HNO2 Example from Lecture

[Gabriel Wolf 3I]

In lecture last week we covered adding .1M nitrous acid and .15M potassium nitrate to a solution, and were asked to find the pH given a KA of 4.3*10^-4. I was wondering how we know that the HNO2 will decrease while the NO2 increases, and not the other way around.

[Jennifer Huynh 3I]

To use the specific Ka given, HNO2 was considered a reactant, and NO2- and H3O+ were products. Thus, as NO2- products were being made (+x), HNO2 reactants were being used up (-x). It is possible to consider the NO2- and H3O+ as reactants to make HNO2 (product), but then the Ka would have to be 1/(the given Ka in the problem) because the reaction is reversed, so we take the inverse of Ka.

HNO2(aq) + H2O(l) ⇌ NO2-(aq) + H3O+(aq)
original Ka = [H3O+][NO2-]/[HNO2]

NO2-(aq) + H3O+(aq) ⇌ HNO2(aq) + H2O(l)
new Ka = [HNO2]/[H3O+][NO2-] = 1/original Ka

[Jason Ho 2L]

I think the one way you can tell is because the Ka given is the forward reaction, so you are expecting HNO3 to decrease and NO2- to increase. Also, we are told that it is a weak acid, so we are expecting it to deprotonate some despite having its salt present.

[Michelle Zhang 1J]

I don't think you have to know which of NO2 or HNO2 increases or decreases to solve the problem. If you do end up mixing the signs up when writing the change in concentration in your ICE box, x will turn out negative, and the final concentrations will still work out.