The value of Ka for nitrous acid (HNO2) at 25 ∘C is 4.5×10−4.
a) Write the chemical equation for the equilibrium that corresponds to Ka. answer: HNO2(aq)⇌H+(aq)+NO−2(aq)
b) By using the value of Ka, calculate ΔG∘ for the dissociation of nitrous acid in aqueous solution.
______ kJ
c) What is the value of ΔG at equilibrium?
______ kJ
d) What is the value of ΔG when [H+] = 5.1×10−2 M , [NO−2] = 6.3×10−4 M , and [HNO2] = 0.21 M ?
______ kJ