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### Calculating pH from weak base and its salt (ex. from class)

Posted: Sun Jan 20, 2019 4:05 pm
In class we were given a problem that said:
What is the pH of a solution with 0.100 M HNO2 and 0.150 M KNO2?
The equation we used was to describe this reaction was
HNO2 + H2O --> H3O+ + NO2-

I understand that K + is not included in the equation because it is on both sides, however, why wouldn't there be an NO2- on the left side from the KNO2?

### Re: Calculating pH from weak base and its salt (ex. from class)

Posted: Sun Jan 20, 2019 4:10 pm
the equation is just the reaction of hno2 with water. We could use a separate one to describe the salt dissociating, but we can just infer that no2- will be there from the salt, too. Thats why for the "I" in the ICE table we put .150M NO2-, instead of 0 like we had been doing previously for reactions without salts.

### Re: Calculating pH from weak base and its salt (ex. from class)

Posted: Sun Jan 20, 2019 4:16 pm
We use the salt, KNO2, to find the pH of nitrous acid.

1. Dissociate the 0.150M of KNO2, which are K+ and NO2-. Since it's a salt, we can use stoichiometry to find the molarity of NO2-. That will be NO2-. You use that to find the pH of nitrous acid.
2. Since HNO2 (nitrous acid) is weak, it'll undergo equilibrium. HNO2 <---> NO2- +H30+
3. Since we figured out the NO2- and you were given HNO2 concentration, you use those values in an ICE box. This will eventually look like (x)(0.150+x)/(0.100-x)=4.3x10^-4.
4. Since the Ka value is less than 10^-3, we can assume that x is a negligible value to add/subtract. (x)(0.150)/(0.100)=4.3x10^-4
5. You have your x value, 2.866x10^-4. Plug that in pH=-log[H30+] and you'll get 3.54 as your pH