A solution is made by dilution 400.mL of 2.0 M HCL to a final volume of 1000. mL. what is the pH of this solution? i'm stuck on this question i have no idea what to do, can anyone help? thanks:)
The easy way to do this problem is:
Starting with 2.0 M, you dilute it from 400 to 1000 mL; therefore, the final concentration is no other than
2.0 x (400/1000) = 0.80 M
The harder way, but the one you've been taught to rely on, ALWAYS, is to use mols (because everything works in mols in chemistry).
How many mols do you have to start with? That is mols = M x L = 2.0 x 0.400 =0.80 mols.
How many mols will you have after it is diluted? Of course, you will have 0.80 mols.
What is the concentration now in mols/L (which of course is molarity) = 0.80/1.0 L = 0.80 mol/L = 0.8 M.
To find the pH of the solution, we need to calculate the concentration of H+ ions in the diluted solution. Here's how you can solve it step by step:
1. Use the dilution formula: C1V1 = C2V2
- C1 is the initial concentration of the solution (2.0 M)
- V1 is the initial volume of the solution (400 mL)
- C2 is the final concentration of the solution (unknown)
- V2 is the final volume of the solution (1000 mL)
Substitute the values into the formula:
(2.0 M)(400 mL) = C2(1000 mL)
2. Solve for C2:
(2.0 M)(400 mL) / 1000 mL = C2
C2 = 0.8 M
After dilution, the final concentration of the HCl solution will be 0.8 M.
3. Convert the concentration to the concentration of H+ ions:
Since HCl is a strong acid, it will fully dissociate in water, forming H+ and Cl- ions with a 1:1 ratio. Therefore, the concentration of H+ ions will also be 0.8 M.
4. Calculate the pH using the formula:
pH = -log[H+]
pH = -log(0.8)
pH ≈ 0.0969
The pH of the diluted HCl solution is approximately 0.0969.