.
To determine the volume of water needed to raise the pH of a solution of HCl in water, we first need to understand the relationship between pH and the concentration of H+ ions in the solution.
The pH scale is a measure of acidity or alkalinity and ranges from 0 to 14. A pH of 7 is considered neutral, below 7 is acidic, and above 7 is alkaline. The pH of a solution is calculated using the formula pH = -log[H+], where [H+] represents the concentration of hydrogen ions in moles per liter.
In this case, the initial pH of the solution is 2.50. To raise the pH to 3.10, we need to decrease the concentration of H+ ions. Since pH is logarithmic, a change of 1 in pH represents a tenfold change in the concentration of H+ ions.
We can calculate the concentration of H+ ions in the initial solution using the formula [H+] = 10^(-pH). In this case, [H+] = 10^(-2.50) = 0.00316 M.
To find the volume of water needed, we need to find the change in concentration required. The change in concentration can be calculated as the difference between the initial concentration and the desired concentration: Δ[H+] = [H+]initial - [H+]desired.
[H+]desired can be found using the desired pH of 3.10: [H+]desired = 10^(-3.10) = 0.000794 M.
Δ[H+] = 0.00316 M - 0.000794 M = 0.002366 M.
Now, we can use the formula for dilution to calculate the volume of water needed. The formula for dilution is:
M1V1 = M2V2
Where M1 is the initial concentration, V1 is the initial volume, M2 is the final concentration, and V2 is the final volume.
Since we are diluting with water, the final concentration is 0 M. We can rearrange the equation to solve for V2:
V2 = (M1V1) / M2
V2 = (0.002366 M × 1.00 L) / 0 M
Since M2 is 0 M, this means we would need an infinite volume of water to completely dilute the solution. However, in practice, it is not possible to completely eliminate the acid concentration. Therefore, a practical answer would be that you would need to add a substantial volume of water to significantly raise the pH of the solution, but an infinite volume is not possible.