C6H5OH + H2O ==> H3O^+ + C6H5O^-
Ka = (H3O^+)(C6H5O^-)/(C6H5OH)
Plug in values for Ka, (H3O^+), (C6H5O^-), solve for (C6H5OH).
Technically, (C6H5OH) = x-(H3O^+) but (H3O^+) is small in comparison to x that this need not be considered.
Phenol C6H5OH partially dissociates in water as shown
C6H5OH + H2O --> H3O+ + C6H5O-
If Ka is 1.6*10^-10 and the concentration of H3O and C6H5O are both 1.0*10^-5 M at equilibrium, what would be the concentration of phenol?
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