Ask questions and get helpful answers.
Ask a New Question
  1. Questions
  2. Chemistry

Calculate the pH at the equivalence point of 25.0 mL of a 0.100 M acetic acid, CH3COOH, titrated with a 0.100 M sodium hydroxide, NaOH. Ka of acetic acid is 1x10^-5

  1. 👍
  2. 👎
  3. 👁
  4. ℹ️
  5. 🚩
Anonymous
2 answers

  1. You must recognize what you have at the equivalence point.
    CH3COOH + NaOH ==> CH3COONa + H2O.
    At the equivalence point, we have converted all of the CH3COOH to CH3COONa (sodium acetate) so we have M x L = 0.1 x 0.025 = 0.0025 moles sodium acetate. That is in a volume of 50 mL (the other 25 came from 25 mL of 0.1 M NaOH). The pH will be determined by the hydrolysis of water. If we call acetate, Ac^-, and acetic acid HAc, then
    Ac^- + HOH ==> HAc + OH^-

    Kb for acetate = (Kw/Ka) = (HAc)(OH^-)/(Ac^-).
    Set up an ICE chart and solve.
    Post your work if you get stuck.

    1. 👍
    2. 👎
    3. ℹ️
    4. 🚩
    DrBob222

  2. help me solve "use the results of your experiment to determine the various endpoints and then plot a titration curves expected of the two indicators, volume of NaOH (x-axis) pH range of indicators (y-axis)
    results are :for bromothymol blue indicator average title volume is27.3 and for phenolphthalein indicator average average titre volume is 27.77

    1. 👍
    2. 👎
    3. ℹ️
    4. 🚩
    kipkemoi

Answer this Question

Related Questions

  1. science

    Which acid-base chemical reaction is irreversible?(1 point) strong acid added to water water on its own weak base added to water weak acid added to water Hydrochloric acid is a strong acid. Acetic acid is a weak acid. Which statement about hydrochloric

  2. Chemistry

    What is the pH of a 0.25 M acetic acid solution if the Ka of acetic acid is 1.8 x 10-5? 2.67 0.60 5.34 11.33 I got A, 2.67. I think this is correct as well since acetic acid is a weaker acid and the ph # would make sense.

  3. Chemistry

    Pure acetic acid, known as glacial acetic acid, is a liquid with a density of 1.049 g/mL at 25°C. Calculate the molarity of a solution of acetic acid made by dissolving 35.00 mL of glacial acetic acid at 25°C in enough water to make 400.0 mL of solution.

  4. Chemistry

    You have the following acids and their conjugate bases available: Acetic Acid: CH 3COOH/CH3COO-; Ka = 1.8 × 10-5 Carbonic Acid: H2CO3/HCO3-; Ka = 4.2 × 10-7 Hydrofluoric acid: HF/F-; Ka = 7.2 × 10-4 Which of these acid/conjugate base pairs would be the

  5. chemistry

    How many grams of acetic acid (C₂H₄O₂) would you use to make 10 L of a 0.1 M aqueous solution of acetic acid? (Note: The atomic masses, in daltons, are approximately 12 for carbon, 1 for hydrogen, and 16 for oxygen.)

  6. Chemistry

    A 30.0 mL sample of 0.165 molL−1 propanoic acid (Ka=1.3×10−5) is titrated with 0.300 molL−1KOH. Calculate the pH at 0, 5,10, at equivalence point, 1/2 of the equivalence point, 20, 25 mL of added base.

  7. chemistry

    If 3.15 mL of vinegar needs 42.5 mL of 0.115 M NaOH to reach the equivalence point in a titration, how many grams of acetic acid are in a 1.40 qt sample of this vinegar?

  8. 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 acid to ~400 mL of water

  9. chemistry

    The pH of a solution that contains 0.818 M acetic acid (Ka = 1.76 x 10-5) and 0.172 M sodium acetate is __________. The Ka of acetic acid is 1.76 × 10-5.

  10. chem.

    A 10.0 mL of vinegar, an aqueous solution of acetic acid (HC2H3O2), is titrated with .5062M NaOH, and 16.58mL is required to reach the equivalence point. A. What is the molarity of the acetic acid? B. If the density of the vinegar is 1.006 g/cm^3, what is

Still need help?

You can ask a new question or browse existing questions.