{"id":248724,"date":"2025-07-08T17:34:45","date_gmt":"2025-07-08T17:34:45","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=248724"},"modified":"2025-07-08T17:34:48","modified_gmt":"2025-07-08T17:34:48","slug":"the-value-of-ka-for-nitrous-acid-hno2-at-25-%e2%88%98c-is-4-5x10%e2%88%924-2","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/08\/the-value-of-ka-for-nitrous-acid-hno2-at-25-%e2%88%98c-is-4-5x10%e2%88%924-2\/","title":{"rendered":"The value of\u00a0Ka\u00a0for nitrous acid (HNO2) at 25\u00a0\u2218C is\u00a04.5\u00d710\u22124."},"content":{"rendered":"\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/07\/image-315.png\" alt=\"\" class=\"wp-image-248725\"\/><\/figure>\n\n\n\n<p class=\"wp-block-paragraph\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-1-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Based on the information provided, a logical derived value is the base dissociation constant (K\u2091) for the conjugate base of nitrous acid.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Correct Answer:<\/strong>&nbsp;The base dissociation constant (K\u2091) for the nitrite ion (NO\u2082\u207b) at 25 \u00b0C is 2.2 \u00d7 10\u207b\u00b9\u00b9.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">The information provided states that the acid dissociation constant (K\u2090) for nitrous acid (HNO\u2082) is 4.5 \u00d7 10\u207b\u2074 at 25 \u00b0C. This value is a quantitative measure of the strength of nitrous acid in a solution. The K\u2090 corresponds to the equilibrium constant for the dissociation of the acid in water, as shown in the reaction:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">HNO\u2082(aq) + H\u2082O(l) \u21cc H\u2083O\u207a(aq) + NO\u2082\u207b(aq)<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">The K\u2090 expression is [H\u2083O\u207a][NO\u2082\u207b] \/ [HNO\u2082]. Since the K\u2090 value of 4.5 \u00d7 10\u207b\u2074 is significantly less than 1, it confirms that nitrous acid is a weak acid. This means that at equilibrium, the majority of the acid remains in its undissociated molecular form (HNO\u2082) rather than splitting into hydronium ions (H\u2083O\u207a) and nitrite ions (NO\u2082\u207b).<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">From the K\u2090 of a weak acid, we can determine the strength of its conjugate base. For nitrous acid (HNO\u2082), the conjugate base is the nitrite ion (NO\u2082\u207b). The strength of this conjugate base is described by its base dissociation constant, K\u2091.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">At a given temperature, the product of the K\u2090 for an acid and the K\u2091 for its conjugate base is equal to the ion-product constant for water, K\u2099. At 25 \u00b0C, K\u2099 is a constant with the value 1.0 \u00d7 10\u207b\u00b9\u2074. The relationship is expressed as:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">K\u2090 \u00d7 K\u2091 = K\u2099<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Using this equation, we can calculate the K\u2091 for the nitrite ion:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">K\u2091 = K\u2099 \/ K\u2090<br>K\u2091 = (1.0 \u00d7 10\u207b\u00b9\u2074) \/ (4.5 \u00d7 10\u207b\u2074)<br>K\u2091 = 2.2 \u00d7 10\u207b\u00b9\u00b9<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">This calculated K\u2091 value of 2.2 \u00d7 10\u207b\u00b9\u00b9 indicates that the nitrite ion is a very weak base, which is expected for the conjugate base of a weak acid. This fundamental relationship allows chemists to understand and predict the behavior of both acidic and basic species in a solution.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner8-302.jpeg\" alt=\"\" class=\"wp-image-248726\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>The Correct Answer and Explanation is: Based on the information provided, a logical derived value is the base dissociation constant (K\u2091) for the conjugate base of nitrous acid. Correct Answer:&nbsp;The base dissociation constant (K\u2091) for the nitrite ion (NO\u2082\u207b) at 25 \u00b0C is 2.2 \u00d7 10\u207b\u00b9\u00b9. Explanation The information provided states that the acid dissociation [&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-248724","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/248724","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=248724"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/248724\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=248724"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=248724"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=248724"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}