chem

posted by .

how many mL of .100 M of acetic acid and .100 M of NaOH, would be need to make the volume 1.0 liters and have a pH7.00 if Ka=1.8x10^-5

please write out the steps or eq for me im gettin stuck

  • chem -

    This is not a good question because the pH of 7.00 is too far from the pKa value to work well as a buffer. However, here is the problem.
    pH = pKa + log (base)/(acid)
    7.00 = 4.74 + log (B/A)
    Solve for (base/(acid). I obtained 181.97 but you need to confirm that. Here base is referring to acetate ion (not NaOH) and acid is referring to acetic acid. I call those HAc for acetic acid and Ac^- for acetate (the base).

    Then you set up another equation.
    moles base = 181.97*moles acid
    For moles base substitute mLbase x 0.1 M. On the right side, substitute for moles acid (1000 - mLbase)*0.1 M
    Solve for mLbase
    I get something like 994.53 for base and 1000-994.53 = 5.47 mL acid. You need to confirm both numbers AND you may need to carry out to more places that you are allowed under significant figure rules BECAUSE it is such a large number for one and a small number for the other one.
    What this means is that you need to add that many mL of 0.1 M NaOH to produce that much Ac^- and since it is neutralizing HAc, the HAc is being used up at the same time. To check if this is correct,
    (Ac^-) formed = 994.53 mL x 0.1M/1000 mL = 0.099453 M.
    (HAc) remaining = 5.465 mL x 0.1 M/1000 = 0.0005465
    pH = 4.74 + log (0.099453/0.0005465) =
    pH = 4.74 + 2.26 = 7.00. I reiterate this is a poor problem AND that I have used more significant figures than allowed. With such a disparity in numbers, that is necessary to make the numbers come out right.

  • chem -

    I don't think this would be correct because The solution you are proposing would have a pH over 12.

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. Chem

    What is the pH of the resulting solution if 30.00 mL of 0.100M acetic acid is added to 10.00mL of 0.100 M NaOH?
  2. college chem

    how many mL of .100 M of acetic acid and .100 M of NaOH, would be need to make the volume 1.0 liters and have a pH7.00 if Ka=1.8x10^-5
  3. college chem

    how many mL of .100 M of acetic acid and .100 M of NaOH, would be need to make the volume 1.0 liters and have a pH7.00 if Ka=1.8x10^-5?
  4. college chem

    how many mL of .100 M of acetic acid and .100 M of NaOH, would be need to make the volume 1.0 liters and have a pH7.00 if Ka=1.8x10^-5 please write out the eq for me
  5. Chem

    25.0 mL of 0.100 M acetic acid (Ka= 1.8 x 10^-5) is titrated with 0.100 M NaOH. Calculate the pH after the addtion of 27.00 mL of 0.100M NaOH. my work CH3COOH + H2O <-> H3O^+ + CH3COO^- 25mL x 0.100 mmol/ml = 2.5mmol CH3COOH …
  6. Chemistry

    25.0 mL of 0.100 M acetic acid (Ka= 1.8 x 10^-5) is titrated with 0.100 M NaOH. Calculate the pH after the addtion of 27.00 mL of 0.100M NaOH. my work CH3COOH + H2O <-> H3O^+ + CH3COO^- 25mL x 0.100 mmol/ml = 2.5mmol CH3COOH …
  7. college chem

    a. According to the Henderson-Hasselbalch equation, how many grams of sodium acetate will you need to add to 125 mL of 0.100 M acetic acid to make a pH=4.74 buffer?
  8. CHEMISTRY HELP!

    a. According to the Henderson-Hasselbalch equation, how many grams of sodium acetate will you need to add to 125 mL of 0.100 M acetic acid to make a pH=4.74 buffer?
  9. Chemistry

    I have no idea where to even start with this problem on my pre-lab! For part of the experiment, you will need to prepare 100 mL of 0.025 M CH3CO2H, starting from commercial "glacial" acetic acid, which has a concentration of 17.4 M. …
  10. Chem 2 lab

    the acid dissociation constant for acetic acid is 1.85x10^-5. Calculate the pH at the equivalent point and the pH at one-half the equivalence point for the titration of 25.00 mL of 0.100 M acetic acid with 0.100 M NaOH?

More Similar Questions