chemistry

posted by .

A 20.1 mL sample of a weak monoprotic acid, HX, requires 50.0 mL of 0.060 M NaOH to reach the equivalence point. After the addition of 30.0 mL of NaOH, the pH is 4.90. What is the Ka of HX?

  • chemistry -

    There are a couple of ways to do this; one is quite straight forward, the other requires a little reasoning power. Here is the straight forward way.
    mLNaOH x MNaOH = mLHX x MHX
    50.0 x 0.060 = 20.1 x MHX
    Solve for Molarity HX. For simplicity I shall call this about 0.15M.
    So millimiles HX to start = 20.1*0.15 = about 3.0
    millimoles NaOH at 30 mL = 30 mL x 0.06 = 1.8 mmoles.

    Set up an ICE chart.
    ...........HX + NaOH ==> NaX + H2O
    initial....3.0...0........0......0
    add.............1.8...............
    react......-1.8..-1.8....1.8.....+1.8
    equilibrium..1.2...0.....1.8......1.8

    Substitute into Ka expression for HX (convert pH to H^+ first), and solve for Ka.
    I worked it out to be about 1.9E-5 for Ka but check my work.
    The more esoteric way of doing it is to use the Henderson-Hasselbalch equation.
    You know, from the equivalence point, that if you add 3.0 millimoles NaOH that you must have started with 3.0 millimoles HX. So you add 1.8 millimoles NaOH at the 30 mL mark which means 1.2 HX is left and 1.8 mmoles were converted to X^-. Then substitute into pH = pKa + log (X^-)/(HX) and solve for pKa, then convert to Ka. I have 4.72 for pKa.

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. Chemistry

    In this case, the inflection point, and equivalence, occurs after 23.25mL of 0.40 M NaOH has been delivered. Moles of base at the equivalence point can be determined from the volume of base delivered to reach the equivalence point …
  2. Chemistry

    A 22.3 mL sample of a weak monoprotic acid, HX, requires 50.0 mL of 0.040 M NaOH to reach the equivalence point. After the addition of 30.0 mL of NaOH, the pH is 5.30. What is the Ka of HX?
  3. Chemistry

    A 22.3 mL sample of a weak monoprotic acid, HX, requires 50.0 mL of 0.040 M NaOH to reach the equivalence point. After the addition of 30.0 mL of NaOH, the pH is 5.30. What is the Ka of HX?
  4. Chemistry

    A 0.225g sample of a weak monoprotic acid requires 37.50 mL of 0.100 M NaOH solution to reach the equivalence point. What is the molar mass of the acid?
  5. chemisry

    A 0.1 molal solution of a weak monoprotic acid was found to depress the freezing point of water 0.1930C. Determine the Ka of the acid. You can assume 0.1 molal and 0.1 molar are equivalent. 100 ml of a 0.1M solution of the above acid …
  6. chemistry

    a. A 0.1 molal solution of a weak monoprotic acid was found to depress the freezing point of water 0.1930C. Determine the Ka of the acid. You can assume 0.1 molal and 0.1 molar are equivalent. b. 100 ml of a 0.1M solution of the above …
  7. Chemistry

    A 20.2 mL sample of a weak monoprotic acid, HX, requires 50.0 mL of 0.060 M NaOH to reach the equivalence point. After the addition of 30.0 mL of NaOH, the pH is 5.50. What is the Ka of HX?
  8. chemistry

    You are titrating an unknown weak acid you hope to identify. Your titrant is a 0.0935 mol/L NaOH solution, and the titration requires 39.9 mL to reach the equivalence point. How many moles of acid were in your sample?
  9. 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?
  10. Chemistry

    A 22.6 mL sample of a weak monoprotic acid, HX, requires 50.0 mL of 0.050 M NaOH to reach the equivalence point. After the addition of 30.0 mL of NaOH, the pH is 4.90. What is the Ka of HX?

More Similar Questions