{"id":190851,"date":"2025-02-13T09:42:50","date_gmt":"2025-02-13T09:42:50","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=190851"},"modified":"2025-02-13T09:42:52","modified_gmt":"2025-02-13T09:42:52","slug":"n-h_4-aq-cn-aq-rightleftharpoons-nh-3-aq-hcnaq","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/02\/13\/n-h_4-aq-cn-aq-rightleftharpoons-nh-3-aq-hcnaq\/","title":{"rendered":"N * H_{4} ^ (aq) +CN^ &#8211; (aq) rightleftharpoons NH 3 (aq)+ HCN(aq)"},"content":{"rendered":"\n<p>Acid<br>C\u2082H\u2082NH<br>CH.CO,H<br>NH<br>HCN<br>CHOH<br>HCO,<br>CH NH<br>HO<br>CH\u2082OH<br>-pK<br>5<br>5<br>9<br>9<br>10<br>10<br>11<br>16<br>16<br>A) NH(aq) + CN-(aq) NH3(aq) +<br>HCN(aq)<br>B) CH3NH3(aq) + CH3CO\u2082(aq) = CH3NH2(aq) + CH3CO\u2082H(aq)<br>C) CHNH(aq) + HO-(aq) C\u2082H\u2083N(aq) + H\u2082O(1)<br>D) CH\u2082OH(aq) + CO3(aq) CH\u2082O-(aq) +<br>HCO3(aq)<br>E) HCN(aq) + CH\u2082O-(aq) = CN-(aq) +<br>CH\u2082OH(aq)<\/p>\n\n\n\n<p><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-6-color\"><strong>The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n<p>The correct answer is <strong>E) HCN(aq) + CH\u2082O\u207b(aq) \u2192 CN\u207b(aq) + CH\u2082OH(aq).<\/strong><\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation:<\/h3>\n\n\n\n<p>To determine the correct answer, we need to analyze the acid-base equilibrium using <strong>pKa values<\/strong>. The general rule is that a <strong>stronger acid donates a proton (H\u207a) to a stronger base<\/strong>.<\/p>\n\n\n\n<h4 class=\"wp-block-heading\">Step 1: Understanding pKa values<\/h4>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The <strong>pKa<\/strong> is the negative logarithm of the acid dissociation constant (<strong>Ka<\/strong>), which determines how strongly an acid donates a proton.<\/li>\n\n\n\n<li>A <strong>lower pKa<\/strong> means a <strong>stronger acid<\/strong> (more likely to donate H\u207a).<\/li>\n\n\n\n<li>A <strong>higher pKa<\/strong> means a <strong>weaker acid<\/strong> (less likely to donate H\u207a).<\/li>\n<\/ul>\n\n\n\n<p>From the table:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>HCN (Hydrocyanic acid)<\/strong> has a <strong>pKa \u2248 9<\/strong>.<\/li>\n\n\n\n<li><strong>CH\u2082OH (Methanol or related hydroxyl group)<\/strong> has a <strong>pKa \u2248 16<\/strong>.<\/li>\n<\/ul>\n\n\n\n<p>Since <strong>HCN (pKa = 9) is a stronger acid than CH\u2082OH (pKa = 16)<\/strong>, HCN will donate a proton to the CH\u2082O\u207b (which is the conjugate base of CH\u2082OH). This results in the formation of <strong>CN\u207b (cyanide ion, the conjugate base of HCN)<\/strong> and <strong>CH\u2082OH (methanol or its equivalent)<\/strong>.<\/p>\n\n\n\n<h4 class=\"wp-block-heading\">Step 2: Checking for Acid-Base Reaction Feasibility<\/h4>\n\n\n\n<ul class=\"wp-block-list\">\n<li>HCN <strong>acts as the acid<\/strong>, donating an H\u207a.<\/li>\n\n\n\n<li>CH\u2082O\u207b <strong>acts as the base<\/strong>, accepting the H\u207a.<\/li>\n\n\n\n<li>The reaction follows the expected acid-base trend, with a <strong>stronger acid (pKa 9) donating a proton to a stronger base<\/strong> (CH\u2082O\u207b, which comes from CH\u2082OH with pKa 16).<\/li>\n<\/ul>\n\n\n\n<p>Since this reaction obeys the acid-base principles and aligns with the pKa values, <strong>E is the correct answer.<\/strong><\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Summary:<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The <strong>acid (HCN, pKa 9) donates a proton<\/strong> to the <strong>base (CH\u2082O\u207b, pKa 16)<\/strong>.<\/li>\n\n\n\n<li>The <strong>conjugate base (CN\u207b) and conjugate acid (CH\u2082OH)<\/strong> are formed.<\/li>\n\n\n\n<li>The reaction is <strong>thermodynamically favorable<\/strong> due to the pKa relationship.<\/li>\n<\/ul>\n\n\n\n<p>Now, I&#8217;ll generate an image illustrating this acid-base reaction.<\/p>\n\n\n\n<p>Here is the chemistry diagram illustrating the acid-base reaction <strong>HCN(aq) + CH\u2082O\u207b(aq) \u2192 CN\u207b(aq) + CH\u2082OH(aq).<\/strong> It highlights the molecular structures, labels the reactants and products, and shows the proton transfer. Let me know if you need any modifications!<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/02\/image-482.png\" alt=\"\" class=\"wp-image-190852\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>AcidC\u2082H\u2082NHCH.CO,HNHHCNCHOHHCO,CH NHHOCH\u2082OH-pK55991010111616A) NH(aq) + CN-(aq) NH3(aq) +HCN(aq)B) CH3NH3(aq) + CH3CO\u2082(aq) = CH3NH2(aq) + CH3CO\u2082H(aq)C) CHNH(aq) + HO-(aq) C\u2082H\u2083N(aq) + H\u2082O(1)D) CH\u2082OH(aq) + CO3(aq) CH\u2082O-(aq) +HCO3(aq)E) HCN(aq) + CH\u2082O-(aq) = CN-(aq) +CH\u2082OH(aq) The Correct Answer and Explanation is : The correct answer is E) HCN(aq) + CH\u2082O\u207b(aq) \u2192 CN\u207b(aq) + CH\u2082OH(aq). Explanation: To determine the correct [&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-190851","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/190851","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=190851"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/190851\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=190851"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=190851"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=190851"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}