Ethanoic acid is a weak acid and dissociates:
#sf(CH_3COOHrightleftharpoonsCH_3COO^(-)+H^+)#
For which #sf(K_a=([CH_3COO^(-)][H^(+)])/([CH_3COOH])=1.8xx10^(-5))#
These are equilibrium concentrations.
The degree of dissociation #sf(alpha)# is the fraction of the amount of electrolyte which dissociates. If C is the initial concentration of the acid then we can write an ICE table based on #sf("mol/l")#:
#sf(color(white)(xx)CH_3COOHcolor(white)(xxxx)rightleftharpoonscolor(white)(xx)CH_3COO^(-)color(white)(xxx)+color(white)(xxxx)H^(+))#
#sf(Icolor(white)(xxxxx)Ccolor(white)(xxxxxxxxxxxxxx)0color(white)(xxxxxxxxxxxxx)0)#
#sf(Ccolor(white)(xx)-Calphacolor(white)(xxxxxxxxxxx)+Calphacolor(white)(xxxxxxxxxxx)+Calpha)#
#sf(Ecolor(white)(xx)C-Calphacolor(white)(xxxxxxxxxxxx)Calphacolor(white)(xxxxxxxxxxxx)Calpha)#
#:.##sf(K_a=(C^2alpha^2)/((C-Calpha))=(C^(2)alpha^2)/(C(1-alpha))=(Calpha^2)/((1-alpha))#
Because the dissociation is small i.e #sf(10^(-9)< K_a<10^(-4))# then we can assume that #sf((1-alpha)rArr1)#.
#:.##sf(K_a=Calpha^2)#
#sf(alpha=sqrt(K_a/C))#
#sf(alpha=sqrt((1.8xx10^(-5))/(0.2))=9.486xx10^(-3))#
Now we double this value to get #sf(alpha^*)#:
#sf(alpha^*=2xx9.486xx10^(-3)=18.97xx10^(-3))#
To get the new concentration we get:
#sf(18.97xx10^(-3)=sqrt((1.8xx10^(-5))/C^*))#
#sf(360xx10^(-6)=(1.8xx10^(-5))/C^*)#
#sf(C^*=(1.8xx10^(-5))/(360xx10^(-6))=0.05color(white)(x)"mol/l")#
This is the new concentration we need. To find how much water we need to add to achieve this dilution we can say:
#sf(C_1V_1=C_2V_2)#
#:.##sf(300xx0.2=0.05xxV_2)#
#sf(V_2=(300xx0.2)/(0.05)=1200color(white)(x)ml)#
This means we need to add a further 1200 - 300 = 900 ml. Which gives option 1.
