To find the value of pKbpK_bpKb\u200b for methylamine (CH3NH2), we need to follow these steps:<\/p>\n\n\n\n
Step 1: Write the equilibrium expression for the reaction.<\/h3>\n\n\n\n
The dissociation of methylamine in water can be represented as:CH3NH2(aq)+H2O(l)\u21ccCH3NH3+(aq)+OH\u2212(aq)\\text{CH}_3\\text{NH}_2 (\\text{aq}) + \\text{H}_2\\text{O} (\\text{l}) \\rightleftharpoons \\text{CH}_3\\text{NH}_3^+ (\\text{aq}) + \\text{OH}^- (\\text{aq})CH3\u200bNH2\u200b(aq)+H2\u200bO(l)\u21ccCH3\u200bNH3+\u200b(aq)+OH\u2212(aq)<\/p>\n\n\n\n
The equilibrium expression for this reaction is:Kb=[CH3NH3+][OH\u2212][CH3NH2]K_b = \\frac{[\\text{CH}_3\\text{NH}_3^+][\\text{OH}^-]}{[\\text{CH}_3\\text{NH}_2]}Kb\u200b=[CH3\u200bNH2\u200b][CH3\u200bNH3+\u200b][OH\u2212]\u200b<\/p>\n\n\n\n
Step 2: Calculate the concentration of OH\u2212\\text{OH}^-OH\u2212.<\/h3>\n\n\n\n
The pH of the solution is given as 10.01. We can first convert pH to pOH:pOH=14\u2212pH=14\u221210.01=3.99\\text{pOH} = 14 – \\text{pH} = 14 – 10.01 = 3.99pOH=14\u2212pH=14\u221210.01=3.99<\/p>\n\n\n\n
Now, we calculate the concentration of OH\u2212\\text{OH}^-OH\u2212 using the relationship:[OH\u2212]=10\u2212pOH=10\u22123.99\u22481.0\u00d710\u22124\u2009M[\\text{OH}^-] = 10^{-\\text{pOH}} = 10^{-3.99} \\approx 1.0 \\times 10^{-4} \\, \\text{M}[OH\u2212]=10\u2212pOH=10\u22123.99\u22481.0\u00d710\u22124M<\/p>\n\n\n\n
Step 3: Use the ICE table to find the equilibrium concentrations.<\/h3>\n\n\n\n
Let\u2019s assume that the initial concentration of CH3NH2 is 0.150 M and the change in concentration at equilibrium due to dissociation is xxx. From the balanced equation, the change in concentrations of CH3NH2, CH3NH3+, and OH- will be as follows:<\/p>\n\n\n\n From the pOH calculation, we know that at equilibrium, [OH\u2212]=1.0\u00d710\u22124\u2009M[\\text{OH}^-] = 1.0 \\times 10^{-4} \\, \\text{M}[OH\u2212]=1.0\u00d710\u22124M, so x=1.0\u00d710\u22124x = 1.0 \\times 10^{-4}x=1.0\u00d710\u22124.<\/p>\n\n\n\n Now we can substitute the equilibrium concentrations into the expression for KbK_bKb\u200b:Kb=(1.0\u00d710\u22124)(1.0\u00d710\u22124)0.150\u22121.0\u00d710\u22124\u22481.0\u00d710\u221280.150K_b = \\frac{(1.0 \\times 10^{-4})(1.0 \\times 10^{-4})}{0.150 – 1.0 \\times 10^{-4}} \\approx \\frac{1.0 \\times 10^{-8}}{0.150}Kb\u200b=0.150\u22121.0\u00d710\u22124(1.0\u00d710\u22124)(1.0\u00d710\u22124)\u200b\u22480.1501.0\u00d710\u22128\u200bKb\u22486.67\u00d710\u22128K_b \\approx 6.67 \\times 10^{-8}Kb\u200b\u22486.67\u00d710\u22128<\/p>\n\n\n\n Finally, the pKbpK_bpKb\u200b is the negative logarithm of KbK_bKb\u200b:pKb=\u2212log\u2061Kb=\u2212log\u2061(6.67\u00d710\u22128)\u22487.18pK_b = -\\log K_b = -\\log(6.67 \\times 10^{-8}) \\approx 7.18pKb\u200b=\u2212logKb\u200b=\u2212log(6.67\u00d710\u22128)\u22487.18<\/p>\n\n\n\n The value of pKbpK_bpKb\u200b for methylamine is approximately 7.18<\/strong>.<\/p>\n\n\n\n Methylamine, CH3NH2, is a weak base. CH3NH2 (aq) + H2O (l) –> CH3NH3+ (aq) + OH- (aq) If the pH of a 0.150 M solution of the amine is 10.01, what is the value of pKb? The Correct Answer and Explanation is: To find the value of pKbpK_bpKb\u200b for methylamine (CH3NH2), we need to follow […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-246180","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/246180","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=246180"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/246180\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=246180"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=246180"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=246180"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Substance<\/th> Initial (M)<\/th> Change (M)<\/th> Equilibrium (M)<\/th><\/tr><\/thead> CH3NH2<\/td> 0.150<\/td> -x<\/td> 0.150 – x<\/td><\/tr> CH3NH3+<\/td> 0<\/td> +x<\/td> x<\/td><\/tr> OH-<\/td> 0<\/td> +x<\/td> x<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n Step 4: Solve for KbK_bKb\u200b.<\/h3>\n\n\n\n
Step 5: Calculate pKbpK_bpKb\u200b.<\/h3>\n\n\n\n
Conclusion:<\/h3>\n\n\n\n
![\"\"](\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner5-807.jpeg\")
<\/figure>\n","protected":false},"excerpt":{"rendered":"