1 a) first start with the forms of asp: pKa's: 2.0 3.9 10.0 H3A+ <====>H2A<====> HA-<=======>A2- 12.0
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![SOLVED: Equations: pH = pKa log ([b] /[a] (Ka) (Kb) = 1x 10-14 Kb = x2/ (y-x) K = xl/ly x) pH (pKa1 pKa2)/2 pK,'s of amino acid side chains: D (3.9),](https://cdn.numerade.com/ask_images/2531a2f0c6754ab3ba0caf17c487c22e.jpg "SOLVED: Equations: pH = pKa log ([b] /[a] (Ka) (Kb) = 1x 10-14 Kb = x2/ (y-x) K = xl/ly x) pH (pKa1 pKa2)/2 pK,'s of amino acid side chains: D (3.9),")
SOLVED: Equations: pH = pKa log ([b] /[a] (Ka) (Kb) = 1x 10-14 Kb = x2/ (y-x) K = xl/ly x) pH (pKa1 pKa2)/2 pK,'s of amino acid side chains: D (3.9),
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Please explain the first step how do we get ph 1 2 pka logc
Expression pKh = pKw - pKa - pKb is not applicable to:
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pH of a solution of salt of weak acid and weak base is : pH=1/2pKw+1/2pKa-1/2pKb and that of weak acid and strong base is pH=1/2pKw+1/2pKa+1/2logc pH of 0.1 M solution of ammonium
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![How is pH = 1/2[pKa - logc] - Chemistry - Chemical and Ionic Equilibrium - 13113047 | Meritnation.com](https://s3mn.mnimgs.com/img/shared/ck-files/ck_57fe3e2aeb864.png "How is pH = 1/2[pKa - logc] - Chemistry - Chemical and Ionic Equilibrium - 13113047 | Meritnation.com")
How is pH = 1/2[pKa - logc] - Chemistry - Chemical and Ionic Equilibrium - 13113047 | Meritnation.com
Acid-base theory pH calculations - ppt download
Match the List - I (solution of salts) with List - II (pH of the solution) and select the correct answer using the codes given below the lists:List - IList - IIA.
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Solved i followed the pH formula which is 1/2(pKw+pKa+logc) | Chegg.com
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List 1 and List 2 contains four entries each. Entries of List 1 are to be matched with some entries of List 2. One or more than one entries of List 1
