A 11.8 mL sample of a HCl solution has a pH of 2.000. What volume of water must be added to change the pH to 4.000?
pH = 2; convert to H^+ (which will be concn in moles/L = M). Convert pH 4 to H^+. Then
mL1 HCl x M1 HCl = mL2HCl x M2HCl
To solve this problem, we need to use the concept of dilution.
Step 1: Calculate the initial concentration of the HCl solution.
Since the pH of the solution is given as 2.000, we can convert it to concentration using the formula:
pH = -log[H+].
Therefore, 2.000 = -log[H+].
Taking the inverse logarithm gives:
[H+] = 10^(-pH) = 10^(-2) = 0.01 mol/L.
Step 2: Calculate the initial number of moles of HCl in the 11.8 mL solution.
Since we have the concentration and the volume, we can use the formula:
moles = concentration x volume.
Applying this formula, we get:
moles = 0.01 mol/L x 0.0118 L = 1.18 x 10^(-4) mol.
Step 3: Calculate the final volume of the solution after dilution.
Since we are diluting the solution by adding water, the number of moles of HCl remains the same after dilution.
Therefore, the final moles of HCl is 1.18 x 10^(-4) mol.
Now, we can use the formula:
final concentration = final moles / final volume.
Since we want to achieve a final pH of 4.000, we can convert it to concentration using the same formula as earlier:
[H+] = 10^(-pH) = 10^(-4) = 0.0001 mol/L.
Therefore, using the formula for concentration, we have:
0.0001 mol/L = 1.18 x 10^(-4) mol / final volume.
Solving for the final volume:
final volume = 1.18 x 10^(-4) mol / 0.0001 mol/L = 1.18 mL.
Step 4: Calculate the volume of water that needs to be added.
The final volume is the sum of the initial volume (11.8 mL) and the volume of water added.
Therefore, the volume of water added is:
Volume of water = final volume - initial volume = 1.18 mL - 11.8 mL = -10.62 mL.
Since a negative volume does not make sense, we conclude that no water needs to be added to change the pH from 2.000 to 4.000.
To find the volume of water that must be added to change the pH of the HCl solution, we can use the equation for pH:
pH = -log[H+]
where [H+] represents the concentration of hydrogen ions in the solution. Since HCl is a strong acid, it dissociates completely in water, resulting in one mole of H+ ions for every mole of HCl.
Step 1: Calculate the initial concentration of H+ ions.
Since the pH is 2.000, the concentration of H+ ions can be calculated as follows:
2.000 = -log[H+]
[H+] = 10^(-pH) = 10^(-2.000)
[H+] = 0.01 M (or 0.01 mol/L)
Step 2: Calculate the volume of the initial solution.
We know the volume of the initial solution is 11.8 mL, which is equal to 0.0118 L.
Step 3: Calculate the number of moles of H+ ions in the initial solution.
To find the number of moles, we multiply the concentration by the volume:
moles of H+ = concentration × volume
moles of H+ = (0.01 mol/L) × (0.0118 L)
moles of H+ = 0.000118 mol
Step 4: Calculate the volume of water required to change the pH to 4.000.
To change the pH from 2.000 to 4.000, the concentration of H+ ions needs to decrease. Since water has a negligible concentration of H+ ions, dilution with water will lower the concentration while keeping the number of moles constant.
Let's denote the final volume of the solution (including water) as Vf.
Since the number of moles of H+ ions must stay the same, we can set up an equation:
initial moles of H+ = final moles of H+
0.000118 mol = (0.01 mol/L) × Vf
Solving for Vf:
Vf = 0.000118 mol / (0.01 mol/L)
Vf = 0.0118 L
This means the volume of the final solution (HCl solution + water) is 0.0118 L. Since we started with an 11.8 mL HCl solution, the volume of water needed to be added to change the pH to 4.000 is equal to the difference in volume between the final solution and the initial HCl solution:
Volume of water = Vf - initial volume of HCl solution
Volume of water = 0.0118 L - 0.0118 L
Volume of water = 0 L
Therefore, no volume of water needs to be added to change the pH of the HCl solution from 2.000 to 4.000.