Question

Questions are based off the following information: An aqueous solution 0.010 M NaNO2 solution at 25 degrees C. Ka for HNO2 is 7.1*10^-4

a) Write down three acid-base reactions that are occurring AND identify the conjugate acid-base pairs

b) Write equilibrium constant expression for all the reactions in part a and give the constant for each/

c) In addition to water, there are five other species in the reaction, identify them, and rank then from highest to lowest (in concentration).

d) What is the pH of this solution?

Answer 1

a.

NaNO2 is a salt of strong base NaOH and weak acid HNO2. Hence, it dissociates to form Na+ and NO2-. Note that Na+ being the cation of a strong base acts as a spectator ion.

Hence, the first acid-base reaction we can write is the protonation of NO2- by water.

Here, is the base and is its conugate acid. Similarly, H2O acts as the acid that protonates and OH- is its conjugate base.

The second acid-base reaction we can write is the deprotonation of the HNO2 formed

Here, HNO2 is the acid and NO2- is its conjugate base. Similalrly, HNO2 protonates H2O which is the base and H3O+ is the conjugate acid of H2O.

The third acid base reaction we can write is the autoprotolysis of water.

Here H2O acts as both acid and base and forms the conjguate acid and base which are H3O+ and OH- respectively.

b)

The equilibrium expression of a generic reaction is expressed as the ratio of concentrations of products raised to their stoichiometric coefficients divided by the concentrations of reactants raised to the power of their stoichiometric coefficients.

For a generic reaction

The equilibrium constant K is given as

For the second acid base reaction:

Its value is given in the question as .

Here also concentration of water do not appear in the denominator.

For the third acid-base reaction

For an aqueous solution at 25 C, .

Note that the denominator is 1 as the equation has pure water as the reactant.

For our first acid base reaction:

Important thing to note is that the concentration of water does not appear in the denominator as it is a pure solvent and its concentration is taken as unity.

Looking at the equation for Kb, Kw and Ka, if we compute  , we get the following

Hence, we can calculate the value of Kb as

c)

From the above equations, the five species other than water are

.

The starting concentration of our NaNO2 solution is 0.010 M.

Since it dissociates completely, the concentration of Na+ in the solution is

Note that concentration of Na+ does not change after equilibrium is established since it acts as a spectator ion.

Now, the initial concentration of NO2- in the solution is also 0.010 M.

It will react with water to form HNO2 as described by our first acid base reaction.

We can construct a an ICE table based on acid base reaction of NO2- with H2O

Initial, M	0.010	0	0
Change, M	-x	+x	+x
Equilibrium, M	0.010-x	x	x

Hence, from the ICE chart, we can set up the following Kb equation

Hence, the equilibrium concentration of different species are

Now, using the Kw equation, we can calculate the H3O+ concentration

Hence, the order of concentration of species from highest to lowest is

Note that Na+ concentration is greater than NO2- as some of the NO2- has formed HNO2.

d)

pH of the solution is the negative logarithm of the H3O+ concentration

Hence,