0.2 M AcOH is ….% dissociation in 0.1 M HCI (Ka of AcOH = 1.8x 10-5).

AcOH⇌AcO−+H+  (weak acid)     C            0           0 (C-Cx)    Cx        CxHCl⇌Cl−+H+(strong acid)   Total H+=Cx+0.1≈0.1M Ka=AcO−H+[AcOH] 1.8×10−5=Cx×0.1C(1−x)≈0.1x ∴x=1.8×10−4 % dissociation =1.8×10−4×100=0.018%