a.) Write the reaction for what happens when KNO2 in water that causes the pH to change.
For this qn:
should we come up with something like pH = pKₐ + log([A⁻]/[HA]).
KNO2+H2O--> KOH+HNO2
NO2-(aq) + H2O(l) ⇌ HNO2(aq) + OH-(aq)
Kb=[HNO2][OH-]/[NO2-]
[OH-] = KA x [NO2-]
pOH = -log[OH-]
pH = 14 - pOH
Please verify if the above is correct and advise.


b.) What information do you need to solve for Kb of NO2- and how would you solve it?
To find the Kb of the NO2- ion, we need to use the relationship: Kw = Ka x Kb where Kw is the ion product constant of water (1.0 x 10^-14 at 25?C). Rearranging this equation to solve for Kb, we get: Kb = Kw / Ka

Quote from: prasamb@yahoo.com on April 13, 2023, 11:17:16 PM
a.) Write the reaction for what happens when KNO2 in water that causes the pH to change.
For this qn:
should we come up with something like pH = pKₐ + log([A⁻]/[HA]).
KNO2+H2O--> KOH+HNO2
NO2-(aq) + H2O(l) ⇌ HNO2(aq) + OH-(aq)
Kb=[HNO2][OH-]/[NO2-]
[OH-] = KA x [NO2-]
pOH = -log[OH-]
pH = 14 - pOH
Please verify if the above is correct and advise.


b.) What information do you need to solve for Kb of NO2- and how would you solve it?
To find the Kb of the NO2- ion, we need to use the relationship: Kw = Ka x Kb where Kw is the ion product constant of water (1.0 x 10^-14 at 25?C). Rearranging this equation to solve for Kb, we get: Kb = Kw / Ka

First answer a) is wrong.
To know pH of KNO₂ solution we need to understand that NO₂^- is the conjugate base of weak acid and it is stronger base than water.
So it will react with water as base and this is called as hydrolysis of salt.
NO₂^- + H₂O <---> OH^-  + HNO₂
Now we can see during hydrolysis of NO₂^- ,OH^- are generated in water which makes pH more than 7.
You can calculate pOH from the [OH-] and convert it into pH.
We need to make RICE table to know the concentration of OH-.NO₂^- is a base stronger than water but still a weak base so RICE table is needed.
Answer b is correct.

Thanks,Uma.

BTW  when we write the below equation:
So it will react with water as base and this is called as hydrolysis of salt.
NO2- + H2O <---> OH-  + HNO2

before this we have the equation: KNO2+H2O--> KOH+HNO2
My Qn: Will we have double arrow or single arrow this above equation. How to decide when to give double arrow. Though equilibrium helps us to decide on that, is there a quick way to get hold this (double VS single arrow)

Thanks
Prasanna

When we have a weak acid or base then we need reversible arrow.NO₂^- or KNO₂ is a salt of weak conjugate base which is just stronger than water. In strong acids and bases it is single headed arrow because it is fully dissociated.