Posted by anonymous on Saturday, January 26, 2013 at 8:44pm.
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?
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 - DrBob222, Saturday, January 26, 2013 at 9:32pm
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-
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.
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