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)rArrc.
:.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)
