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Calculate the [OH-] in a solution where 0.050 mole CH3NH3Cl is added to 200.0 mL of 0.50 M in CH3NH2. CH3NH₂ + H₂O = CH3NH + OH- KbCH3NH₂ = 4.38 x 10-4 [OH-] = [?] x 10¹²) M Coefficient (green) -4 Exponent (yellow) Enter

Calculate the [OH-] in a solution where 0.050 mole CH3NH3Cl is added to 200.0 mL of 0.50 M in CH3NH2. CH3NH₂ + H₂O = CH3NH + OH- KbCH3NH₂ = 4.38 x 10-4 [OH-] = [?] x 10¹²) M Coefficient (green) -4 Exponent (yellow) Enter

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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So my answers, 5.0(10^-4) and 1.752 E -4, are wrong. What is an alternative solution? (The exponent -4 is correct, just not the coefficient).
Calculate the [OH-] in a solution where 0.050 mole CH3NH3Cl is added to 200.0 mL of 0.50 M in CH3NH2. CH3NH₂ + H₂O ⇒ CH3NH + OH- Кьснӡин2 = 4.38 x 10–4 [OH-] = [?] x 102) M Coefficient (green) Exponent (yellow) Enter
Transcribed Image Text:Calculate the [OH-] in a solution where 0.050 mole CH3NH3Cl is added to 200.0 mL of 0.50 M in CH3NH2. CH3NH₂ + H₂O ⇒ CH3NH + OH- Кьснӡин2 = 4.38 x 10–4 [OH-] = [?] x 102) M Coefficient (green) Exponent (yellow) Enter
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