What is the pH of the solution created by combining 2.70 mL of the 0.10 M NaOH(aq) with 8.00 mL of the 0.10 M HCl(aq)? with 8.00 mL of the 0.10 M HC2H3O2(aq)?
What are the pH values if you take into account that the 8.00 mL of 0.10 M Acid was first diluted with 100 mL of water (like it will be in the experiment you perform in lab)?
they ask you to find the pH for both the HCl and the HC2H3O2.
i understand the first question its just the second question that is confusing me
i keep trying to do it and i don't understand what i am doing wrong
If you would like to post your work, I'm sure I can find the error. Since this is a multipart question. you may feel better about doing it piece-meal although that will take longer because it takes time for you to post and me to read, repost etc etc.
actually i understand it now. i looked at some of the other responses you gave to other similar questions. Thanks tho. :)
You're quite welcome!
To find the pH for both the HCl and the HC2H3O2 solutions after dilution, we need to calculate the concentration of the acid after dilution.
For the HCl solution:
First, let's calculate the concentration of HCl in the original solution.
Given:
Volume of HCl solution = 8.00 mL
Concentration of HCl solution = 0.10 M
Using the formula:
Concentration (C) = moles (n) / volume (V)
Rearranging the formula, we get:
n = C x V
n = 0.10 M x 8.00 mL
= 0.80 mmol
Now, let's consider the dilution. We're diluting the original HCl solution by adding 100 mL of water to it. This means the final volume of the solution will be 8.00 mL (original volume of HCl) + 100 mL (volume of water) = 108.00 mL = 0.108 L.
Using the dilution formula:
C1V1 = C2V2
Where:
C1 = initial concentration
V1 = initial volume
C2 = final concentration
V2 = final volume
We can rearrange the formula to solve for the final concentration:
C2 = (C1V1) / V2
C2 = (0.80 mmol) / 0.108 L
= 7.41 x 10^-3 M
Now that we have the final concentration of HCl (after dilution), we can calculate the pH using the formula for strong acids:
pH = -log[H+]
Since HCl is a strong acid, it completely dissociates into H+ ions and Cl- ions. Therefore, the concentration of H+ ions is the same as the concentration of HCl.
Hence, the pH of the HCl solution, after dilution, is:
pH = -log(7.41 x 10^-3)
≈ 2.13
Now, let's calculate the pH for the HC2H3O2 solution after dilution.
Using the same process as above, the final concentration of HC2H3O2 after dilution is 7.41 x 10^-3 M.
However, HC2H3O2 is a weak acid, and it does not completely dissociate. Therefore, we need to consider the dissociation reaction of acetic acid:
HC2H3O2 ⇌ H+ + C2H3O2-
The equilibrium constant for this reaction is represented by Ka. At equilibrium, the concentration of H+ ions and C2H3O2- ions will be equal.
For weak acids, Ka is used instead of concentration to determine the acidity. The expression for Ka is:
Ka = [H+] x [C2H3O2-] / [HC2H3O2]
Since [H+] = [C2H3O2-], we can rewrite the expression as:
Ka = [H+]^2 / [HC2H3O2]
Given that Ka for acetic acid is 1.8 x 10^-5 M, we can rearrange the equation and solve for [H+]:
[H+] = √(Ka x [HC2H3O2])
[H+] ≈ √(1.8 x 10^-5 M x 7.41 x 10^-3 M)
≈ 1.51 x 10^-3 M
Now, we can calculate the pH using the same formula:
pH = -log[H+]
pH = -log(1.51 x 10^-3)
≈ 2.82
Therefore, the pH of the HC2H3O2 solution after dilution is approximately 2.82.