Describe how you would prepare 2.00 L of an aqueous solution having a pH=3.60, if you had a supply of 0.100 M HCl available.
Convert pH to H^+.
pH = -log(H^+), then
c1v1 = c2v2
c = concn
v = volume
To prepare a 2.00 L aqueous solution with a pH of 3.60 using a supply of 0.100 M HCl, we need to calculate the volume of HCl required and adjust the solution volume accordingly. Here's how you can do it:
Step 1: Determine the concentration of H+ ions needed:
The pH value is a measure of the concentration of H+ ions in a solution. Since pH is given as 3.60, we can convert this into H+ ion concentration on a logarithmic scale using the formula:
[H+] = 10^(-pH)
[H+] = 10^(-3.60)
[H+] = 2.51 × 10^(-4) M
This means our desired concentration of H+ ions is 2.51 × 10^(-4) M.
Step 2: Determine the volume of 0.100 M HCl required:
We have a supply of 0.100 M HCl, which means it contains 0.100 moles of HCl per liter. To calculate the volume required, we can use the equation:
Moles = Molarity × Volume
Rearranging the equation, we have:
Volume = Moles / Molarity
To achieve a concentration of 2.51 × 10^(-4) M, we need to calculate the moles of HCl required:
Moles = Concentration × Volume
Moles = (2.51 × 10^(-4) M) × (2.00 L)
Moles = 5.02 × 10^(-4) mol
Now, we can calculate the volume of 0.100 M HCl required:
Volume = Moles / Molarity
Volume = (5.02 × 10^(-4) mol) / (0.100 M)
Volume = 5.02 × 10^(-3) L or 5.02 mL
Therefore, we need to measure out 5.02 mL of 0.100 M HCl solution.
Step 3: Adjust the solution volume:
You have calculated the volume of 0.100 M HCl required, but we need a final volume of 2.00 L. So, you would add the calculated volume of HCl to a container and then add sufficient water to reach a final volume of 2.00 L.
This process will result in a 2.00 L aqueous solution with a pH of 3.60, prepared using the supply of 0.100 M HCl.