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K1 and K2 for oxalic acid C2H2O4 are 6.5×102 and 6.1×105 respectively. What will be [OH] in a 0.01 M solution of sodium oxalate?
  1. 9.6×1016
  2. 1.4×101
  3. 1.26×106
  4. 3.3×109


Solution

The correct option is C 1.26×106
Concentration of oxalate ion= 0.01 M
The hydrolysis of C2O24 is represented by:
C2O24+H2OHC2O4+OHKh=[HC2O4][OH][C2O4]2Also,HC2O4H++C2O24K2=[C2O24][H+][HC2O4]Kw=[H+][OH] We see that, Kw=Kh×K2Kh=KwK2=1.0×10146.1×105=1.6×1010At equilibrium, [HC2O4]=[OH] Kh=[OH]2[C2O24]=1.6×1010[C2O4]=0.01 M[OH]=1.6×1012=16×1012104×10610=1.26×106

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