Calculate z pH of a buffer solun containing o.1 M acetic acid & 0.1 M solun of sodiumacetate when 1 mL of 0.1M HCl is added to 100 mL of buffer .

100 mL buffer x 0.1M CH3COOH = 10 mmoles.

100 mL buffer x 0.1M CH3COONa = 10 mmoles.
add 1 mL x 0.1M HCl = 0.1mmole.

Let's call acetic acid, CH3COOH, HAc.

...........Ac^- + H^+ ==> HAc
initial.....10.....0.......10
add...............0.1..............
change.....-0.1..-0.1.....+0.1
equil.......9.9.....0.....10.1

Plug into the Henderson-Hasselbalch equation and solve for pH. Using 1.75E-5 for Ka I obtained a little over 4.7 but you need to do it more accurately AND your text/notes may list a different value for Ka.

To calculate the pH of a buffer solution, you need to know the concentrations of both the weak acid (acetic acid, CH3COOH) and its conjugate base (sodium acetate, CH3COONa), as well as their dissociation constants (Ka and Kb).

The dissociation of acetic acid in water can be represented by the equation:
CH3COOH ⇌ CH3COO- + H+

The dissociation constant (Ka) for acetic acid is 1.8 x 10^-5.

The dissociation of sodium acetate in water can be represented by the equation:
CH3COONa ⇌ CH3COO- + Na+

Since sodium acetate is the salt of a weak acid (acetic acid), it will completely dissociate into its ions and create additional CH3COO- ions in solution. Therefore, the concentration of the CH3COO- ions will be equal to the initial concentration of sodium acetate (0.1 M).

When a strong acid (HCl) is added to the buffer solution, it reacts with the CH3COO- ions to form acetic acid molecules. This shifts the equilibrium towards the left, reducing the concentration of CH3COO- ions.

In this case, 1 mL of 0.1 M HCl is added to 100 mL of the buffer. First, we need to calculate the amount of moles of HCl added:
moles HCl = volume (L) x concentration (M)
moles HCl = 0.001 L x 0.1 M = 0.0001 mol

Since HCl is a strong acid, it will completely dissociate in water to form H+ ions. Therefore, the concentration of H+ ions added to the buffer is also 0.0001 M.

Next, we need to calculate the new concentration of the CH3COO- ions.

Let's assume the change in concentration of CH3COO- ions is x. This is because, when HCl reacts with CH3COO- ions, an equal amount of acetic acid is formed. So, the decrease in concentration of CH3COO- ions is exactly compensated by the increase in concentration of acetic acid.

The initial concentration of CH3COO- ions is 0.1 M, and the change in concentration is x. Therefore, the final concentration of CH3COO- ions will be (0.1 - x) M.

Using the equation for acetic acid dissociation, you can set up an expression for the equilibrium between acetic acid and its conjugate base:
Ka = [CH3COO-][H+] / [CH3COOH]

Substituting the values, we have:
Ka = (0.1 - x) * (0.0001) / x

Now, we solve this equation to find the value of x, which represents the amount of CH3COO- ions consumed.

After finding the value of x, substitute it back into the expression (0.1 - x) to get the new concentration of CH3COO- ions.

Finally, you can calculate the pH using the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
where pKa is the negative logarithm of the dissociation constant (pKa = -log(Ka)), [A-] is the concentration of the conjugate base (CH3COO-) and [HA] is the concentration of the weak acid (acetic acid).