Post a New Question

chemistry

posted by .

Assignment 1

Question

Consider a monohydrogen phosphate ( HPO42-) and dihydrogen phosphate (H2PO4-) buffer solution.

[HPO42-] = 0.063M

[H2PO4-] = 0.10M

What happens when you add 1.0 ml of 0.10 M HCl to the a 99ml solution?

What would the pH of the solution be without the buffer?

What would the pH of the solution be with the buffer?

Answer

For your answer you will need to go beyond simply filling in the formula and giving me a number.


You will need to convince me that you know what is going on here. Words are required for the explanation not just numbers.

  • chemistry -

    You fill in the words. I will get you started on the numbers.
    pH without the buffer? Is that the pH of 1 mL of 0.1M HCl? If so, then

    (HCl) = 0.1M and pH = -log(0.1) = 1

    pH buffer before addition of HCl.
    pH = pK2 + log (base)/(acid). Substitute and solve for pH.

    pH with 1 mL 0.1M HCl added.
    millimols H2PO4^- = 0.1M x 99 mL = 9.90
    millimols HPO4^2- = 0.063 x 99 = 6.24
    millimols HCl = 0.1M x 1 mL = 0.1

    ..........HPO4^- + H^+ ==> H2PO4^2-
    I.........6.24.....0........9.90
    add...............0.1...........
    C........-0.1....-0.1........+0.1
    E.........6.14.....0.........10.0

    Substitute E line into H-H equation and solve for pH.
    Note: Some profs are picky and will count off for using millimoles and not mols/L in the above ICE chart and the following HH equation. That's because the HH equation is ....(base)/(acid) and not mmols/mmols. However, the answer comes out the same BECAUSE concn = mols/L = mmols/mL and since you are using base = mmols/mL and acid = mmols/mL, the mL is ALWAYS the same so they cancel and you are left with mmols/mmols and things work out fine. However, if your prof is picky (I counted off) you have two choices.
    1. Divide EACH of the mmols by the total volume (in this case it is 100 mL) and use the concn then in mmols/mL = M or
    2. Use mmols as shown in the ICE table above BUT divide each by V (for volume). It would look like this
    pH = pK2 + log (10.0/V)/(5.14/V), cancel the Vs and continue with the problem.
    Hope this helps. Post any further questions; however, explain in detail what you don't understand.

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. To Jessica

    Sorry, but I made two errors. Monocalcium phosphate is monocalcium monohydrogen phosphate which is CaHPO4. The Ca is +2, H is +1, and PO4 is -3 so the +3 and -3 balance. And Bob Pursley did NOT suggest that name of monocalcium dihydrogen …
  2. naming chemical compounds

    what is the dihydrogen phosphate ion? and the answer is not H2PO4-, H2PO4, HPO4... =(
  3. chemistry

    Given the table of Ka values on the right below, arrange the conjugate bases in order from strongest to weakest. Acid Ka HClO 3.5 e-8 HClO2 1.2 e-2 HCN 6.2 e-10 H2PO4- 6.2 e-8 A. ClO2-, ClO¬-, HPO42-, CN- B. ClO2-, HPO42-, ClO¬-, …
  4. Chemistry

    Superphosphate, a water soluble fertilizer, is sometimes marked as a "triple phosphate". It is a mixture of Ca(H2PO4)2 + 2CaSO4. Ca3(PO4)2 + 2H2SO4 --> Ca(H2PO4)2 + 2CaSO4 If you treat 400g of calcium phosphate with 267g of sulfuric …
  5. chemistry

    For the reaction: H2PO4- + HAsO42- ---> HPO42- + H2AsO4- Which pair exists as base in the equilibrium?
  6. Chemistry(Please check)

    Which of the following equations correspond to the Ka2 for phosphoric acid?
  7. Science

    The Ka for the H2PO4–/HPO42– buffer is given by the equation: Ka= ([HPO4^-2]gamma [H^+])/[H2PO4^-]=10^-7.20 What is the pH of a 2:3 mixture of H2PO4– to HPO42– at 0.10 M ionic strength?
  8. chemistry

    The total phosphate concentration in a blood sample is determined by spectrophotometry to be 3.0×10−3 M. The pH of the blood sample is 7.45, What are the concentrations of H2PO4− and HPO42- in the blood?
  9. Chemistry

    A pH = 7.6 buffer is needed in the lab. This buffer is made by first dissolving 17.42 g K2HPO4 in 600 mL of water. What is the pH of this salt solution?
  10. Chemistry

    A phosphate buffer solution contain 0.08 M K2HPO4 and 0.12 M KH2PO4. pKa values for H3PO4 are 2, 6.8, and 12.2. Calculate the concentrations of a) H3PO4 b) H2PO4- c) HPO42- d) H+ e) OH- f) K+

More Similar Questions

Post a New Question