In: Chemistry
A 0.100 M solution of nitrous acid has a PH ==2.10 . Determine the value of the ionization constant for nitrous acid . Determine the present ionization for this acid.
PH = 2.1
-log[H^+] = 2.1
[H^+] = 10^-2.1 = 0.00794M
at equilibrium [H^+] = [NO2^-] = 0.00794M
HNO2(aq) -------------------> H^+ (aq) + NO2^- (aq)
Ka = [H^+][NO2^-]/[HNO2]
= 0.00794*0.00794/0.1
= 0.00063 = 6.3*10^-4
percent ionization of acid = final conc of H^+ *1000/initial conc of acid
= 0.00794*100/0.1 = 7.94% >>>>answer