The ammonium ion ionises:
#sf(NH_4^+rightleftharpoonsNH_3+H^+)#
For which:
#sf(K_a=([NH_3][H^+])/([NH_4^+])=5.6xx10^(-10)color(white)(x)"mol/l")#
When #sf(OH^-)# ions are added the following takes place:
#sf(NH_4^++OH^(-)rarrNH_3+H_2O)#
The no. of moles added is given by:
#sf(n_(OH^-)=cxxv=0.75xx20.0/1000=0.015)#
This means, from the equation, that the number of moles of #sf(NH_4^+)# consumed = 0.015
#:.# The number of moles remaining is given by:
#sf(n_(NH_4^+)=0.025-0.015=0.01)#
From the equation you can see that the number of moles of #sf(NH_3)# formed = 0.015.
The total moles of #sf(NH_3)# is given by:
#sf(n_(NH_3)=0.04+0.015=0.055)#
The small value of #sf(K_a)# means we can assume that these are the moles present at equilibrium.
Rearranging the expression for #sf(K_a)# we get:
#sf([H^+]=K_axx([NH_4^+])/([NH_3]))#
Since the total volume is common to both salt and base we can use moles directly:
#sf([H^+]=5.6xx10^(-10)xx0.01/0.055color(white)(x)"mol/l")#
#sf([H^+]=1.018xx10^(-10)color(white)(x)"mol/l")#
#sf(pH=-log[H^+]=-log[1.018xx10^(-10)])#
#sf(pH=9.99)#
