#1.
C6H5COOH ==> C6H5COO^- + H^+ k1 = above.
#2. Reverse HNO2 ionization which is k2 = above.
NO2^- + H^+ ==> HNO2 k' = 1/k2
Add #1 to #2 and you obtain
C6H5COOH + NO2^- ==> C6H5COO^- + NO2^- and Keq for that reaction is k1*k'
C6H5COOH(aq) + NO2-(aq) <=> C6H5COO-(aq) + HNO2(aq)
A table of ionizations is given to calculate the equilibrium constant for this overall reaction.. but when i looked onto the table the only thing given was the Ka for C6H5COOH [6.3×10^–5] and HNO2 [5.6×10^–4]
it doesn't show me the Ka for NO2- or C6H5COO- ... is there a way I can solve K without knowing the Ka for those last two..? please help! it'll be greatly appreciated!
2 answers
Thanks DrBob!