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.