{"id":218331,"date":"2025-05-23T16:20:09","date_gmt":"2025-05-23T16:20:09","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=218331"},"modified":"2025-05-23T16:20:12","modified_gmt":"2025-05-23T16:20:12","slug":"identify-the-conjugate-acid-base-pairs-in-the-following-reactions","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/23\/identify-the-conjugate-acid-base-pairs-in-the-following-reactions\/","title":{"rendered":"Identify the conjugate acid\u2013base pairs in the following reactions."},"content":{"rendered":"\n<p>Identify the conjugate acid\u2013base pairs in the following reactions. (a) HNO3 + N2H4 ? NO3- + N2H5+ (b) NH3 + N2H5+ ? NH4+ + N2H4 (c) H2PO4- + CO3^2- ? HPO4^2- + HCO3- (d) HIO3 + HC2O4- ? IO3- + H2C2O4<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-0-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">(a)<\/h3>\n\n\n\n<p><strong>Reaction:<\/strong><br>HNO\u2083 + N\u2082H\u2084 \u2192 NO\u2083\u207b + N\u2082H\u2085\u207a<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>HNO\u2083<\/strong> loses a proton (H\u207a) and becomes <strong>NO\u2083\u207b<\/strong>.<\/li>\n\n\n\n<li><strong>N\u2082H\u2084<\/strong> gains a proton and becomes <strong>N\u2082H\u2085\u207a<\/strong>.<\/li>\n<\/ul>\n\n\n\n<p><strong>Conjugate acid-base pairs:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>HNO\u2083 \/ NO\u2083\u207b (acid\/base)<\/li>\n\n\n\n<li>N\u2082H\u2084 \/ N\u2082H\u2085\u207a (base\/acid)<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">(b)<\/h3>\n\n\n\n<p><strong>Reaction:<\/strong><br>NH\u2083 + N\u2082H\u2085\u207a \u2192 NH\u2084\u207a + N\u2082H\u2084<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>NH\u2083<\/strong> gains a proton to become <strong>NH\u2084\u207a<\/strong>.<\/li>\n\n\n\n<li><strong>N\u2082H\u2085\u207a<\/strong> loses a proton to become <strong>N\u2082H\u2084<\/strong>.<\/li>\n<\/ul>\n\n\n\n<p><strong>Conjugate acid-base pairs:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>NH\u2083 \/ NH\u2084\u207a (base\/acid)<\/li>\n\n\n\n<li>N\u2082H\u2085\u207a \/ N\u2082H\u2084 (acid\/base)<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">(c)<\/h3>\n\n\n\n<p><strong>Reaction:<\/strong><br>H\u2082PO\u2084\u207b + CO\u2083\u00b2\u207b \u2192 HPO\u2084\u00b2\u207b + HCO\u2083\u207b<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>H\u2082PO\u2084\u207b<\/strong> loses a proton and becomes <strong>HPO\u2084\u00b2\u207b<\/strong>.<\/li>\n\n\n\n<li><strong>CO\u2083\u00b2\u207b<\/strong> gains a proton and becomes <strong>HCO\u2083\u207b<\/strong>.<\/li>\n<\/ul>\n\n\n\n<p><strong>Conjugate acid-base pairs:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>H\u2082PO\u2084\u207b \/ HPO\u2084\u00b2\u207b (acid\/base)<\/li>\n\n\n\n<li>CO\u2083\u00b2\u207b \/ HCO\u2083\u207b (base\/acid)<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">(d)<\/h3>\n\n\n\n<p><strong>Reaction:<\/strong><br>HIO\u2083 + HC\u2082O\u2084\u207b \u2192 IO\u2083\u207b + H\u2082C\u2082O\u2084<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>HIO\u2083<\/strong> loses a proton and becomes <strong>IO\u2083\u207b<\/strong>.<\/li>\n\n\n\n<li><strong>HC\u2082O\u2084\u207b<\/strong> gains a proton and becomes <strong>H\u2082C\u2082O\u2084<\/strong>.<\/li>\n<\/ul>\n\n\n\n<p><strong>Conjugate acid-base pairs:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>HIO\u2083 \/ IO\u2083\u207b (acid\/base)<\/li>\n\n\n\n<li>HC\u2082O\u2084\u207b \/ H\u2082C\u2082O\u2084 (base\/acid)<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h2 class=\"wp-block-heading\"><\/h2>\n\n\n\n<p>In acid-base chemistry, <strong>conjugate acid-base pairs<\/strong> are two species that differ by one proton (H\u207a). The acid donates a proton, and the base accepts that proton. After donation, the acid becomes its conjugate base, and after acceptance, the base becomes its conjugate acid.<\/p>\n\n\n\n<p>For each reaction, identify which species loses a proton (acid \u2192 conjugate base) and which gains a proton (base \u2192 conjugate acid). This change defines the pairs.<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>In <strong>reaction (a)<\/strong>, nitric acid (HNO\u2083) acts as an acid by donating a proton to hydrazine (N\u2082H\u2084), which acts as a base by accepting the proton. After donation, HNO\u2083 becomes NO\u2083\u207b (its conjugate base), and N\u2082H\u2084 becomes N\u2082H\u2085\u207a (its conjugate acid).<\/li>\n\n\n\n<li>In <strong>reaction (b)<\/strong>, ammonia (NH\u2083) acts as a base accepting a proton from N\u2082H\u2085\u207a, which is the acid here. Their conjugates are NH\u2084\u207a (acid form of ammonia) and N\u2082H\u2084 (base form of hydrazine).<\/li>\n\n\n\n<li>In <strong>reaction (c)<\/strong>, dihydrogen phosphate (H\u2082PO\u2084\u207b) donates a proton, making it an acid, and carbonate ion (CO\u2083\u00b2\u207b) accepts a proton, acting as a base. Their conjugate pairs are H\u2082PO\u2084\u207b \/ HPO\u2084\u00b2\u207b and CO\u2083\u00b2\u207b \/ HCO\u2083\u207b.<\/li>\n\n\n\n<li>In <strong>reaction (d)<\/strong>, iodic acid (HIO\u2083) donates a proton, becoming IO\u2083\u207b, and hydrogen oxalate (HC\u2082O\u2084\u207b) accepts a proton, forming oxalic acid (H\u2082C\u2082O\u2084). These species form the conjugate acid-base pairs.<\/li>\n<\/ul>\n\n\n\n<p>Recognizing these pairs helps understand proton transfer reactions, which is central in acid-base chemistry and many biological and chemical processes.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/05\/image-213.png\" alt=\"\" class=\"wp-image-218332\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Identify the conjugate acid\u2013base pairs in the following reactions. (a) HNO3 + N2H4 ? NO3- + N2H5+ (b) NH3 + N2H5+ ? NH4+ + N2H4 (c) H2PO4- + CO3^2- ? HPO4^2- + HCO3- (d) HIO3 + HC2O4- ? IO3- + H2C2O4 The Correct Answer and Explanation is: (a) Reaction:HNO\u2083 + N\u2082H\u2084 \u2192 NO\u2083\u207b + N\u2082H\u2085\u207a [&hellip;]<\/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-218331","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/218331","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=218331"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/218331\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=218331"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=218331"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=218331"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}