Question #76220

1 Answer
May 28, 2017

Answer:

#sf(pH=2.36)#

Explanation:

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)#