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 valuesf(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")
