A beaker with 150 mL 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 7.90 mL of a 0.480 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.760.

I can get you started. Let's call acetic acid HAc and the anion Ac.

pH = pKa + log (Ac/HAc)
You know pH = 5 and you know pKa, you can calculate (Ac)/(HAc). Do that first.
Then you know from the problem that
(HAc) + (Ac) = 0.1
That give you two equations and two unknowns. Solve for (HAc) and (Ac).

Next, you add 7.90 mL x 0.480 M H^+ or 3.79 millimoles acid. That will use up 3.79 millimoles Ac and form 3.79 millimoles of additional HAc. That will let you calculate a new (HAc) and a new (Ac). Plug those numbers into the HH equation and solve for the new pH.
Post your work if you get stuck.

When I plugged the pH and the pKa into the pH = pKa + log [Ac/HAc] I got [Ac/HAc] to be 0.575.

I then rearranged the equation of [Ac/HAc] = 0.575 and solved for Ac, which gave me the following equation: Ac = 0.575HAc.

I then plugged that equation into HAc + Ac = 0.1 and found that HAc = 0.06 and that Ac = 0.04.

Then I multiplied 7.90 x 0.480 and got 3.79 millimoles.

I then found the new value for HAc: 0.06 + 3.79x10^-6 = 0.06000379

I then found the new value for Ac: 0.04 - 3.79x10^-6 = 0.03999621

I then plugged those of those values into the HH equation as follows:

pH = 4.760 + log [0.03999621/0.06000379] = 4.58

Therefore the pH change would be: 5.00 - 4.58 = 0.42

The only problem is that it said my answer was wrong...

When I plugged the pH and the pKa into the pH = pKa + log [Ac/HAc] I got [Ac/HAc] to be 0.575.

I don't get that.
5.00 = 4.76 + log(B/A)
0.24 = log(B/A)
(B)/(A) = 1.7378 and
(B) = 1.7378*(A) which still gives values for (Ac) and (HAc) very close but reversed(see below).


I then rearranged the equation of [Ac/HAc] = 0.575 and solved for Ac, which gave me the following equation: Ac = 0.575HAc.

I then plugged that equation into HAc + Ac = 0.1 and found that HAc = 0.06 and that Ac = 0.04.
Using my value for (B/A), I came up with (HAc) = 0.0365 M and (Ac) = 0.0635 M

Then I multiplied 7.90 x 0.480 and got 3.79 millimoles.

I then found the new value for HAc: 0.06 + 3.79x10^-6 = 0.06000379
This step isn't correct. What you must do is add moles (or millimoles), not molarity.
mmoles HAc in the 150 mL buffer is 150 mL x 0.0365 M = 5.475 mmoles HAc.
mmoles Ac = 150 mL x 0.0635 M = 9.52 mmoles Ac.
Now add 3.79 mmoles to HAc and subtract 3.79 mmoles from Ac and recalculate pH and the difference.


I then found the new value for Ac: 0.04 - 3.79x10^-6 = 0.03999621

I then plugged those of those values into the HH equation as follows:

pH = 4.760 + log [0.03999621/0.06000379] = 4.58

Therefore the pH change would be: 5.00 - 4.58 = 0.42
If I didn't make an error, and you should check it closely to confirm that I did not, I found
pH = 4.76 + log(5.73/9.265)
pH = 4.55 which isn't all that far away from your final value. You should realize, also, that I plugged in millimoles for base and acid and not concentration. That is, I didn't go through th extra step of dividing base and acid millimoles by 150 mL to obtain molarity BECAUSE both numerator and denominator would be divided by 150 mL so that plus the conversion from millimoles to moles cancel. I just saved a little work, that's all.

I also got the new pH to be 4.55. Thank you so much. This helps out a lot.

I trust that's the correct answer.

To determine how much the pH will change after adding the HCl solution, we need to calculate the new concentration of acid and conjugate base in the buffer, as well as the change in the amounts of these species.

First, let's calculate the number of moles of acetic acid and conjugate base in the original buffer. The total volume of the buffer is 150 mL or 0.150 L, and the total molarity is given as 0.100 M. Therefore, the number of moles of acid and conjugate base in the original buffer is:

moles of acid = volume of buffer (L) * concentration of acid (M)
= 0.150 L * 0.100 M
= 0.015 mol

Since this is a buffer solution, the original concentration of acid is equal to the concentration of conjugate base. Therefore, the number of moles of conjugate base is also 0.015 mol.

Now, let's calculate the number of moles of HCl added to the beaker. The volume of HCl solution added is given as 7.90 mL or 0.00790 L, and the concentration of HCl is 0.480 M. Therefore, the number of moles of HCl added is:

moles of HCl = volume of HCl solution (L) * concentration of HCl (M)
= 0.00790 L * 0.480 M
= 0.00379 mol

Since HCl is a strong acid, it is completely dissociated in solution. Therefore, the number of moles of H+ ions added to the beaker is also 0.00379 mol.

The addition of H+ ions from the HCl solution will react with the conjugate base (acetate ion, CH3COO-) in the buffer to form acetic acid (CH3COOH). The reaction can be represented as:

H+ + CH3COO- -> CH3COOH

Since the concentrations of acetic acid and conjugate base are equal in the buffer, and the reaction consumes 1 mole of CH3COO- for every 1 mole of H+, the change in the number of moles of acid and conjugate base can be calculated as follows:

change in moles of acid = moles of H+
= 0.00379 mol

change in moles of conjugate base = -moles of H+
= -0.00379 mol

The new number of moles of acid and conjugate base in the buffer is given by:

new moles of acid = moles of acid in buffer + change in moles of acid
= 0.015 mol + 0.00379 mol
= 0.01879 mol

new moles of conjugate base = moles of conjugate base in buffer + change in moles of conjugate base
= 0.015 mol - 0.00379 mol
= 0.01121 mol

Finally, we can calculate the new concentration of acid and conjugate base in the buffer using the new number of moles and the total volume:

new concentration of acid = new moles of acid / total volume
= 0.01879 mol / 0.150 L
= 0.125 M

new concentration of conjugate base = new moles of conjugate base / total volume
= 0.01121 mol / 0.150 L
= 0.0747 M

Now, let's determine the new pH of the buffer using the Henderson-Hasselbalch equation:

pH = pKa + log10(concentration of conjugate base / concentration of acid)

Using the given pKa of acetic acid (4.760) and the new concentrations we calculated:

new pH = 4.760 + log10(0.0747 M / 0.125 M)
= 4.760 + log10(0.5976)
= 4.760 - 0.2249
= 4.5351

Therefore, the pH of the buffer will change by approximately -0.465 units (from 5.00 to 4.5351) after adding 7.90 mL of a 0.480 M HCl solution.