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Question

The number of S2 ions present in 1 L of 0.1 M H2S  [Ka(H2S)=1021] solution having [H+]=0.1 M is:  
  1. 6.023×103
  2. 6.023×104
  3. 6.023×105
  4. 6.023×106


Solution

The correct option is A 6.023×103
Given reaction is,
H2S2H++S2
Ka=[H+]2[S2][H2S]
1021=(0.1)2×[S2]0.1
[S2]=1020 M
So, total number of ions = 1020×6.023×1023=6.023×103 ions

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