You should have been provided with the acid dissociation constant #sf(K_a)#.
Vinegar is essentially a solution of ethanoic acid. It dissociates:
#sf(CH_3COOHrightleftharpoonsCH_3COO^(-)+H^+)#
#sf(K_a=([CH_3COO^-][H^+])/([CH_3COOH])#
Looking up #sf(pK_a)# I get 4.76. This =#sf(-logK_a)#
If the concentration of the acid = c we can set up an ICE table in mol/l:
#sf(" "CH_3COOH" "rightleftharpoons" "CH_3COO^(-)+H^+)#
#sf(I" "c" "0" "0)#
#sf(C" "-x" "+x" "+x)#
#sf(E" "(c-x)" "x" "x)#
#:.##sf(K_a=x^2/((c-x))#
Because of the small value of #sf(K_a)# (it is less than #sf(10^-4)#) we can assume that (c-x)#rArr#c.
#:.##sf(K_a=x^2/c)#
#sf(x^2=K_ac)#
#sf(x=sqrt(K_ac)=(K_ac)^(1/2)=[H^+])#
#:.##sf(-log[H^+]=1/2[pK_a-logc]=pH)#
You have calculated c = 1.1 mol/l so:
#sf(pH=1/2[4.76-0.04139])#
#sf(pH=2.36)#
