moles CH_(3)COOH=0.04 molesvolume of solution =131ml=0.131 lmolarity of CH_(3)coOH=(0.04" moles ")/(0.232" lifers ")Initial conc =0.3053mCH_(3)COOH+H_(2)O⇌H_(3)O^(+)+CH_(3)COOthetaI 0.3053c -x quad+x quad+xE 0.3053-x quad x quad xka=([n_(3)0^(+)]["nn "_(3)coc])/([" cnscoon "])=(x^(2))/(0.30 ... See the full answer