Question

The nitrite ion, a weak base, establishes the following equilibrium in water:


NO2–(aq) + H2O(l) qwwe HNO2(aq) + OH–(aq)


Calculate the [OH – ] in a 1.75 M solution of NO2–. What is the pH of the solution?

Answer 1

The nitrite ion is a weak base so its conjugate acid will be strong acid which means 100 dissociation will take place

pH=-log[H⁺]=-log[1.75]= -0.24

$[H^+]\times[OH^-]=10^{-14}$

$[OH^-]=10^{-14}/1.75=5.7\times10^{-15}$