A beaker with 100 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 6.10 mL of a 0.260 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.760.

I got 0.27 as the change, but it says I made a rounding error, and I can't find my rounding error. I also got 0.26 and 0.25, but they didn't work.

  1. 👍 0
  2. 👎 0
  3. 👁 283
  1. I would assume three significant digits on this.

    work it out to three digits.

    1. 👍 0
    2. 👎 0
  2. Sorry, I forgot to say that it told me to round it to two decimal places.

    1. 👍 0
    2. 👎 0
  3. recheck how you rounded the antilogs.

    1. 👍 0
    2. 👎 0
  4. I did everything from the beginning and I didn't round anything at all until my last step where it told me to round to two decimal places. I still can't get the answer, though

    1. 👍 0
    2. 👎 0
  5. I can't get any of your answers.
    Post your work and let me look for the error. I think the answer is close to 0.7 if I didn't make an error.

    1. 👍 0
    2. 👎 0
  6. Here's what I did:

    So I set values for:
    [Acetic Acid] = y
    [H3O+] = 10^-5
    [Acetate] = 0.1M - y
    Ka = 10^-4.76

    And from the equation: Ka = [Acetate][H3O+]/[Acetic Acid], I found y, or the molarity of acetic acid, to be 10^-6/(10^-5 + 10^-7.46). And from this, I also got the molarity of acetate: 0.1 - y.

    Then, I found the total molarity. I converted 10^-5 M acetic acid to 10^-6 M/0.1 L. I then took 0.260 M HCl and converted it to 0.001586 M/0.0061 L. I added the top and the bottom and it turned out to be 0.001587 moles/.1061 L. Let this value be A.

    And since the Ka value is low, it goes to completion to the left. So, [Acetic Acid] = y + A. And [Acetate] = 0.1 - y -A. And now for the Henderson/Hasselbach equation:

    pH = 4.76 + log((y + A)/(0.1 - y - A))
    And I plugged in:
    y = 10^-6/(10^-5 + 10^-7.46)
    A = 0.001587/.1061
    Then, I subtracted 5 from this and got my answer.

    1. 👍 0
    2. 👎 0
  7. I said I would find your error but I've changed my mind after I see how you worked the problem. Have you seen the Henderson-Hasselbalch equation? Have you used it? Why not use it here? It is MUCH simpler and it uses exactly the same chemistry; just a different form of it.
    pH = pKa + log[(base)/(acid)] and all of that comes from Ka = (H^+)(acetate)/(acetic acid) and pH = -log(H^+).
    From pH = 5.00 and pKa = 4.760, find (base)/(acid) ratio. (Two unknowns there). Then from the problem you know base + acid = 0.1 (two of the same unknowns). Solve for (base) and (acid). Then set up an ICE chart, add the 6.10 mL and plug that information into a new HH equation and solve for pH. calculate delta pH.

    1. 👍 0
    2. 👎 0
  8. Okay, so I found the ion concentrations through the Henderson-Hasselbalch equation, and the concentrations I got:

    [Acetic Acid] = 0.036525666
    [Acetate] = 0.063474334

    I got the same concentrations as when I did my ice chart, and I also got the same delta pH, which was -0.266.

    1. 👍 0
    2. 👎 0
  9. I still don't get it.
    pH = pKa + log(B/A)
    5.00 = 4.76 + log(B/A)
    B/A = 1.7378
    A + B = 0.100
    Two equation in two unknowns.
    A = your number
    B = your number.
    You started with 100 mL; therefore, you started with 3.653 mmoles HAc (which may be the difference) and 6.347 mmoles Ac^-

    When I did the ice chart I ended up with this.

    ...........Ac^- + H^+ ==> HAc

    pH = 4.760 + log(4.761/5.239)
    pH = 4.71845
    delta pH =0.28155 to 2 s.f. is 0.28 and you data base MAY want to see -0.28
    I did find an error in my original calculation in which I thought it was closer to 0.7. Interestingly enough, I check every one of my steps but one and I didn't check that one because it was too simple (just before I posted the 0.7 number). Guess where the error was? I subtracted incorrectly. But I think 0.28 is right. I hope this helps.

    1. 👍 0
    2. 👎 0
  10. Thanks a lot! I realized that I forgot to account for the volume change when I calculated my molarities for the base and the acid, and that was enough to get the pH off by 0.01.

    1. 👍 0
    2. 👎 0

Respond to this Question

First Name

Your Response

Similar Questions

  1. Chemistry

    An acetic acid buffer solution is required to have a pH of 5.27. You have a solution that contains 0.01 mol of acetic acid. How many moles of sodium acetate will you need to add to the solution? The pKa of acetic acid is 4.74.

  2. Chemistry

    You are asked to prepare 500. mL of a 0.150 M acetate buffer at pH 5.00 using only pure acetic acid, 3.00 M NaOH, and water. Calculate the quantities needed for each of the following steps in the buffer preparation. 1. Add acetic

  3. Quantitive Analysis

    You are asked to prepare 500. mL of a 0.250 M acetate buffer at pH 4.90 using only pure acetic acid (MW=60.05 g/mol, pKa=4.76), 3.00 M NaOH, and water. Answer the following questions regarding the preparation of the buffer.

  4. Chemistry (buffers)

    Say, for example, that you had prepared a Buffer C, in which you mixed 8.203 g of sodium acetate, NaC2H3O2, with 100.0 mL of 1.0 M acetic acid. a. What would be the initial pH of Buffer C? HOW???????

  1. school

    What is the pH of 0.1 M formic acid solution? Ka=1.7„e10-4? What is the pH value of buffer prepared by adding 60 ml of 0.1 M CH3COOH to 40 ml of a solution of 0.1 M CH3COONa? What is the pH value of an acetate buffer (pK=4.76)

  2. Chemistry - Buffers

    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.

  3. Chemistry

    some of the acetic acid splashed out the beaker when you dropped the tablets into the acetic acid. How would this affect your measurement? Would your final calculated mass of sodium bicarbonate in the tablet be artificially high

  4. Chemistry

    A chemist wants to be a supercook and wants to create 8%(m/v) acetic acud solution to be "extra strength" vinegar. if pure acetic acid Is a liquid density of 1.049 g/ml. how many ml of acetic acid should be dissolved to male 500

  1. chemistry

    Prepare 500mL of a 0.200 M acetate buffer at pH 4.90 using only pure acetic acid, 3M NaOH, and water. 1) Add acetic acid to ~400mL of water in a 500 mL beaker. How many grams of acetic acid are needed? 2)Add 3 M NaOH solution

  2. chemistry

    Calculate the volume of 1.00 M acetic acid solution and mass of sodium acetate needed to make 250 mL of a 0.450 M acetic acid/acetate buffer solution with a pH of 4.00. (The pKa is 4.74 for acetic acid).

  3. chemistry

    If you add 5.0 mL of 0.50 M NaOH solution to 20.0 mL to Buffer C, what is the change in pH of the buffer? (where buffer C is 8.203 g sodium acetate with 100.0 mL of 1.0 M acetic acid) I have calculated the pH of buffer C to be

  4. chemistry

    If you add 5.0 mL of 0.50 M HCl solution to 20.0 mL to Buffer C, what is the pH of the buffer? (where buffer C is 8.203 g sodium acetate with 100.0 mL of 1.0 M acetic acid)

You can view more similar questions or ask a new question.