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Question

Calculate [H+] from a CH3COOH in a solution which is 101 M in HCl and 103 M in CH3COOH;  [Ka=2×105]
  1. 2×104
  2. 4×107
  3. 2×107
  4. None of the above


Solution

The correct option is C 2×107
               CH3COOHCH3COO+H+At equ.  C(1α)               Cα               (103+Cα)
H+ ion can be considered completely from HCl, due to less dissociation of CH3COOH (because of common ion effect by H+ of HCl,α<<1) and its low concentration, So, [H+]=101M  pH=log [H+]=
pH=log 101=1
From above equilibrium,
K=[CH3COO][H+][CH3COOH]2×105=Cα×101C
α=2×104
[H+]from CH3COOH=Cα=103×2×104
[H+]=2×107M

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