chemistry

19 mL of 0.50 mool/L NaOH which is standardized becomes titrated alongside 24 mL of 0.44 mol/L acetic acid. Determine the pH of the solution

Please judge my work:

Becasue NaOH and acetic acid react in a 1:1 ratio,

initital moles of acetic acid
=0.0019 L x 0.44M
= 0.000836 mols

mols of acetic acid neutralized is
= 0.0024 L x 0.44 M
=0.0019 L

molfs of acetic acid is 0.0019-0.000836 =0.001064

Whats my next steps?

asked by Joseph
  1. 19 mL of 0.50 mool/L NaOH which is standardized becomes titrated alongside 24 mL of 0.44 mol/L acetic acid. Determine the pH of the solution

    Please judge my work:

    Becasue NaOH and acetic acid react in a 1:1 ratio, OK to here but check your numbers. They don't make sense.

    initital moles of acetic acid
    =0.0019 L x 0.44M It's 19 mL of NaOH which is 0.44 M. You've mixed up mL and component, I believe.
    = 0.000836 mols

    mols of acetic acid neutralized is
    = 0.0024 L x 0.44 M The one above also says acaeitic acid so which is which?
    =0.0019 L Liters???. You started out by saying this was mols. Is molarity measured in L?

    molfs of acetic acid is 0.0019-0.000836 =0.001064

    Whats my next steps?

    posted by DrBob222

Respond to this Question

First Name

Your Response

Similar Questions

  1. Chemistry

    In this experiment NaOH is standardized to find out the percent by mass of the acetic acid in a sample of vinegar. A student had not allowed the NaOH pellets to dissolve completely before standardizing it with KHP, but when the
  2. Chemistry

    Commercial vinegar was titrated with NaOH solution to determine the content of acetic acid, HC2H3O2. For 20.0 milliliters of the vinegar 26.7 milliliters of 0.600-molar NaOH solution was required. What was the concentration of
  3. chemistry

    Please judge my answer: Question: 24 mL of 0.39 mol/L acetic acid is titrated with a standardized 0.33 mol/L KOH solution. Calculate the pH of the solution after 17 mL of the KOH solution has been added. Assume the Ka of acetic
  4. chemistry

    hi, I am trying to find the solubility of Sodium Acetate by using 0.100 mol Hydroxide(base) and 0.837 mol acetic acid( vinegar). The theoretical solubility of sodium acetate is 82.3g/100ml at 25C. Doing calculations we would want
  5. Chemistry

    Titration of 50.0 ml of acetic acid reaches equivalence after delivery of 22.5 ml of standardized NaOH 0.21 M. What is the initial concentration of acetic acid and what is the pH of the solution? What is the pH at equivalence?
  6. chemistry 2

    Titration of 50.0 mL of acetic acid reaches equivalence after delivery of 22.5mL of standardized NaOH 0.21 M. What is the initial concentration of acetic acid and what is the pH of the solution? What is the pH at equivalence? What
  7. chemistry

    25 mL of standardized 0.45 mol/L NaOH is titrated with 21 mL of 0.35 mol/L acetic acid. Calculate the pH of the solution
  8. Chemistry

    Consider a weak acid-strong base titration in which 25.0 mL of 0.100 M acetic acid is titrated with 0.100 M NaOH. a) Calculate the pH after the addition of 3.00mL of NaOH. b) What is the pH of the solution before the addition of
  9. Chemistry

    I did an experiment using 20ml of 0.1M acetic acid and added 8ml of 0.1M NaOH. pH obtained after adding NaOH was 5.2. How do i calculate the mols of NaOH that reacted? mols of acetate formed, acetic acid initially present and
  10. For Dr.Bob222 (chem work)

    What is the pH of a solution of 120 ml 0.15M acetic acid to which we add 30mL 0.2M NaOH? CAN YOU CHECK MY WORK PLEASE? 0.15 mol/L * 0.12 L = 0.018 mol acetic acid 0.2 mol/L * 0.03 L = 0.006 mol NaOH 0.018 - 0.006 = 0.012 mol of

More Similar Questions