(a)
Hypobromous acid is a weak acid and dissociates:
#sf(HBrOrightleftharpoonsH^(+)+OBr^(-))#
For which:
#sf(K_a=([H^+][OBr^(-)])/([HOBr])=2.5xx10^(-9)color(white)(x)"mol/l"" "color(red)((1)))#
Please note that these refer to equilibrium concentrations and not initial concentrations.
The initial number of moles of #sf(HOBr)# is given by:
#sf(n_(HOBr)=cxxv=0.025xx2=0.050)#
Since #sf(pH=9)# then #sf([H^+]=10^(-pH)=10^(-9)color(white)(x)"mol/l")#
Rearranging #sf(color(red)((1))# gives:
#sf([H^(+)]=K_axx([HBrO])/([BrO^(-)])" "color(red)((2))#
At this point I will make the important assumption that, because the value of #sf(K_a)# is so small, then the equilibrium concentrations are a good approximation to the initial concentrations.
The fact that you are asked to assume that the volume change is negligible is actually irrelevant. Since #sf([color(white)(x)]=n/v)# you can see that the volume is common to both acid and co - base so will cancel in #sf(color(red)((2))#.
This means we can write out #sf(color(red)((2))# using moles#sf(rArr#
#:.##sf(10^(-9)=2.5xx10^(-9)xx(0.050)/(nBrO^(-))#
#:.##sf(nBrO^(-)=(2.5xxcancel(10^(-9))xx0.050)/(cancel(10^(-9)))=0.125)#
(b)
From #sf(color(red)((2))# you can see that the pH of the buffer depends on the ratio of acid to co - base concentrations, whereas the amount of added #sf(H^+)# or #sf(OH^-# which the buffer can cope with will depend on the individual concentrations of acid and co - base.
This is referred to as the buffer capacity.
In (b) to maintain a pH of 9 the concentration of #sf(HOBr)# and of #sf(OBr^(-))# must have both been increased, so we would expect a greater buffer capacity.