What is the pH of 0.15 M aqueous nitrite ion? (Kb of NO2– = 1.7 × 10–11) NO2–(aq) + H2O(l) HNO2(aq) + OH-(aq)
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Answer:-
This question is solved by using simple concept of hydrolysis of salt of weak acid and then determination of pH.
The answer is given in the image,
![Answer: CNO2-J = 0.15 Kb = lo 7x10-11 Nog Cont 490 LUSZ HNO2009) + Onions Kb = CHNO3) (04-) [Mon] 107x10-11 = 2. 10.15 x2 = 2](http://img.homeworklib.com/questions/2f8394c0-d5f1-11eb-96fa-b1240d18c349.png?x-oss-process=image/resize,w_560)
What is the pH of 0.15 M aqueous nitrite ion? (Kb of NO2– = 1.7 ×...
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Question 14
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