(A)
A quick way of finding the pH of a weak acid is to use the expression:
sf(pH=1/2(pK_a-loga))
sf(pK_a=-logK_a=-log(3xx10^(-8))=7.523)
sf(a) is the concentration of the acid.
:.sf(pH=1/2[7.523-(-1)]=4.26)
(B)
You need to do the volume first:
The equation is:
sf(HClO+KOHrarrKClO+H_2O)
sf(n_(HClO)=cxxv=0.1000xx100/1000=0.01)
Since they react in a 1:1 molar ratio, the number of moles of base must be the same:
sf(n_(KOH)=0.01)
sf(c=n/v)
:.sf(v=n/c=0.01/0.01=1color(white)(x)L)
This is a bad question as you would never set up a titration which requires 1L for an end-point.
sf(ClO^-) is the co - base and is hydrolysed by water:
sf(ClO^(-)+H_2OrightleftharpoonsHClO+OH^-)
sf(K_(b)=([HClO][OH^(-)])/([ClO^(-)])
These are equilibrium concentrations.
We can use this expression to find sf([OH^-]) hence sf([H^+]) and the pH.
We can find sf(K_b) using the expression:
sf(K_bxxK_a=K_w=10^(-14)color(white)(x)"mol"^2."l"^(-2)) at sf(25^@C)
:.sf(K_b=10^(-14)/(3.0xx10^(-8))=0.333xx10^(-6)color(white)(x)"mol/l")
Now we can set up an ICE table based on concentrations in mol/l:
The total volume = 100 ml + 1000 ml = 1.1 L
:. sf([ClO^-]=n/v=0.01/1.1=0.00909color(white)(x)"mol/l")
" "sf(ClO^(-)" "+" "H_2O" "rightleftharpoons" "HClO" "+" "OH^-)
sf(color(red)(I)" "0.00909" "0" "0)
sf(color(red)(C)" "-x" "+x" "+x)
sf(color(red)(E)" "(0.00909-x)" "x" "x)
:.sf(K_b=x^2/((0.00909-x))=0.333xx10^(-6)color(white)(x)"mol/l")
Because the dissociation is so small I will make the assumption that sf((0.00909-x)rArr0.00909).
:.sf(x^2=0.333xx10^(-6)xx0.00909=0.0032xx10^(-6))
:.sf(x=sqrt(0.0032xx10^(-6))=5.5xx10^(-5)color(white)(x)"mol/l")
This is equal to sf([OH^-]).
The ionic product of water is given by:
sf(K_w=[H^+][OH^-]=10^(-14)color(white)(x)"mol"^2."l"^-2) at sf(25^@C)
:.sf([H^+]=10^(-14)/[[OH^-]]=10^(-14)/(5.5xx10^(-5))=1.818xx10^(-10)color(white)(x)"mol/l")
sf(pH=-log[H^+]=-log[1.818xx10^(-10)]=9.7)
This is the salt of a weak acid and a strong base so you can see that, at the equivalence point, the solution is slightly alkaline.
