What is the pH at the equivalence point of the titration of 30 mL of 0.2 M nitrous acid with 0.1 M NaOH at 25 °C?

The concentration of #"0.1 M"# #"NaOH"# is half the concentration of the monoprotic #"HNO"_2#, so twice the volume is needed, i.e. #"60 mL"# of #"NaOH"#. At the equivalence point, only #"NO"_2^(-)# is left.

Hence, the pH must be slightly basic, and this calculation requires that you give the #K_a#... The #K_a# of #"HNO"_2# MIGHT be #4.5 xx 10^(-4)#, but you'll have to verify that.

Thus, at #25^@ "C"#,

#K_b = K_w/K_a = 10^(-14)/(4.5 xx 10^(-4)) = 2.22 xx 10^(-11)#

You should be able to explain why we must use #K_b#, the base association constant, for

#"NO"_2^(-)(aq) + "H"_2"O"(l) rightleftharpoons "HNO"_2(aq) + "OH"^(-)(aq)#

The concentration of #"NO"_2^(-)# was diluted by a factor of

#("30 mL HNO"_2 + "60 mL NaOH")/"30 mL HNO"_2 = 3#,

i.e. #"0.200 M HNO"_2# before mixing #-># #"0.066 M NO"_2^(-)# after mixing.

For this the mass action expression becomes:

#K_b = (["HNO"_2]["OH"^(-)])/(["NO"_2^(-)])#

#= x^2/(0.066 - x)#

We can clearly see that #K_b# is very small, so #x "<<" "0.200 M"#. Therefore,

#2.22 xx 10^(-11) = x^2/0.066#

#=> color(blue)(x = 1.21 xx 10^(6) "M")#

Knowing what #x# represents (which is?), you should then fill in the steps to get that the #"pH"# is #8.08#.