Correct Answers:<\/h3>\n\n\n\n
a. CH\u2082FCH\u2082CO\u2082H<\/strong> Part a:<\/strong> Part b:<\/strong> Part c:<\/strong> Part d:<\/strong> In acid-base chemistry, the strength of Br\u00f8nsted acids is often predicted using factors such as electronegativity<\/strong>, inductive effects<\/strong>, resonance stabilization<\/strong>, and atom size<\/strong>. Electronegativity and inductive effects are especially important in carboxylic acids, as electron-withdrawing groups stabilize the negative charge of the conjugate base, enhancing acid strength. Understanding conjugate pairs also helps in determining acid-base reactions and buffer behavior.<\/p>\n\n\n\n Which of the following compounds is the strongest Bronsted acid? CH2BrCH2CO2H CH2FCH2CO2H b. Which of the following compounds is the strongest Bronsted acid? CH2ClCO2H CH2ICO2H c. What is the conjugate base of H3S+? S2- HS- H4S2+ H3S+ H2S None of these choices is correct. d. What is the conjugate acid of H2S? H2S H4S2+ HS- […]<\/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-238142","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/238142","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=238142"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/238142\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=238142"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=238142"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=238142"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
b. CH\u2082ClCO\u2082H<\/strong>
c. H\u2082S<\/strong>
d. H\u2083S\u207a<\/strong><\/p>\n\n\n\n
\n\n\n\nExplanation <\/h3>\n\n\n\n
Between CH\u2082BrCH\u2082CO\u2082H<\/strong> and CH\u2082FCH\u2082CO\u2082H<\/strong>, the stronger Br\u00f8nsted acid is CH\u2082FCH\u2082CO\u2082H<\/strong>. This is due to the inductive effect<\/strong> of fluorine, which is more electronegative<\/strong> than bromine. Electronegative atoms withdraw electron density through sigma bonds, stabilizing the negative charge on the conjugate base (the carboxylate anion). Since fluorine withdraws more electron density than bromine, it stabilizes the conjugate base better, increasing the acidity of the acid.<\/p>\n\n\n\n
Comparing CH\u2082ClCO\u2082H<\/strong> and CH\u2082ICOOH<\/strong>, the stronger Br\u00f8nsted acid is CH\u2082ClCO\u2082H<\/strong>. Chlorine is more electronegative than iodine. As a result, it exerts a stronger electron-withdrawing effect via induction, which stabilizes the conjugate base more effectively. Greater stability of the conjugate base correlates with a stronger acid.<\/p>\n\n\n\n
The conjugate base of an acid is formed by removing one proton (H\u207a). The species H\u2083S\u207a<\/strong> is a protonated form of hydrogen sulfide. Removing a proton from H\u2083S\u207a gives H\u2082S<\/strong>. So, H\u2082S<\/strong> is the correct conjugate base of H\u2083S\u207a.<\/p>\n\n\n\n
The conjugate acid of a base is formed by adding one proton (H\u207a)<\/strong>. The molecule H\u2082S<\/strong>, when it gains a proton, forms H\u2083S\u207a<\/strong>. Thus, H\u2083S\u207a<\/strong> is the conjugate acid of H\u2082S.<\/p>\n\n\n\n![\"\"](\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner5-693.jpeg\")
<\/figure>\n","protected":false},"excerpt":{"rendered":"