A buffer was prepared by mixing 0.1 M ch3cooh with 0.1M ch3cooNa in the ratio 8:2. Calculate ph after the addition of 2mL of 0.1 M HCl?

When you do the 8:2 ratio and add 2 mL of 0.1 M HCl, the buffer is gone and you are left with 1*10^-3 moles of ch3cooh. How would you calculate the pH after that?
the total volume should be 10 mL so the ratio 8mL:2mL before the addition of HCl

You didn't read my earlier response? I guess it's just easier to repost and hope someone will work the problem for you.

You are correct about the buffer salt being consumed and that there are 0.001 mole of HOAc remaining. Now put this in 12 ml solution and calculate the [HOAc] for the resulting solution. [HOAc] = (0.003mol/0.012L) = 0.083M.

Given Ka = 1.8x10^-5 & [HOAc] = 0.083M
=> [H^+] = Sqr-Root(Ka x [HOAc])
pH = -log[H^+]. Good Luck

To calculate the pH after the addition of 2 mL of 0.1 M HCl, we need to consider the reaction that occurs between the acid (HCl) and the base (CH3COONa) in the buffer solution.

First, let's determine the moles of CH3COOH and CH3COONa initially present in the buffer solution. Since the ratio of CH3COOH to CH3COONa is 8:2, we can assume that the total volume of the buffer solution is 10 mL (8 mL of CH3COOH + 2 mL of CH3COONa).

Moles of CH3COOH = (0.1 M) x (8 mL / 1000 mL/mL) = 0.0008 moles
Moles of CH3COONa = (0.1 M) x (2 mL / 1000 mL/mL) = 0.0002 moles

Next, let's determine the number of moles of HCl added to the buffer solution. Since 2 mL of 0.1 M HCl is added, we can calculate the moles of HCl as follows:

Moles of HCl = (0.1 M) x (2 mL / 1000 mL/mL) = 0.0002 moles

Now, let's consider the reaction between CH3COOH and HCl. The balanced equation for this reaction is:

CH3COOH + HCl -> CH3COOH2+ + Cl-

From the equation, we can see that 1 mole of CH3COOH reacts with 1 mole of HCl to form 1 mole of CH3COOH2+ and 1 mole of Cl-.

Since the number of moles of CH3COOH and HCl are equal (0.0002 moles), they will react completely, and the resulting solution will contain 0.0002 moles of CH3COOH2+ and 0.0002 moles of Cl-. The moles of CH3COOH remaining will be 0.0008 - 0.0002 = 0.0006 moles.

Now, let's calculate the concentration (M) of CH3COOH2+ and CH3COOH in the final solution. The total volume of the final solution will be 10 mL (8 mL of CH3COOH + 2 mL of CH3COONa) + 2 mL of HCl = 12 mL.

Concentration (M) of CH3COOH2+ = 0.0002 moles / (12 mL / 1000 mL/mL) = 0.0167 M
Concentration (M) of CH3COOH = 0.0006 moles / (12 mL / 1000 mL/mL) = 0.0500 M

Now that we have the concentration of CH3COOH and CH3COOH2+, we can calculate the pH of the solution using the Henderson-Hasselbalch equation, which is given by:

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

The pKa value for the acetic acid (CH3COOH) is 4.76, so we can substitute the values into the equation:

pH = 4.76 + log(0.0167 / 0.0500)
pH = 4.76 - 0.7936
pH = 3.9664

Therefore, the pH of the solution after the addition of 2 mL of 0.1 M HCl is approximately 3.97.