13. NEED HELP FAST

Question

Calculate the pH of a  0.123 M  solution of  NaNO2.  The ionization constant,  𝐾a,  for the acid,  HNO2,  is  4.60×10−4.

pH=

Answer 1

NO₂^- acts as the conjugate base of HNO₂. Thus, from the Ka for HNO₂ we can find the Kb for NO₂^-

KaKb = Kw = 1x10^-14

Kb = 1x10^-14/4.6x10^-4 = 2.17x10^-11

NO₂^- + H₂O ==> HNO₂ + OH^-

Kb = [HNO₂][OH^-]/[NO₂^-]

2.17x10^-11 = (x)(x)/0.123

x² = 2.67x10^-12

x = 1.63x10^-6 = [OH^-]

pOH = -log 1.63x10^-6 = 5.79

pH = 14 - pOH = 8.21