Chemistry

Determine how many mL of solution A (acetic acid-indicator solution) must be added to solution B (sodium acetate-indicator solution) to obtain a buffer solution that is equimolar in acetate and acetic acid.

Solution A:
10.0 mL 3.0e-4M bromescol green solution
25.0mL 1.60M acetic acid (HAc)
10.0mL .200M KCl solution

Solution B: 10.0mL 3.0e-4M bromescol green solution
10.0mL of .160M sodiuma cetate solution

a.Calculate moles Ac- in solution B
b. Calculate the molarity of HAc in solution A.
c. Calculate the volume of solution A needed when the moles of Ac- equals moles of HAc
--------------------------------------
K HAc is approximately 2e-5 under your experimental conditions. Use pK HAc = pH + log(N HAc/ N Ac-)
a. Calculate pKa
b. calculate moles HAc in 5mL solution A added to solution B
c.Calculate moles Ac- in solution B
d. Rearrange equation to solve for pH

  1. 👍
  2. 👎
  3. 👁
  1. Determine how many mL of solution A (acetic acid-indicator solution) must be added to solution B (sodium acetate-indicator solution) to obtain a buffer solution that is equimolar in acetate and acetic acid.

    Solution A:
    10.0 mL 3.0e-4M bromescol green solution
    25.0mL 1.60M acetic acid (HAc)
    10.0mL .200M KCl solution

    Solution B: 10.0mL 3.0e-4M bromescol green solution
    10.0mL of .160M sodiuma cetate solution

    The following three seem clear enough.
    a.Calculate moles Ac- in solution B
    (Note: mols = M x L = ? but I like to work in millimoles). So 10 mL x 0.160M = 1.6 mmols = 0.0016 mols.

    b. Calculate the molarity of HAc in solution A.
    1.6M x ((25 mL of HAc/45 mL total) = 0.888M HAc.

    c. Calculate the volume of solution A needed when the moles of Ac- equals moles of HAc
    mLA x MA = mmols B
    mLA x 0.8889M = 1.6
    mL A = 1.6/0.8889 = about 1.799 which I would round to 1.8 mL. You see why I like to work in mmols. mL x M = mmols AND M = mmols/mL.)


    --------------------------------------
    K HAc is approximately 2e-5 under your experimental conditions. Use pK HAc = pH + log(N HAc/ N Ac-)
    a. Calculate pKa
    pKa = -log Ka = -log 2E-5 = about 4.70 but you need to confirm that and round to the right number of s.f.

    b. calculate moles HAc in 5mL solution A added to solution B
    mL A + M B = mmols HAc in B = 5 mL x 0.8889 = 4.45 millimols = 0.00445 mols HAc.

    c.Calculate moles Ac- in solution B
    Isn' that done above in the first part a? 0.0016 mols Ac^- (or 1.6 mmols).

    d. Rearrange equation to solve for pH
    The rearranged equation is pH = pKa + log(base)/(acid)
    pH = 4.70 + log (1.60/4.45) = ? or if you want to use mols it is
    pH = 4.80 + log(0.0016/0.00445) = ?

    Check all of this carefully. Post any questions but show your work and explain exactly what your trouble is.

    1. 👍
    2. 👎
  2. The molarity of HAc in solution A is wrong according to my TA.

    1. 👍
    2. 👎

Respond to this Question

First Name

Your Response

Similar 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

  2. 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

  3. Chemistry

    I found the following procedure to prepare starch indicator solution on a website:- To prepare starch indicator solution, add 1 gram of starch (either corn or potato) into 10 mL of distilled water, shake well, and pour into 100 mL

  4. chemistry

    Lab: Determining Ka of Acetic Acid Purpose: The purpose of this experiment is to determine the molar concentration of a sample of acetic acid and to calculate its Ka. Materials: phenolphthalein 125 mL Erlenmeyer flask 25 mL pipet

  1. science

    Hydrochloric acid is a strong acid. Acetic acid is a weak acid. Which statement about hydrochloric acid and acetic acid is correct?(1 point) The dissociation constant for hydrochloric acid is greater than the dissociation constant

  2. CHEMISTRY

    HOW CAN U TELL IF HNO3 +KNO3 IS A BUFFER SOLUTION A buffer solution must contain a weak acid and its conjugate base OR a weak base and its conjugate acid. HNO3 is a strong base and KNO3 is the salt of a strong base (KOH) and a

  3. Chemistry

    A solution buffered at a pH of 5.00 is needed in an experiment. Can we use acetic acid and sodium acetate to make it? If so, how many moles of NaC2H3O2 must be added to a 1.0L of a solution that contains 1.0 mol HC2H3O2 to prepare

  4. Chemistry

    A solution is 40% acetic acid by mass. The density of this solution is 1.049 g/mL. Calcu- late the mass of pure acetic acid in 120 mL of this solution at 20 C.

  1. chemistry

    Lab: Determining Ka of Acetic Acid Purpose: The purpose of this experiment is to determine the molar concentration of a sample of acetic acid and to calculate its Ka. Materials: 25 mL pipet and bulb burette 2x150 mL beaker 125 mL

  2. Chemistry

    40.0 ml of an acetic acid of unknown concentration is titrated with 0.100 M NaOH. After 20.0 mL of the base solution has been added, the pH in the titration flask is 5.10. What was the concentration of the original acetic acid

  3. Chemistry

    1)A solution prepared by adding 0.400 mol of acetic acid (pKa = 4.74) and 0.400 mol of sodium acetate to 100.0 mL of water. The pH of this buffer solution is initially 4.74. Predict the final pH when 55.0 mL of 1.10 M NaOH is

  4. Chemistry

    A buffer contains 5.00 M acetic acid and 5.00 M acetate anion. Gaseous HCl (0.010 mole) is added to 1.00 L of this buffer solution (the total volume does not change). For this buffer solution, the initial pH is_______and the final

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