Calculate the pH of a solution prepared by mixing 25.0 mL of 0.60 M HC2H3O2 and 15.0 mL of 0.60 M NaOH?
duplicate post.
One is using NaOH and the other NaC2H3O2, but will I still get the same pH for both?
I apologize. I didn't see the difference at first glance. No, the answer is not the same.
The first one is a buffered solution and the pH = 4.52 but check my arithmetic.
The second one is 25 x 0.6 = 9 millimoles acetic acid.
Add 15 mL x 0.6 NaOH = 9 mmoles NaOH.
You will form 9 mmoles acetate ion and that will neutralize 9 mmoles of acetic acid leaving 15-9 = 6 mmoles acetic acid. Redo the HH equation and solve for pH.
To calculate the pH of a solution, we need to determine the concentration of the hydronium ion (H3O+). In this case, we will use the concept of acid-base titration to find the concentration of H3O+.
Step 1: Determine the moles of HC2H3O2 and NaOH used.
To find the moles of HC2H3O2 and NaOH, we can use the formula:
moles = concentration (M) × volume (L)
For HC2H3O2:
moles of HC2H3O2 = 0.60 M × 0.025 L = 0.015 moles
For NaOH:
moles of NaOH = 0.60 M × 0.015 L = 0.009 moles
Step 2: Determine the limiting reactant.
To find the limiting reactant, we compare the moles of HC2H3O2 and NaOH. The reactant that produces the lowest number of moles is the limiting reactant. In this case, NaOH produces the lowest number of moles (0.009 moles).
Step 3: Determine the excess reactant.
Since NaOH is the limiting reactant, HC2H3O2 is in excess. This means that only a portion of HC2H3O2 can react with NaOH, and the excess will remain unchanged.
Step 4: Determine the moles of excess HC2H3O2.
To find the moles of excess HC2H3O2, subtract the moles of NaOH (limiting reactant) from the total moles of HC2H3O2 used.
moles of excess HC2H3O2 = 0.015 moles - 0.009 moles = 0.006 moles
Step 5: Determine the concentration of the excess HC2H3O2.
To find the concentration of the excess HC2H3O2, divide the moles of excess HC2H3O2 by the total volume of the solution.
concentration of excess HC2H3O2 = moles of excess HC2H3O2 / total volume (L)
total volume = 0.025 L + 0.015 L = 0.04 L
concentration of excess HC2H3O2 = 0.006 moles / 0.04 L = 0.15 M
Step 6: Determine the concentration of H3O+.
Since HC2H3O2 is a weak acid, it partially dissociates in water. The balanced equation for the dissociation of HC2H3O2 is:
HC2H3O2 + H2O ⇌ H3O+ + C2H3O2-
From the equation, we can see that the concentration of H3O+ is equal to the concentration of the excess HC2H3O2.
concentration of H3O+ = 0.15 M
Step 7: Calculate the pH.
The pH is defined as the negative logarithm (base 10) of the concentration of H3O+.
pH = -log[H3O+]
pH = -log(0.15)
Using a scientific calculator, we find that the pH of the solution is approximately 0.82.