Consider the titration of `50.0 mL` of `0.20 M` `NH_3` Kb `1.8x10^-5` with `0.20 M` `HNO_3`. How do you calculate the pH after addition of 50.0mL of the titrant?

The equation for the neutralisation is:

`sf(NH_3+H^+rarrNH_4^+)`

We can find the number of moles using `sf(n=cxxv)`:

`sf(n_(NH_3)=0.20xx50.0/1000=0.01)`

`sf(n_(H^+)=0.20xx50.0/1000=0.01)`

This tells us that:

`sf(n_(NH_4^+)=0.01)`

`sf(c=n/v)`

The total volume is 50.0 + 50.0 = 100.0 ml

`:.``sf([NH_4^+]=(0.01)/(100.0/1000)=0.10color(white)(x)"mol/l")`

The ammonium ion partially dissociates:

`sf(NH_4^+rightleftharpoonsNH_3+H^+)`

For which:

`sf(K_a=([NH_3][H^+])/([NH_4^+])`

To find the `sf(pH)` we need to know the value of `sf(K_a)`. We can get this by using the expression:

`sf(K_axxK_b=K_w=10^(-14)color(white)(x)"mol"^2."l"^-2)` at `sf(25^@C)`

`:.``sf(K_a=10^(-14)/(1.8xx10^(-5))=5.55xx10^(-10)color(white)(x)"mol/l")`

Now we set up an ICE table based on concentrations:

`sf(" "NH_4^+" "rightleftharpoons" "NH_3" "+" "H^+)`

`sf(color(red)(I)" "0.10" "0" "0")`

`sf(color(red)(C)" "-x" "+x" "+x)`

`sf(color(red)(E)" "(0.1-x)" "x" "x)`

`:.``sf(K_a=x^2/(0.1-x))`

Because the value of `sf(K_a)` is so small we can assume that the size of `sf(x)` is negligible compared with `sf(0.1)` so we can write:

`sf(K_a=x^2/0.1=5.55xx10^(-10))`

`:.``sf(x^2=5.55xx10^(-10)xx0.1=5.55xx10^(-11))`

`:.``sf(x=[H^+]=sqrt(5.55xx10^(-11))=7.449xx10^(-6)color(white)(x)"mol/l")`

`sf(pH=-log[H^+]=-log[7.449xx10^(-6)]=5.12)`

`sf(color(red)(pH=5.12))`

This shows that the pH is slightly acidic at the equivalence point which is typical of a salt produced from a strong acid and a weak base.