At the equivalence point the reaction is complete:
#sf(NH_(3(aq))+HNO_(3(aq))rarrNH_4NO_(3(aq))#
The number of moles of #sf(NH_3)# reacting = #sf(0.3xx300/1000=0.09)#
So the number of moles of #sf(NH_4^+)# formed = #sf(0.09)#
Now we can set up an ICE table:
#" "##sf(NH_4^(+)" "rightleftharpoons" "NH_3" "+" "H^+)#
#sf(color(red)(I) " " 0.09" "0" "0 )#
#sf(color(red)(C)" " -x" "+x" "+x)#
#sf(color(red)(E)" "(0.09-x)" "x" "x)#
#sf(K_a=([NH_3][H^+])/([NH_3])#
We can find #sf(K_a)# using:
#sf(K_w=K_axxK_b)#
#:.##sf(K_a=(10^(-14))/(1.8xx10^(-5))=5.55xx10^(-10)" ""mol/l")# at #sf(25^@C)#
The concentrations are at equilibrium so we can write:
#sf(K_a=(x.x)/((0.09-x))=5.55xx10^(-10))#
Because the value of #sf(K_a)# is so small we can assume that #sf((0.09-x)rArr0.9)#.
#:.##sf((x^2)/0.09=5.55xx10^(-10))#
#:.##sf(x^2=4.99xx10^(-11))#
#sf(x=7.07xx10^(-6)=[H^+]_(eqm))#
#sf(pH=-log[H^+]_(eqm)=-log[7.07xx10^(-6)])#
#sf(pH=5.15)#
This is an example of "salt hydrolysis". When a strong acid and a weak base produce a salt the final product is slightly acidic.