posted by Katherine on .
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!