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Question 7 1 pts A 100 mL sample of 0.25 M CH3NH2(aq) is titrated with a 100 ml. of 0.25 M HNO3(aq). Select ALL main componen
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page-1 solution: Given * data- -> m moles of CH3NH2 = 0.25 X 100 = 25 1 - m moles of HNO3 = 0.25 X 100 = 25 - Kb = 4,48104 =)page-2 >> Salt concentration = - total volume 25 (100+600) = 0.125 M ::PH = 7 - Ź (PKb + log c) =) PH = 7-Ź (3.36 + dog 0.125

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