What is the value of the dissociation equilibrium constant K of water (considered as an acid) at 25oC :
a) K = [ H2O ] = 55.5 M b) K = Ke = 1.0 x 10-14 c) K = Ke * [ H2O ] = 5.6 x 10-13 d) K = Ke / [ H2O ] = 1,8 x 10-16
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d) is correct , $K$ $ =$ $\frac{K_e}{[H_2O]}$ $=$ $1,8\cdot 10^{-16}$