Adding 10.0 mL of 0.500 M HCl to 100.0 mL of pH = 7 water produces a solution with a pH of 1.34. What is the pH of the resulting solution when 10.0 mL of 0.500 M HCl is added to 100.0 mL of the buffer solution in the introductory example? That buffer contained 0.100 moles of both CH3COOH and CH3COO- in 0.100 L of water?

Question

Adding 10.0 mL of 0.500 M HCl to 100.0 mL of pH = 7 water produces a solution with a pH of 1.34. What is the pH of the resulting solution when 10.0 mL of 0.500 M HCl is added to 100.0 mL of the buffer solution in the introductory example? That buffer contained 0.100 moles of both CH3COOH and CH3COO- in 0.100 L of water?

The Ka of acetic acid is 1.8105

Answer 1

millimols CH3COOH = 100

millimols CH3COO- = 100
millimols HCl = 0.5M x 10 mL = 5 mn.
-----------------------------
The added HCl will add to the base (acetatae ion) to make CH3COOH.
........CH3COO- + H+ ==> CH3COOH
I.........100......0..........100
add...............5
C.........-5.....-5..........+5
E.........95......0...........105

Substitute the E line into the HH equation and solve for pH of the final solution. Compare that with 1.34 in an un-buffered solution.

Answer 2

To solve this problem, we need to consider the reaction between the acetic acid (CH3COOH) and the added HCl.

The balanced chemical equation for the reaction between acetic acid and hydrochloric acid (HCl) is:

CH3COOH + HCl → CH3COO- + H3O+

We have the initial concentrations of acetic acid and its conjugate base, which are both 0.100 M. We also have the concentration of HCl, which is 0.500 M, and the volumes of the two solutions.

To find the concentration of the resulting solution, we can use the Henderson-Hasselbalch equation, which relates the pH of a buffer solution to the ratio of the concentrations of the acid and its conjugate base:

pH = pKa + log([A-]/[HA])

Where pKa is the negative logarithm of the acid dissociation constant (Ka), and [A-] and [HA] are the concentrations of the conjugate base and acid, respectively.

In this case, the conjugate base is CH3COO-, and the acid is CH3COOH.

The pKa value for acetic acid (CH3COOH) is given as 1.810^-5.

To find the concentration ([A-]/[HA]), we need to calculate the moles of each component:

Moles of CH3COOH = 0.100 moles
Moles of CH3COO- = 0.100 moles
Total moles = 0.100 + 0.100 = 0.200 moles

Now, we can calculate the volume of the resulting solution:

Volume of resulting solution = 100 mL + 10 mL = 110 mL = 0.110 L

We can use the moles and volume values to calculate the concentrations of CH3COOH and CH3COO-:

Concentration of CH3COOH = moles of CH3COOH / volume of resulting solution
= 0.100 moles / 0.110 L
= 0.909 M

Concentration of CH3COO- = moles of CH3COO- / volume of resulting solution
= 0.100 moles / 0.110 L
= 0.909 M

Now, we can calculate the pH of the resulting solution using the Henderson-Hasselbalch equation:

pH = pKa + log([A-]/[HA])
= -log(1.810^-5) + log(0.909/0.909)
= -log(1.810^-5)
= 4.78

Therefore, the pH of the resulting solution when 10.0 mL of 0.500 M HCl is added to 100.0 mL of the buffer solution is approximately 4.78.