Chemistry - Adding Strong Acid to Buffer

In summary, a student is trying to determine the change in pH when 7.80 mL of a 0.400 M solution is added to a beaker containing 135 mL of an acetic acid buffer with a pH of 5.00 and a total molarity of 0.1 M. Using the Henderson-Hasselbach equation, the student incorrectly calculated the ratio of [base]/[acid] and obtained a pH change of -0.40. The correct calculation involves finding the initial value of [A-] from the given information and using the fact that [A-](initial) + [HA](initial) = 0.1.
  • #1
sam.
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Homework Statement


A beaker with 135 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 0.1 M. A student adds 7.80 mL of a 0.400 M solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.76.


Homework Equations



Henderson-Hasselbach equation: pH = pKa + log([base]/[acid])

The Attempt at a Solution



hat I did was find the total number of moles of acid + base = 0.0135. Then using the pH = pKa + log([base]/[acid]) I found the ratio of [base]/[acid] to be 1.74. Then using these two equations I found the moles of acid to be 4.93*10^-3 and base = 5.45*10^-3. Next I found the moles of HCl = 0.00312 mol and substracted that from the base and added it to the acid so: acid = 8.05*10^-3 and base = 5.45*10^-3. Now I put these values into pH = pKa + log([base]/[acid]) and got the pH to be 4.59...therefore the change would be -0.40...But I think this value is too high...I think it's supposed to be lower. Can anyone tell me where I went wrong? Any help is appreciated.
 
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  • #2
sam. said:

Homework Statement


A beaker with 135 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 0.1 M. A student adds 7.80 mL of a 0.400 M solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.76.

7.80 mL of a 0.400 M solution of what? Possibly a strong acid? I'll assume it is a strong acid (HCl).

The Attempt at a Solution



What I did was find the total number of moles of acid + base = 0.0135.
No. What does your question say about the total concentration of the acid [HA] and conjugate base[A-]? It isn't 0.0135.

Next I found the moles of HCl = 0.00312 mol
This part is right.

... and substracted that from the base and added it to the acid so: acid = 8.05*10^-3 and base = 5.45*10^-3. Now I put these values into pH = pKa + log([base]/[acid]) and got the pH to be 4.59...therefore the change would be -0.40...But I think this value is too high...I think it's supposed to be lower. Can anyone tell me where I went wrong? Any help is appreciated.
Your logic is sound here but you used the wrong value for the ratio of [A-]/[HA]. Can you calculate initial [A-] given only the initial pH? You should also keep in mind that [A-](initial) + [HA](initial) = 0.1 and that total doesn't change throughout the problem, only the individual values of [A-] and [HA].

Hope it helps.
 
  • #3


I would like to commend you for your attempt to solve this problem using the Henderson-Hasselbach equation and your understanding of buffer solutions. However, I believe your calculation for the moles of HCl added may be incorrect. You stated that you found the moles of HCl to be 0.00312 mol, but this would only be true if all of the HCl reacted with the base in the buffer. In reality, only a portion of the HCl will react and the rest will remain in its original form.

To find the change in pH, you can use the Henderson-Hasselbach equation again, but this time using the new concentrations of acid and base after the addition of HCl. This will give you a more accurate result. Additionally, you can also use the equation pH = -log[H+], where [H+] is the concentration of hydrogen ions in the solution. Using this equation, you can find the concentration of hydrogen ions before and after the addition of HCl and calculate the change in pH.

It is important to keep in mind that when a strong acid is added to a buffer solution, the pH will decrease, but it may not change significantly depending on the buffer capacity. In this case, the buffer capacity is quite high due to the relatively high concentrations of acid and base, so the change in pH may be minimal.

In conclusion, your understanding of buffer solutions and the Henderson-Hasselbach equation is commendable, but it is important to carefully consider the amount of HCl that will actually react with the buffer and to use the correct equations to calculate the change in pH.
 

1. What is a buffer?

A buffer is a solution that resists changes in pH when small amounts of acid or base are added to it.

2. What happens when a strong acid is added to a buffer?

When a strong acid is added to a buffer, the buffer solution will resist large changes in pH by neutralizing the added acid with a weak base present in the buffer solution.

3. Can a buffer be overwhelmed by a strong acid?

Yes, a buffer can be overwhelmed by a strong acid if the amount of acid added is significantly larger than the amount of weak base present in the buffer solution.

4. How do you calculate the pH of a buffer after adding a strong acid?

The Henderson-Hasselbalch equation is used to calculate the pH of a buffer after adding a strong acid. The equation is pH = pKa + log [base]/[acid], where pKa is the acid dissociation constant of the weak acid and [base] and [acid] are the concentrations of the weak base and strong acid, respectively.

5. What is the purpose of adding a strong acid to a buffer?

The purpose of adding a strong acid to a buffer is to maintain a stable pH in a solution. This can be useful in many laboratory experiments and industrial processes that require a specific pH in order to function properly.

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