LaGuardia Community Co x CChegg Study IGuide php?id 2575999 pling Sapling Learni
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LaGuardia Community Co x CChegg Study IGuide php?id 2575999 pling Sapling Learning I La Guardia Community College Scc 202 Spring16 (Session I) DUTTA I Activities and Due Dates I HW7-Chapter 17 My Assig O 5/20/2016 06:00 PM 9.8/10 35/13/2016 08:33 PM Gradebook Attempts score Print Calculator Periodic Table 27 27 University Science Boo General Chemistr Consider an amphoteric hydroxide, M(OH)2(s), where M is a generic metal. K -8 x 10 M (OH), (s)+20H (aq) Kr 0.03 M (OH Estimate the solubility of M (OH in a solution buffered at pH 7.0, 10.0 and 14.0. pH 4.0 pH 7.0 pH 0.0 Number 0.04 A View Previous View Next Exit Explanation d Wind Resources O Assignment Information Available From: 4/14/2016 06:00 PM 5/20/2016 06:00 PM Due Date: Points Possible: 10 Grade Category: HW 7 Description Policies: Practice You can check your answers. You can view solutions when you complete or give up on any question. You can keep trying to answer each question until you get it right or give up There is no penalty for incorrect answers. O eTextbook O Help With This Topic O Web Help & Videos O Technical Support and Bug Reports O Logout HelpExplanation / Answer
we have the two reactions where
M(OH)2 <----------> M+2 + 2OH- K1 = 8x10-6M2 [M+2][OH-] = K1
M(OH)2 + 2OH <--------> M(OH)4-2 K2 = 0.03 [M(OH)4-2] / [OH-] = K2
[M+2][OH-] = K1 =) [M+2] = K1/ [OH-]
[M(OH)4-2] / [OH-] = K2 =) [M(OH)4-2] = K2 [OH-]
so the solubility is S and S = [M+2] + [M(OH)4-2] =) S = K1/ [OH-] + K2 [OH-] =)
S = 8x10-6M2/ [OH-] +0.03[OH-] =
as [OH] = 10-14+pH
for pH = 7 [OH] = 10-14+7 = 1x10-7M
S = K1/ [OH-] + K2 [OH-] = 8x10-6M2/ [OH-] +0.03[OH-] = 8x10-6M2/ 1x10-7M +0.03x1x10-7M = 80M
for pH = 10 [OH] = 10-14+10 = 1x10-4M
S = K1/ [OH-] + K2 [OH-] = 8x10-6M2/ [OH-] +0.03[OH-] = 8x10-6M2/ 1x10-4M +0.03x1x10-4M = 0.08M
for pH = 14 [OH] = 10-14+14 = 1M
S = K1/ [OH-] + K2 [OH-] = 8x10-6M2/ [OH-] +0.03[OH-] = 8x10-6M2/ 1M +0.03x1M = 0.03M
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