{"id":227264,"date":"2025-06-06T03:57:08","date_gmt":"2025-06-06T03:57:08","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=227264"},"modified":"2025-06-06T03:57:19","modified_gmt":"2025-06-06T03:57:19","slug":"ethanamine-ch3ch2nh2-is-a-weak-base","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/06\/ethanamine-ch3ch2nh2-is-a-weak-base\/","title":{"rendered":"Ethanamine, CH3CH2NH2, is a weak base."},"content":{"rendered":"\n

Ethanamine, CH3CH2NH2, is a weak base. pKa(CH3CH2NH3+) = 10.6 (a) (i) Write an equation to show the reaction of ethanamine CH3CH2NH2 with water. CH3CH2NH2 + H2O \u00e2\u2021\u0152 CH3CH2NH3+ + OH- (ii) Explain, with reference to the equation and the species in solution, what is meant by the terms weak and base. In the equation, ethanamine (CH3CH2NH2) reacts with water to form ethanammonium ion (CH3CH2NH3+) and hydroxide ion (OH-). A weak base is a substance that partially dissociates in water, meaning it does not completely ionize. In this case, ethanamine only partially reacts with water to form the ethanammonium ion and hydroxide ion. (b) (i) Determine the pKb for ethanamine. pKb = 14 – pKa pKb = 14 – 10.6 pKb = 3.4 (ii) Calculate the pH of a 0.109 mol L-1 solution of ethanamine. To calculate the pH, we need to find the pOH first. pOH = -log10[OH-] pOH = -log10(0.109) pOH = 0.962 pH + pOH = 14 pH = 14 – pOH pH = 14 – 0.962 pH = 13.038<\/p>\n\n\n\n

The Correct Answer and Explanation is:<\/mark><\/strong><\/em><\/p>\n\n\n\n

Corrected Answer:<\/strong><\/p>\n\n\n\n

(a)<\/h3>\n\n\n\n

(i)<\/strong>
Equation:<\/strong>
CH\u2083CH\u2082NH\u2082 + H\u2082O \u21cc CH\u2083CH\u2082NH\u2083\u207a + OH\u207b<\/p>\n\n\n\n

(ii)<\/strong>
Ethanamine (CH\u2083CH\u2082NH\u2082) acts as a weak base<\/strong>, meaning it only partially reacts<\/strong> with water to produce its conjugate acid (ethanammonium ion, CH\u2083CH\u2082NH\u2083\u207a) and hydroxide ions (OH\u207b). A base<\/strong> is a substance that accepts protons (H\u207a), and in this reaction, ethanamine accepts a proton from water. Because the reaction does not go to completion, and a significant amount of unreacted CH\u2083CH\u2082NH\u2082 remains in solution, ethanamine is classified as a weak base.<\/p>\n\n\n\n

\n\n\n\n

(b)<\/h3>\n\n\n\n

(i)<\/strong>
pKa(CH\u2083CH\u2082NH\u2083\u207a) = 10.6
pKb = 14 \u2212 pKa = 14 \u2212 10.6 = 3.4<\/strong><\/p>\n\n\n\n

(ii) Calculate the pH of a 0.109 mol L\u207b\u00b9 ethanamine solution<\/strong><\/p>\n\n\n\n

We start by calculating the concentration of OH\u207b using the base dissociation constant (Kb):<\/p>\n\n\n\n

Step 1: Calculate Kb from pKb<\/strong> pKb=3.4\u21d2Kb=10\u22123.4\u22483.98\u00d710\u22124\\text{pKb} = 3.4 \\Rightarrow K_b = 10^{-3.4} \\approx 3.98 \\times 10^{-4}<\/p>\n\n\n\n

Step 2: Set up ICE table for CH\u2083CH\u2082NH\u2082 \u21cc CH\u2083CH\u2082NH\u2083\u207a + OH\u207b<\/strong><\/p>\n\n\n\n

Let initial [CH\u2083CH\u2082NH\u2082] = 0.109 mol L\u207b\u00b9
At equilibrium:
[OH\u207b] = x
CH\u2083CH\u2082NH\u2082 \u21cc CH\u2083CH\u2082NH\u2083\u207a + OH\u207b
Initial: 0.109 0 0
Change: -x +x +x
Equilibrium: 0.109\u2212x x x Kb=x20.109\u2212x\u2248x20.109K_b = \\frac{x^2}{0.109 – x} \\approx \\frac{x^2}{0.109}<\/p>\n\n\n\n

Assuming x is small, 3.98\u00d710\u22124=x20.109\u21d2x2=(3.98\u00d710\u22124)(0.109)\u21d2x2=4.34\u00d710\u22125\u21d2x=4.34\u00d710\u22125\u22486.59\u00d710\u221233.98 \\times 10^{-4} = \\frac{x^2}{0.109} \\Rightarrow x^2 = (3.98 \\times 10^{-4})(0.109) \\Rightarrow x^2 = 4.34 \\times 10^{-5} \\Rightarrow x = \\sqrt{4.34 \\times 10^{-5}} \\approx 6.59 \\times 10^{-3}<\/p>\n\n\n\n

So [OH\u207b] \u2248 6.59 \u00d7 10\u207b\u00b3 mol L\u207b\u00b9<\/p>\n\n\n\n

Step 3: Calculate pOH and then pH<\/strong> pOH=\u2212log\u2061(6.59\u00d710\u22123)\u22482.18\u21d2pH=14\u22122.18=11.82\\text{pOH} = -\\log(6.59 \\times 10^{-3}) \\approx 2.18 \\Rightarrow \\text{pH} = 14 – 2.18 = \\boxed{11.82}<\/p>\n\n\n\n

\n\n\n\n

Summary<\/h3>\n\n\n\n

Ethanamine (CH\u2083CH\u2082NH\u2082) is a weak base that reacts with water by accepting a proton to form the ethanammonium ion (CH\u2083CH\u2082NH\u2083\u207a) and hydroxide ion (OH\u207b). The equilibrium lies significantly to the left, indicating only partial ionization, a hallmark of weak bases. The extent of this ionization is quantified by the base dissociation constant (Kb<\/strong>), which is derived from the pKa of its conjugate acid. Since pKa = 10.6, the corresponding pKb = 14 \u2212 10.6 = 3.4, and Kb = 10\u207b\u00b3.\u2074 \u2248 3.98 \u00d7 10\u207b\u2074.<\/p>\n\n\n\n

To determine the pH of a 0.109 mol L\u207b\u00b9 ethanamine solution, we use the Kb expression and solve for [OH\u207b], assuming the change in concentration is small compared to the initial base concentration. The resulting hydroxide concentration is found to be approximately 6.59 \u00d7 10\u207b\u00b3 mol L\u207b\u00b9. Using this, the pOH is calculated as \u2212log(6.59 \u00d7 10\u207b\u00b3) \u2248 2.18, leading to a final pH of 14 \u2212 2.18 = 11.82<\/strong>.<\/p>\n\n\n\n

The original approach incorrectly used the full base concentration (0.109 mol L\u207b\u00b9) as the [OH\u207b] concentration, which is only valid for strong bases<\/strong> that fully dissociate. Since ethanamine is a weak base<\/strong>, this assumption leads to a significant overestimation of pH. The correct method involves using the equilibrium expression for weak base dissociation and solving for the hydroxide ion concentration using Kb.<\/p>\n\n\n\n

\"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

Ethanamine, CH3CH2NH2, is a weak base. pKa(CH3CH2NH3+) = 10.6 (a) (i) Write an equation to show the reaction of ethanamine CH3CH2NH2 with water. CH3CH2NH2 + H2O \u00e2\u2021\u0152 CH3CH2NH3+ + OH- (ii) Explain, with reference to the equation and the species in solution, what is meant by the terms weak and base. In the equation, ethanamine […]<\/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-227264","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/227264","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=227264"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/227264\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=227264"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=227264"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=227264"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}