What is the pH of 0.15 M aqueous nitrite ion? (Kb of NO2– = 1.7 ×...

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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)

PLEASE SHOW ALL WORK and GIVE ANSWERS WITH THE CORRECT NUMBER OF SIGNIFICANT FIGURES AND UNITS :) 2. (10 pts) What is the pH of 0.15 M aqueous nitrite ion? (Kh of NO2 = 1.7 x 10) NO2 (aq) + H2O(1) = HNO2(aq) + OH(aq)

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Answer 1

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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

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Answer 2

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