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)`