A beaker with 120mL of an acetic acid buffer with a pH of 5.00 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.100 M. A student adds 5.70mL of a 0.450M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.760.

Use the Henderson-Hasselbalch equation and substitute pH, pKa, and solve for ratio of (b ase)/(acid) which I will call b/a. That's equation 1.

The problem tells you that
b + a = 0.1M; that's equation 2.
Solve eqn 1 and eqn 2 simultaneously for b and for a. You should get approximately 1.74 for b/a (but you need to do it more accurately than that) which makes b = 1.74a
and b = 0.0635M with a = 0.0365 (again you should be more accurate).

Then
millimols HAc = 120 mL x 0.0365 = about 4.4
mmols base(Ac^-) = 120 x 0.0635 = about 7.6
mmols HCl added = 5.70 x 0.450M = about 2.6.
........HAc + H^+ ==> Ac^- + H2O
I.......4.4..0.......7.6.......
add..........2.6.............
C......-2.6..-2.6....+2.6
E.......1.8...0.....10.2

Now substitute the E line into a new HH equation and solve for pH.
I love these problems.

To find out how much the pH will change after adding the HCl solution, we need to calculate the change in the concentration of acetic acid and its conjugate base, as well as calculate the new concentrations of acid and base.

First, let's start by calculating the concentration of acetic acid and the conjugate base in the original buffer solution.

Molarity (M) is defined as the moles of solute per liter of solution. Since we have 120 mL (0.120 L) of the buffer solution and we know the total molarity is 0.100 M, we can calculate the moles of acetic acid and its conjugate base.

Moles of acetic acid = molarity × volume
Moles of acetic acid = 0.100 M × 0.120 L
Moles of acetic acid = 0.012 moles

Since acetic acid is a weak acid, it partially dissociates in water to form its conjugate base (acetate ion, C2H3O2-).

By using the Henderson-Hasselbalch equation, which relates the pH to the concentrations of the acid and its conjugate base, we can find the ratio of acetic acid to its conjugate base in the solution.

pH = pKa + log([conjugate base]/[acid])

Given that the pKa of acetic acid is 4.760 and the pH of the buffer is 5.00, we can substitute these values into the equation to find the ratio.

5.00 = 4.760 + log([conjugate base]/[acid])

Solving for [conjugate base]/[acid]:
0.24 = log([conjugate base]/[acid])

Now, let's calculate the change in moles of acetic acid and its conjugate base after adding 5.70 mL (0.00570L) of HCl solution to the beaker.

Change in moles of HCl = molarity × volume
Change in moles of HCl = 0.450 M × 0.00570 L
Change in moles of HCl = 0.00257 moles

To maintain the 0.100 M total molarity of the buffer, the HCl will react with a portion of the acetic acid to form more acetate ions (conjugate base).

Since the stoichiometric ratio between HCl and acetic acid is 1:1, the change in moles of acetic acid and acetate ions will be equal to the change in moles of HCl.

Change in moles of acetic acid = Change in moles of acetate ions = 0.00257 moles

Now, let's calculate the new concentrations of acetic acid and its conjugate base in the solution after the HCl is added.

New moles of acetic acid = initial moles of acetic acid - change in moles of acetic acid
New moles of acetic acid = 0.012 moles - 0.00257 moles
New moles of acetic acid = 0.00943 moles

New moles of acetate ions = initial moles of acetate ions + change in moles of acetate ions
New moles of acetate ions = 0 moles + 0.00257 moles
New moles of acetate ions = 0.00257 moles

Finally, we can calculate the new concentrations of acetic acid and its conjugate base by dividing the moles by the final volume of the solution.

New concentration of acetic acid = new moles of acetic acid / volume
New concentration of acetic acid = 0.00943 moles / 0.120 L
New concentration of acetic acid = 0.0786 M

New concentration of acetate ions = new moles of acetate ions / volume
New concentration of acetate ions = 0.00257 moles / 0.120 L
New concentration of acetate ions = 0.0214 M

Now we can use the Henderson-Hasselbalch equation again to calculate the new pH of the buffer solution after the addition of HCl.

New pH = pKa + log([new concentration of acetate ions] / [new concentration of acetic acid])
New pH = 4.760 + log(0.0214 M / 0.0786 M)
New pH = 4.760 + log(0.272)

Using logarithmic properties, we can simplify this equation and find the final pH:

New pH = 4.760 - log(1/0.272)
New pH = 4.760 - log(3.676)
New pH = 4.760 - 0.565
New pH = 4.195

Therefore, the pH of the buffer solution will change from 5.00 to 4.195 after adding 5.70 mL of 0.450 M HCl solution.