{"id":221148,"date":"2025-05-29T08:11:06","date_gmt":"2025-05-29T08:11:06","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=221148"},"modified":"2025-05-29T08:29:05","modified_gmt":"2025-05-29T08:29:05","slug":"what-is-the-equivalent-weight","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/29\/what-is-the-equivalent-weight\/","title":{"rendered":"Molecular Weight Of HCO3 Is 61 G\/Mole."},"content":{"rendered":"\n<p>Molecular Weight Of HCO3 Is 61 G\/Mole. What Is The Equivalent Weight? Og<\/p>\n\n\n\n<p><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><strong>Correct Answer:<\/strong><br>The <strong>equivalent weight of HCO\u2083\u207b (bicarbonate)<\/strong> is <strong>61 g\/eq<\/strong>.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Explanation:<\/strong><\/h3>\n\n\n\n<p>To determine the <strong>equivalent weight<\/strong> of a substance, we use the formula:<\/p>\n\n\n\n<blockquote class=\"wp-block-quote is-layout-flow wp-block-quote-is-layout-flow\">\n<p><strong>Equivalent weight = Molecular weight \/ n<\/strong>,<br>where <strong>n<\/strong> is the number of <strong>replaceable hydrogen ions (H\u207a)<\/strong> the substance can donate or accept in a chemical reaction, or the number of <strong>charges<\/strong> it can carry in a redox or acid-base reaction.<\/p>\n<\/blockquote>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Step-by-Step Analysis:<\/strong><\/h3>\n\n\n\n<p><strong>1. Given:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Molecular weight (MW) of <strong>HCO\u2083\u207b<\/strong> = <strong>61 g\/mol<\/strong><\/li>\n<\/ul>\n\n\n\n<p><strong>2. Determine the number of equivalents (n):<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>HCO\u2083\u207b<\/strong> is the <strong>bicarbonate ion<\/strong>, which can <strong>accept 1 H\u207a<\/strong> to form <strong>H\u2082CO\u2083 (carbonic acid)<\/strong>, or <strong>donate 1 H\u207a<\/strong> to form <strong>CO\u2083\u00b2\u207b (carbonate ion)<\/strong>.<\/li>\n\n\n\n<li>In <strong>acid-base reactions<\/strong>, it acts as a <strong>monovalent ion<\/strong>, meaning <strong>n = 1<\/strong> equivalent per mole.<\/li>\n<\/ul>\n\n\n\n<p><strong>3. Apply the formula:<\/strong> Equivalent&nbsp;weight=61&nbsp;g\/mol1=61&nbsp;g\/eq\\text{Equivalent weight} = \\frac{61\\ \\text{g\/mol}}{1} = 61\\ \\text{g\/eq}Equivalent&nbsp;weight=161&nbsp;g\/mol\u200b=61&nbsp;g\/eq<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Why This Is Important:<\/strong><\/h3>\n\n\n\n<p>In <strong>clinical settings<\/strong>, especially in <strong>acid-base balance and metabolic panel interpretations<\/strong>, the <strong>bicarbonate ion (HCO\u2083\u207b)<\/strong> plays a critical role. It is a <strong>buffer<\/strong> that helps maintain <strong>physiological pH<\/strong>.<\/p>\n\n\n\n<p>Understanding its equivalent weight helps in:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Calculating <strong>normality (N)<\/strong> of bicarbonate solutions.<\/li>\n\n\n\n<li>Interpreting <strong>anion gap<\/strong> or <strong>buffer capacities<\/strong>.<\/li>\n\n\n\n<li>Preparing reagents in <strong>titration or dialysis solutions<\/strong>.<\/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\"><strong>Summary:<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Molecular weight of HCO\u2083\u207b = <strong>61 g\/mol<\/strong><\/li>\n\n\n\n<li>Number of equivalents (n) = <strong>1<\/strong><\/li>\n\n\n\n<li><strong>Equivalent weight<\/strong> = 61 g\/eq<\/li>\n<\/ul>\n\n\n\n<p>Thus, the equivalent weight of <strong>HCO\u2083\u207b is 61 grams per equivalent<\/strong>, the same as its molecular weight due to its <strong>monovalent charge<\/strong><\/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\/learnexams-banner8-75.jpeg\" alt=\"\" class=\"wp-image-221150\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Molecular Weight Of HCO3 Is 61 G\/Mole. What Is The Equivalent Weight? Og The Correct Answer and Explanation is: Correct Answer:The equivalent weight of HCO\u2083\u207b (bicarbonate) is 61 g\/eq. Explanation: To determine the equivalent weight of a substance, we use the formula: Equivalent weight = Molecular weight \/ n,where n is the number of replaceable [&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-221148","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/221148","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=221148"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/221148\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=221148"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=221148"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=221148"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}