chemistry

posted by .

calculate the pH of the solution when 100mL of .05M KF is mixed with 200mL of .05M HF; Ka (HF) = 3.467x10^-4. I tried to use the Henderson Hasselbach equation to get the correct answer (pH=3.16) but can't seem to get it correct. Any help would be AMAZING! Thank you in advance!

  • chemistry -

    Why not show your work with the HH equation and let me find the error.

  • chemistry -

    I tried to find the numbers I need for the HH equation, and I can't seem to find the correct way to use my chemical equation to find them.
    H+ + KF -> HF + K+
    .01 .005 0 0
    -.005 -.005 +.005 +.005
    .005 0 .005 .005 <moles left at equilibrium
    KF <-> F- + K+
    0 0 .005
    ^I believe there is an error with the chemical equations I used because the equilibrium should shift to my products in the <-> equation.

  • chemistry -

    the formatting is a little off, I'm so sorry. Let me know if you need me to re-type it.

  • chemistry -

    Oh! I think I just caught my mistake...HF is a weak acid, so that reaction does not go to completion....

  • chemistry -

    And then the KF reaction will go to completion because it is completely soluable!

  • chemistry -

    I got it! Nevermind! I just was assuming the wrong equation went to completion! Thank you for asking me to type it out! It definitely helped me! :)

  • chemistry -

    You may be ok by now but all you need to do is to substitute HF and KF. There is no reaction.
    (KF) = 0.05 x (100/300) = ?
    (HF) = 0.05 x (200/300) = ?
    Then HF goes in for the acid concn and KF goes in for the base concn.

    Actually, there is an easier way to do it I think. Work with millimoles.
    millimols HF = 0.05 x 200 mL = ?
    millimoles KF = 0.05 x 100 = ?
    Then mmols KF substitutes for the base
    mmols HF substitutes for the acid.
    Technically, substituting mmols is not substituting concn BUT mmols/mL = concn and since you divide both acid and base by the same number (300 in this case), the 300 cancels and you never have to bother with it.

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. chemistry

    What is the finaly concentration of barium ions, Ba2+, in soulution with 100 ml of .1 M BaCl2 is mixed with 100ml of .05M H2SO4?
  2. chemistry

    We're studying acids and bases; the question states that 50mL of 0.1M HNO3 solution is mixed with 25mL of 0.1M NaOH. What is the pH?
  3. chemistry

    Which of the following mixtures would made the most effective buffer?
  4. chemistry

    2CO2(G)--->2CO(g)+O2(g), deltaH=566kj what is Kc and state whether the reaction favors products or reactants?
  5. chemistry

    2CO2(G)--->2CO(g)+O2(g), deltaH=566kj what is Kc and state whether the reaction favors products or reactants?
  6. Chemistry

    A solution is prepared by dissolving 50.0 g of pure HC2H3O2 and 20.0 g of NaC2H3O2 in 975 mL of solution (the final volume). what is the ph?
  7. chemistry

    1L of a buffer composed of acetic acid and sodium acetate has a pH of 4.3 (total molarity=0.0774) what is the capacity of 200mL of the buffer to survive an addition of HCl without changing by more than 0.4 pH units?
  8. chemistry

    1L of a buffer composed of acetic acid and sodium acetate has a pH of 4.3 (total molarity=0.0774) what is the capacity of 200mL of the buffer to survive an addition of HCl without changing by more than 0.4 pH units?
  9. Chemistry - Solubility

    Given: Concentration of HCl is 0.1388M 0.5 g of Ca(OH)2 was placed in a flask. 100mL of 0.05M NaOH was poured into the flask. 25mL aliquot was filtrated and used for titration. Suppose to calculate the OH^- equilibrium concentration …
  10. Chemistry

    Calculate the pH of a solution made by mixing 60.0 ml of 0.80 M N2H4 (a weak base) with 50.0 ml of 0.50 M HCl. (Kb for N2H4=1.7 x 10^-6) Can I use the Henderson Hasselbach equation here?

More Similar Questions