To find the value of the base dissociation constant (Kb) for methylamine (CH3NH2), we can use the given data and apply the equilibrium principles.<\/p>\n\n\n\n
Step 1: Write the equilibrium expression<\/h3>\n\n\n\n
The equilibrium reaction for methylamine is:<\/p>\n\n\n\n
CH3NH2(aq) + H2O(l) \u21cc CH3NH3+(aq) + OH-(aq)<\/strong><\/p>\n\n\n\n The equilibrium constant expression for Kb is:Kb=[CH3NH3+][OH\u2212][CH3NH2]K_b = \\frac{[CH3NH3^+][OH^-]}{[CH3NH2]}Kb\u200b=[CH3NH2][CH3NH3+][OH\u2212]\u200b<\/p>\n\n\n\n Given that the pH of the solution is 11.70, we can first calculate the pOH using the formula:pOH=14\u2212pH=14\u221211.70=2.30\\text{pOH} = 14 – \\text{pH} = 14 – 11.70 = 2.30pOH=14\u2212pH=14\u221211.70=2.30<\/p>\n\n\n\n Then, we can calculate the concentration of OH- ions using the relationship between pOH and [OH-]:[OH\u2212]=10\u2212pOH=10\u22122.30=5.012\u00d710\u22123 M[\\text{OH}^-] = 10^{-\\text{pOH}} = 10^{-2.30} = 5.012 \\times 10^{-3} \\text{ M}[OH\u2212]=10\u2212pOH=10\u22122.30=5.012\u00d710\u22123 M<\/p>\n\n\n\n Next, we will use an ICE (Initial, Change, Equilibrium) table to calculate the concentrations of all species at equilibrium. Initially, before any dissociation:<\/p>\n\n\n\n At equilibrium:<\/p>\n\n\n\n From the pOH calculation, we know that at equilibrium, [OH-] = 5.012\u00d710\u22123 M5.012 \\times 10^{-3} \\text{ M}5.012\u00d710\u22123 M, so:x=5.012\u00d710\u22123 Mx = 5.012 \\times 10^{-3} \\text{ M}x=5.012\u00d710\u22123 M<\/p>\n\n\n\n Now, we can substitute these equilibrium concentrations into the Kb expression:Kb=(5.012\u00d710\u22123)(5.012\u00d710\u22123)0.065\u22125.012\u00d710\u22123K_b = \\frac{(5.012 \\times 10^{-3})(5.012 \\times 10^{-3})}{0.065 – 5.012 \\times 10^{-3}}Kb\u200b=0.065\u22125.012\u00d710\u22123(5.012\u00d710\u22123)(5.012\u00d710\u22123)\u200b<\/p>\n\n\n\n Simplify the expression:Kb=(5.012\u00d710\u22123)20.065\u22125.012\u00d710\u22123=2.512\u00d710\u221250.05999K_b = \\frac{(5.012 \\times 10^{-3})^2}{0.065 – 5.012 \\times 10^{-3}} = \\frac{2.512 \\times 10^{-5}}{0.05999}Kb\u200b=0.065\u22125.012\u00d710\u22123(5.012\u00d710\u22123)2\u200b=0.059992.512\u00d710\u22125\u200bKb\u22484.18\u00d710\u22124K_b \\approx 4.18 \\times 10^{-4}Kb\u200b\u22484.18\u00d710\u22124<\/p>\n\n\n\n The value of KbK_bKb\u200b for methylamine is approximately 4.18 \u00d7 10\u207b\u2074<\/strong>.<\/p>\n\n\n\n In this calculation, we first used the pH to find the concentration of OH- ions in the solution. Then, we applied the equilibrium principles by setting up an ICE table to find the concentration of CH3NH2, CH3NH3+, and OH- at equilibrium. Using the known equilibrium concentrations, we calculated Kb, which reflects the base’s strength. The higher the Kb value, the stronger the base.<\/p>\n\n\n\n Methylamine, CH3NH2, is a weak base. CH3NH2(aq) + H2O(l) \u00e2\u2020\u2019 CH3NH3+(aq) + OH-(aq) If the pH of a 0.065 M solution is 11.70, what is the value of Kb? Show the steps please. The Correct Answer and Explanation is: To find the value of the base dissociation constant (Kb) for methylamine (CH3NH2), we can use […]<\/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-246176","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/246176","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=246176"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/246176\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=246176"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=246176"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=246176"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Step 2: Use the pH to find [OH-]<\/h3>\n\n\n\n
Step 3: Set up an ICE table<\/h3>\n\n\n\n
\n
\n
Step 4: Substitute into the Kb expression<\/h3>\n\n\n\n
Final Answer:<\/h3>\n\n\n\n
Explanation:<\/h3>\n\n\n\n
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<\/figure>\n","protected":false},"excerpt":{"rendered":"