Calculate the pH of 75 ml of the buffer composed of .1M HC2H3O2 with .2M NaC2H3O2; to which 9.5 ml of .1M HCl is added.

millimols HAc = 75 mL x 0.1M 7.5.

mmols NaAc = 75 mL x 0.2M = 15.
9.5 mL x 0.1M HCl = 0.95 mmols.

........Ac^- + H^+ ==> HAc
I.......15.....0........7.5
add...........0.95............
C....-0.95..-0.95.......+0.95
E.....14.05...0........8.45

Substitute the E line into the HH equation and solve for H.

To calculate the pH of the buffer solution after adding HCl, we need to consider the Henderson-Hasselbalch equation. The Henderson-Hasselbalch equation is given by:

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

Where:
pH = the pH of the solution
pKa = the acid dissociation constant of the acid component of the buffer
[A-] = the concentration of the conjugate base component of the buffer
[HA] = the concentration of the acid component of the buffer

Given:
Volume of the buffer solution (V1) = 75 ml
Concentration of HC2H3O2 (acid component) = 0.1 M
Concentration of NaC2H3O2 (conjugate base component) = 0.2 M
Volume of HCl added (V2) = 9.5 ml
Concentration of HCl = 0.1 M

First, let's find the moles of HC2H3O2 and NaC2H3O2 in the buffer solution:

Number of moles of HC2H3O2 = Concentration * Volume = 0.1 M * 0.075 L = 0.0075 moles
Number of moles of NaC2H3O2 = Concentration * Volume = 0.2 M * 0.075 L = 0.015 moles

Next, let's calculate the moles of HCl that reacted with the buffer:

Number of moles of HCl reacted = Concentration * Volume = 0.1 M * 0.0095 L = 0.00095 moles

Since HCl reacts with HC2H3O2, the moles of HC2H3O2 will decrease, while the moles of HC2H3O2- (conjugate base) will increase by the same amount. The moles of NaC2H3O2 will remain unchanged.

New moles of HC2H3O2 = initial moles - moles of HCl reacted = 0.0075 moles - 0.00095 moles = 0.00655 moles
New moles of HC2H3O2- = initial moles + moles of HCl reacted = 0.015 moles + 0.00095 moles = 0.01595 moles

Now, let's calculate the concentrations of HC2H3O2 and HC2H3O2- in the new solution:

New concentration of HC2H3O2 = moles / volume = 0.00655 moles / 0.075 L = 0.0873 M
New concentration of HC2H3O2- = moles / volume = 0.01595 moles / 0.075 L = 0.2127 M

Finally, let's calculate the pH using the Henderson-Hasselbalch equation:

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

The pKa value of HC2H3O2 is 4.74 (you can find this value in a table).

pH = 4.74 + log(0.2127 M / 0.0873 M) = 4.74 + log(2.435) = 4.74 + 0.385 = 5.125

Therefore, the pH of the buffer solution after adding 9.5 ml of 0.1 M HCl is approximately 5.125.

To calculate the final pH of the buffer solution after adding the HCl, you will need to consider the Henderson-Hasselbalch equation. The Henderson-Hasselbalch equation relates the pH of a buffer solution to the pKa of the acid and the ratio of the concentrations of the acid and its conjugate base.

The pKa of acetic acid (HC2H3O2) is 4.74. Therefore, we can use this value in the Henderson-Hasselbalch equation along with the concentration ratio of acetic acid (HC2H3O2) and sodium acetate (NaC2H3O2).

The Henderson-Hasselbalch equation is given by:

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

Where:
- pH is the logarithmic scale that indicates acidity or alkalinity
- pKa is the negative logarithm of the acid dissociation constant (Ka)
- [A-] is the concentration of the conjugate base
- [HA] is the concentration of the acid

Let's calculate step-by-step:

Step 1: Calculate the moles of acetic acid (HC2H3O2):
Moles of HC2H3O2 = concentration (M) x volume (L)
Moles of HC2H3O2 = 0.1 M x 0.075 L
Moles of HC2H3O2 = 0.0075 mol

Step 2: Calculate the moles of sodium acetate (NaC2H3O2):
Moles of NaC2H3O2 = concentration (M) x volume (L)
Moles of NaC2H3O2 = 0.2 M x 0.075 L
Moles of NaC2H3O2 = 0.015 mol

Step 3: Calculate the change in moles of acetic acid (HC2H3O2) after adding HCl:
Change in moles of HC2H3O2 = concentration (M) x volume (L)
Change in moles of HC2H3O2 = 0.1 M x 0.0095 L
Change in moles of HC2H3O2 = 0.00095 mol

Step 4: Calculate the new concentration of acetic acid (HC2H3O2) after adding HCl:
New moles of HC2H3O2 = initial moles of HC2H3O2 - change in moles of HC2H3O2
New moles of HC2H3O2 = 0.0075 mol - 0.00095 mol
New moles of HC2H3O2 = 0.00655 mol

New concentration of HC2H3O2 = moles of HC2H3O2 / volume (L)
New concentration of HC2H3O2 = 0.00655 mol / 0.075 L
New concentration of HC2H3O2 = 0.0873 M

Step 5: Calculate the concentration of sodium acetate (NaC2H3O2) after dilution (no change in moles):
Concentration of NaC2H3O2 = moles of NaC2H3O2 / volume (L)
Concentration of NaC2H3O2 = 0.015 mol / 0.075 L
Concentration of NaC2H3O2 = 0.2 M

Step 6: Calculate the pH using the Henderson-Hasselbalch equation:
pH = pKa + log ([A-]/[HA])
pH = 4.74 + log (0.2 M / 0.0873 M)
pH = 4.74 + 0.4953
pH = 5.2353

Therefore, the pH of the buffer solution after adding 9.5 ml of 0.1 M HCl to 75 ml of the buffer composed of 0.1 M HC2H3O2 and 0.2 M NaC2H3O2 is approximately pH = 5.24.