The #sf(HSO_3^-)# ion dissociates:
#sf(HSO_3^(-)rightleftharpoonsSO_3^(2-)+H^+)#
#sf(K_(a2)=([SO_3^(2-)][H^+])/([HSO_3^(2-)])=6.6xx10^(-8))#
These are equilibrium concentrations. Rearranging gives:
#sf([H^+]=K_(a2)xx([HSO_3^-])/([SO_3^(2-)])#
#sf(pH=6.83)# from which #sf([H^+]=1.48xx10^(-7)color(white)(x)"mol/l")#
#:.##sf(1.48xx10^(-7)=6.6xx10^(-8)xx([HSO_3^-])/([SO_3^(2-)])#
#sf({[HSO_3^-])/([SO_3^(2-)])=(1.48xx10^(-7))/(6.6xx10^(-8))=2.242)#
This is the ratio of acid to co - base we need to achieve.
Because of the small value of #sf(K_(a2))# we will assume that the initial moles approximate to the equilibrium moles.
#sf(c=n/v)#
#:.##sf(n=cxxv)#
#:.# the no. of moles of #sf(OH^(-))# added is given by:
#sf(n_(OH^-)=3.81xxV)#
where V is the volume which we are asked to find.
The #sf(OH^-)# ions are neutralised as follows:
#sf(HSO_3^(-)+OH^(-)rarrSO_3^(2-)+H_2O)#
The initial moles of #sf(HSO_3^-)# is given by:
#sf(n_(HSO_3^(-))=0.655xx67.2/1000=0.044016)#
The equation tells us that they react with #sf(OH^-)# in a 1:1 molar ratio so the no. moles of #sf(HSO_3^-)# remaining is given by:
#sf(0.044016-3.81V)#
The equation also tells us that the no. of moles of #sf(SO_3^(2-))# formed will be the same as the no. of moles of #sf(OH^-)# added.
#:.##sf(n_(SO_3^(2-))=3.81V)#
Since the total volume is common to both acid and co - base we can use moles in the expression for their ratio:
#sf((n_(HSO_3^-))/(n_(SO_3^(2-)))=(0.044016-3.81V)/(3.81V)=2.242)#
#:.##sf(0.044016-3.81V=2.241xx3.81V)#
#sf(V=0.044016/12.352=0.00356color(white)(x)L)#
#sf(V=3.56color(white)(x)ml)#
