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

bonnie_schmitz_1F
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### Calculating pH from weak base and its salt (ex. from class)

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?

Kathryn 1F
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Joined: Fri Sep 28, 2018 12:19 am

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

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.

305113590
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Joined: Fri Sep 28, 2018 12:28 am

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

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