{"id":37788,"date":"2025-06-26T06:03:19","date_gmt":"2025-06-26T06:03:19","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=37788"},"modified":"2025-06-26T06:03:20","modified_gmt":"2025-06-26T06:03:20","slug":"aniline-c6h5nh2-has-a-base-dissociation-constant-kb-of-4-3-x-10-10","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/aniline-c6h5nh2-has-a-base-dissociation-constant-kb-of-4-3-x-10-10\/","title":{"rendered":"Aniline (C6H5NH2) has a base dissociation constant (Kb) of 4.3 x 10^-10."},"content":{"rendered":"\n<p>Aniline (C6H5NH2) has a base dissociation constant (Kb) of 4.3 x 10^-10. The conjugate acid of aniline is C6H5NH3+ and its Ka value is 9.7 x 10^-4.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"700\" height=\"301\" src=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/image-352.png\" alt=\"\" class=\"wp-image-37789\" srcset=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/image-352.png 700w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/image-352-300x129.png 300w\" sizes=\"auto, (max-width: 700px) 100vw, 700px\" \/><\/figure>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-vivid-cyan-blue-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>The correct conjugate acid of aniline (C\u2086H\u2085NH\u2082) is <strong>C\u2086H\u2085NH\u2083\u207a<\/strong>, and its <strong>Ka value is 2.3 \u00d7 10\u207b\u2075<\/strong>.<\/p>\n\n\n\n<p>Here&#8217;s a breakdown and explanation:<\/p>\n\n\n\n<p>Aniline is a weak base and its base dissociation constant, <strong>Kb<\/strong>, is given as 4.3 \u00d7 10\u207b\u00b9\u2070. The conjugate acid of aniline is formed when the lone pair on the nitrogen atom in the NH\u2082 group accepts a proton (H\u207a), resulting in the positively charged <strong>C\u2086H\u2085NH\u2083\u207a<\/strong> ion.<\/p>\n\n\n\n<p>To find the acid dissociation constant (<strong>Ka<\/strong>) for the conjugate acid, we use the following relationship, which is derived from the ionization of water:<\/p>\n\n\n\n<p><strong>Ka \u00d7 Kb = Kw<\/strong><\/p>\n\n\n\n<p>Kw is the ionization constant of water, which is <strong>1.0 \u00d7 10\u207b\u00b9\u2074<\/strong> at 25 \u00b0C. We rearrange the equation to solve for Ka:<\/p>\n\n\n\n<p><strong>Ka = Kw \/ Kb<\/strong><\/p>\n\n\n\n<p>Substituting the known values:<\/p>\n\n\n\n<p>Ka = (1.0 \u00d7 10\u207b\u00b9\u2074) \/ (4.3 \u00d7 10\u207b\u00b9\u2070) Ka \u2248 2.3 \u00d7 10\u207b\u2075<\/p>\n\n\n\n<p>So the conjugate acid of aniline is <strong>C\u2086H\u2085NH\u2083\u207a<\/strong>, and its Ka is <strong>2.3 \u00d7 10\u207b\u2075<\/strong>.<\/p>\n\n\n\n<p>Among the multiple-choice options provided in the image, this corresponds to the fourth option: <strong>C\u2086H\u2085NH\u2083\u207a, 2.3 \u00d7 10\u207b\u2075<\/strong><\/p>\n\n\n\n<p>This result not only checks out mathematically, but it also reinforces the fundamental concept that a weak base has a strong conjugate acid, and vice versa. The weaker the base, the smaller the Kb, which leads to a relatively larger Ka for its conjugate acid. Understanding this balance is central to acid\u2013base equilibrium problems in aqueous solutions.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"852\" height=\"1024\" src=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-973.jpeg\" alt=\"\" class=\"wp-image-37790\" srcset=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-973.jpeg 852w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-973-250x300.jpeg 250w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-973-768x923.jpeg 768w\" sizes=\"auto, (max-width: 852px) 100vw, 852px\" \/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Aniline (C6H5NH2) has a base dissociation constant (Kb) of 4.3 x 10^-10. The conjugate acid of aniline is C6H5NH3+ and its Ka value is 9.7 x 10^-4. The Correct Answer and Explanation is: The correct conjugate acid of aniline (C\u2086H\u2085NH\u2082) is C\u2086H\u2085NH\u2083\u207a, and its Ka value is 2.3 \u00d7 10\u207b\u2075. Here&#8217;s a breakdown and explanation: [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-37788","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/37788","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=37788"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/37788\/revisions"}],"predecessor-version":[{"id":37791,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/37788\/revisions\/37791"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=37788"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=37788"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=37788"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}