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)
