#sf((1)).#
The auto - ionisation of water gives us:
#sf(pH+pOH=14)# at #sf(25^@C)#
From this we get:
#sf(pH=14-pOH=14-11=3)#
This means that #sf([H^+]=10^(-pH)=10^(-3)color(white)(x)"mol/l")#
#sf(HA)# dissociates:
#sf(HArightleftharpoonsH^(+)+A^(-))#
For which:
#sf(K_a=([H^+][A^(-)])/([HA])=10^(-5)color(white)(x)"mol/l")#
The concentrations refer to those at equilibrium.
To find #sf([HA])# I will set up an ICE table based on concentrations. Let #sf(C)# be the initial concentration of #sf(HA):#
#sf(color(white)(xxxx)HAcolor(white)(xxxx)rightleftharpoonscolor(white)(xxxx)H^+color(white)(xxx)+color(white)(xxxx)A^-)#
#sf(color(red)(I)color(white)(xxxx)Ccolor(white)(xxxxxxxxxxxx)0color(white)(xxxxxxxxxx)0)#
#sf(color(red)(C)color(white)(xx)-xcolor(white)(xxxxxxxxxx)+xcolor(white)(xxxxxxxx)+x)#
#sf(color(red)(E)color(white)(xx)(C-x)color(white)(xxxxxxxxxx)xcolor(white)(xxxxxxxxxx)x)#
From the expression for #sf(K_a)# I can write:
#sf(K_a=(x^2)/((C-x))=10^(-5)color(white)(x)"mol/l")#
I have already found the value of #sf(x)# which = #sf([H^+]=0.001color(white)(x)"mol/l")#.
Putting in that value#sf(rArr)#
#sf((0.001^2)/((C-0.001))=10^(-5))#
With a small value of #sf(K_a)# like this it is common to assume that #sf(x)# is much smaller than #sf(C)# such that #sf(C-x)# approximates to #sf(C)#.
However, in this case I don't know #sf(C)# so I won't make that assumption.
So:
#sf((10^(-6))/((C-10^(-3))##sf(=10^-5)#
#:.##sf(10^(-5)(C-10^(-3))=10^(-6))#
#:.##sf((C-10^(-3))=10^(-6)/10^(-5)=0.1)#
#:.##sf(C=0.1+0.001=0.1001color(white)(x)"mol/l")#
#sf(C=[HA]=0.1color(white)(x)"mol/l"" "(1"sig.fig")#
#sf((2)).#
Since we are told that:
#sf(pOH=11)#
This means that:
#sf(-log[OH^-]=11)#
This gives:
#sf([OH^-]=10^(-11)color(white)(x)"mol/l")#
