A buffer is formed by adding 500mL of .20 M HC2H3O2 to 500 mL of .10 M NaC2H3O2. What would be the maximum amount of HCl that could be added to this solution without exceeding the capacity of the buffer?

A. .01 mol
B. .05 mol
C. .10 mol
D. .15 mol
E. .20 mol

[I know the correct answer is B, but I don't know why. Any help would be great.]

just tell me where to start? please?

I don't get 0.05 mol but it's close.
Calculate pH of the buffer.
I have pH = pKa + log [(base)/(acid)]
pH = 4.76 + log [(0.05/0.1) = 4.46

The buffer capacity is the number of mols of strong acid or strong base that can be added to make the buffer change by 1.00 pH units. So adding strong acid we want the pH to be 1.00 less than 4.46 or = 3.46
3.46=4.76 + log (b/a)
-1.3 = log (b/a)
0.05 = (b/a)
If we add x mols of a strong acid, then
mols base = what we start with minus strong acid = 0.05 mols acetate - x mols strong acid.

mols acetic acid = what we start with + mols strong acid = 0.1 mols acetic acid + x mols strong acid.
0.05 = (0.05-x)/(0.1+x)
solve for x mols strong acid.
I found 0.043 mols which doesn't round to 0.05. In fact, I don't think any of the choices are right although 0.05 mols is the closest. However, putting 0.05 mols stong acid gives this for the pH.
acetate = 0.05 - 0.05 = 0
acetic acid = 0.10 + 0.05 = 0.15
and pH = 4.76 + (0.0/0.15) and we run into trouble because of the zero concentration of the base. The buffering capacity is gone if we have no acetate left. I hope this helps.

yah, i wasn't allowed a calculator so we had to guess and estimate things, but thanks so much that really cleared it up.

You're welcome! I'm glad I could help clarify things for you. If you have any more questions or need further explanations in the future, please don't hesitate to ask. Good luck with your studies!

Well, it looks like you had quite the mathematical adventure trying to figure out the answer! I guess you could say that buffer calculation turned into a bit of a circus. But hey, at least you were able to narrow it down to the closest option of 0.05 mol. It's like you were playing a game of "guess the mol" without a calculator. Keep up the great work, my math-savvy friend!

To determine the maximum amount of HCl that can be added without exceeding the capacity of the buffer, you need to calculate the pH of the buffer first.

The pH of the buffer can be calculated using the Henderson-Hasselbalch equation:

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

In this case, the acetic acid (HC2H3O2) is the acid and the sodium acetate (NaC2H3O2) is the base. The pKa of acetic acid is typically around 4.76.

Plugging in the values, you get:

pH = 4.76 + log([0.10]/[0.20])
pH = 4.76 + log(0.5)
pH = 4.76 - 0.30
pH = 4.46

Next, you want to find the pH at which the buffer capacity would be exceeded, which is 1.00 pH unit lower than the original pH.

pH = 4.46 - 1.00
pH = 3.46

To calculate the amount of HCl that can be added, you need to determine the amount of acetic acid and sodium acetate remaining in the buffer after the addition of HCl.

Let's assume x moles of HCl are added.

The moles of sodium acetate remaining = initial moles of sodium acetate - moles of HCl added
= 0.05 moles - x moles

The moles of acetic acid remaining = initial moles of acetic acid + moles of HCl added
= 0.1 moles + x moles

To ensure that the buffer capacity is not exceeded, the ratio of sodium acetate to acetic acid should be maintained:

0.05 moles - x moles / 0.1 moles + x moles = 0.05

Solving the above equation, you find x to be approximately 0.043 moles.

Therefore, the maximum amount of HCl that can be added without exceeding the capacity of the buffer is about 0.043 moles.

To determine the maximum amount of HCl that could be added to the buffer solution without exceeding its capacity, we need to calculate the pH of the buffer first using the Henderson-Hasselbalch equation: pH = pKa + log[(base)/(acid)].

1. Start by calculating the pKa of acetic acid (HC2H3O2). The pKa value for acetic acid is given as 4.76.

2. Calculate the ratio of the base (NaC2H3O2) to the acid (HC2H3O2). Since both solutions have the same volume (500 mL), the ratio is simply the concentration of the base divided by the concentration of the acid:

[(0.10 mol/L)/(0.20 mol/L)] = 0.5

3. Substitute the pKa value (4.76) and the ratio (0.5) into the Henderson-Hasselbalch equation:

pH = 4.76 + log(0.5)
pH ≈ 4.46

So the pH of the buffer solution is approximately 4.46.

Next, we want to determine the buffer capacity, which is the number of moles of a strong acid or base that can be added to the buffer solution without causing a drastic change in its pH.

4. We want the pH to be 1.0 unit lower than the initial value (4.46-1.0 = 3.46).

5. Substitute this new pH value and the pKa (4.76) into the Henderson-Hasselbalch equation:

3.46 = 4.76 + log[(base)/(acid)]
-1.3 = log[(base)/(acid)]

6. Take the antilog of both sides to eliminate the logarithm:

10^(-1.3) = (base)/(acid)
0.050 = (base)/(acid)

This means that the ratio of the base to the acid should be approximately 0.050 in order to achieve a pH of 3.46.

7. Now, consider that we are adding x moles of HCl to the buffer solution. The amount of base (NaC2H3O2) remaining after the addition would be (0.05 mol - x) and the amount of acid (HC2H3O2) would be (0.10 mol + x).

8. Use the ratio obtained in step 6 to set up an equation:

0.050 = (0.05 mol - x) / (0.10 mol + x)

9. Solve for x, the moles of HCl added, by cross-multiplication:

0.050 * (0.10 mol + x) = (0.05 mol - x)
0.005 mol + 0.050x = 0.05 mol - x
0.050x + x = 0.05 mol - 0.005 mol
1.050x = 0.045 mol
x ≈ 0.043 mol

Therefore, the maximum amount of HCl that can be added to the buffer solution without exceeding its capacity is approximately 0.043 moles. Note that this does not exactly match any of the given answer choices (A, B, C, D, E), but option B (0.05 mol) is the closest choice. However, it's important to note that when exactly 0.05 moles of HCl is added, the pH calculation becomes invalid due to the resulting zero concentration of the base, and the buffer capacity is lost.

I apologize for any confusion caused by the discrepancy between the calculated value and the given answer choices.