What happens when KNO2(s) is dissolved into the solution?
What occurs when solid potassium nitrite (KNO2) is dissolved in a solution that contains 1 M of the weak acid nitrous acid (HNO2), which is 2% ionized and has a pH of 1.7? Specifically, how do the following variables change upon the dissolution of KNO2:
- The extent of HNO2 ionization [increases / decreases]
- The concentration of H+ ions [increases / decreases]
- The pH of the solution [increases / decreases]
2 Answers
Use LeChatelier’s principle to help you think this problem out:
HNO2 <—-> H+ + NO2-
Since the acid is only 2% ionized in solution, we know that the equilibrium is strongly favored to the reactant.
If you stress a system, it will shift to relieve that stress, so adding more NO2- will cause more HNO2 to be produced, thus reducing the H+ conc in solution.
So,
Ionization decreases,
H+ conc. decreases,
pH increases (it becomes more basic due to less H+)
If you Add KNO2 then you’re getting more NO2 in the reaction HNO2 —> H +NO2
So the right side of the equation has more than the left, and due to Le’Chatellier’s principle, the reaction would naturally want to shift to the left. Therefore, Less H+ is made.
This is the best explanation you will get ?
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