12 mL of 0.0100 M HCl are added to 25.0 mL of a buffer solution that is 0.010 M HC2H3O2 and 0.08 M C2H3O2-. What is the pH of the resulting solution?
To find the pH of the resulting solution after adding the HCl, we need to consider the changes in the concentrations of the components in the buffer solution. Let's go through the steps:
Step 1: Calculate the moles of HCl added:
Moles of HCl = (volume of HCl) x (molarity of HCl)
= 12 mL x 0.0100 M
= 0.12 mmol (millimoles)
Step 2: Calculate the moles of HC2H3O2 and C2H3O2- in the buffer:
Moles of HC2H3O2 = (volume of HC2H3O2) x (molarity of HC2H3O2)
= 25.0 mL x 0.010 M
= 0.25 mmol
Moles of C2H3O2- = (volume of C2H3O2-) x (molarity of C2H3O2-)
= 25.0 mL x 0.08 M
= 2.00 mmol
Step 3: Determine the change in moles of HC2H3O2 and C2H3O2- after the HCl is added:
Change in moles of HC2H3O2 = 0 mmol (since no HCl is added to this component)
Change in moles of C2H3O2- = (moles of HCl added)
= 0.12 mmol
Step 4: Calculate the new concentrations of HC2H3O2 and C2H3O2-:
Concentration of HC2H3O2 = (moles of HC2H3O2) / (total volume of solution)
= (0.25 mmol) / (25.0 mL + 12 mL)
= 0.00794 M
Concentration of C2H3O2- = (moles of C2H3O2-) / (total volume of solution)
= (2.00 mmol + 0.12 mmol) / (25.0 mL + 12 mL)
= 0.0874 M
Step 5: Calculate the new pH using the Henderson-Hasselbalch equation:
pH = pKa + log10 ([C2H3O2-] / [HC2H3O2])
where pKa is the dissociation constant of HC2H3O2.
The pKa value for acetic acid (HC2H3O2) is approximately 4.76.
pH = 4.76 + log10 (0.0874 M / 0.00794 M)
= 4.76 + log10 (11.00)
= 4.76 + 1.04
= 5.8
Therefore, the pH of the resulting solution is approximately 5.8.
To find the pH of the resulting solution, we need to calculate the concentration of H+ ions in the solution. In this case, we can start by writing down the chemical equation representing the dissociation of acetic acid (HC2H3O2) in water:
HC2H3O2 ⇌ H+ + C2H3O2-
The given buffer solution contains HC2H3O2 and C2H3O2- ions, which can accept or release H+ ions to resist changes in pH. When HCl is added, it will provide additional H+ ions to the solution, leading to a change in the equilibrium.
First, let's calculate the moles of HCl added to the solution:
Moles of HCl = volume of HCl (in L) × concentration of HCl (in M)
= 12 mL × 0.0100 M
= 0.012 moles
Next, let's determine the new concentrations of HC2H3O2 and C2H3O2- after the addition of HCl. The initial moles of HC2H3O2 and C2H3O2- in the buffer can be calculated as follows:
Moles of HC2H3O2 = volume of buffer (in L) × concentration of HC2H3O2 (in M)
= 25.0 mL × 0.010 M
= 0.250 moles
Moles of C2H3O2- = volume of buffer (in L) × concentration of C2H3O2- (in M)
= 25.0 mL × 0.08 M
= 2.00 moles
After adding HCl, the moles of HC2H3O2 will decrease by the moles of HCl added, while the moles of C2H3O2- will remain unchanged. Therefore, the new moles of HC2H3O2 and C2H3O2- can be calculated as follows:
New moles of HC2H3O2 = initial moles of HC2H3O2 - moles of HCl added
= 0.250 moles - 0.012 moles
= 0.238 moles
New moles of C2H3O2- = initial moles of C2H3O2-
= 2.00 moles
Now, we can calculate the new concentrations of HC2H3O2 and C2H3O2-:
New concentration of HC2H3O2 = new moles of HC2H3O2 / volume of buffer (in L)
= 0.238 moles / 25.0 mL
= 0.00952 M
New concentration of C2H3O2- = new moles of C2H3O2- / volume of buffer (in L)
= 2.00 moles / 25.0 mL
= 0.080 M
Finally, we can calculate the concentration of H+ ions in the solution using the Henderson-Hasselbalch equation:
pH = pKa + log10(C2H3O2-/HC2H3O2)
The pKa for acetic acid (HC2H3O2) is approximately 4.76. Using the given concentrations of C2H3O2- and HC2H3O2, we can substitute the values into the equation:
pH = 4.76 + log10(0.080/0.00952)
Now, using a calculator, we can evaluate the expression:
pH = 4.76 + log10(8.40)
pH ≈ 4.76 + 0.924 = 5.68
Therefore, the pH of the resulting solution is approximately 5.68.
12 mL x 0.01M HCl = 0.12 mmoles.
25 mL x 0.08M Ac^0 = 2.00 mmoles.
25 mL x 0.01M HAc = 0.25 mmoles.
...........Ac^- + H^+ ==> HAc
initial....2.00....0.......0
added.............0.12...........
change....-0.12...-0.12....0.12
equil.....1.88......0.....0.12
Substitute the ICE values into the Henderson-Hasselbalch equation and solve for pH.