Consider the titration of 50.0 mL of 0.20 M NH3 (Kb=1.8×10−5) with 0.20 M HNO3. Calculate the pH after addition of 50.0 mL of the titrant at 25 ∘C?

I got #"pH" = 4.98#. This agrees with the fact that strong acid + weak base = acidic pH.

If you notice, the volume and concentration of both #"NH"_3# and #"HNO"_3# are identical, and #"HNO"_3# only has one #"H"^+#. Therefore, you should convince yourself that these neutralize each other exactly because the number of mols of each are equal.

This leaves the same number of mols of #"NH"_4^+# as #"NH"_3#, in TWICE the volume (i.e. don't forget to account for dilution!):

#["NH"_4^+]_i = "0.20 mol"/"L" xx ("50.0 mL")/("50.0 mL" + "50.0 mL")#

#=# #"0.10 M"#

This #"NH"_4^+# then dissociates in water again, so at #25^@ "C"#, the ACID dissociation constant becomes;

#K_a = K_w/K_b = 10^(-14)/(1.8 xx 10^(-5)) = 5.56 xx 10^(-10)#

And this dissociation is:

#"NH"_4^(+)(aq) + "H"_2"O"(l) rightleftharpoons "NH"_3(aq) + "H"_3"O"^(+)(aq)#

#"I"" "0.20" "" "" "-" "" "" "0.00" "" "" "0.00#
#"C"" "-x" "" "" "-" "" "" "+x" "" "" "+x#
#"E"" "0.20-x" "-" "" "" "" "x" "" "" "" "x#

Therefore:

#5.56 xx 10^(-10) = x^2/(0.20 - x)#

#"NH"_3# is already a weak base, and #"NH"_4^(+)# is quite a weak acid. We can easily justify that #x# is small compared to #0.20#.

#5.56 xx 10^(-10) = x^2/0.20#

#=> x = sqrt(0.20 cdot 5.56 xx 10^(-10))#

#= 1.05 xx 10^(-5) "M"#

As a result,

#color(blue)("pH") = -log(1.05 xx 10^(-5)) = color(blue)(4.98)#