{"id":229714,"date":"2025-06-08T11:38:48","date_gmt":"2025-06-08T11:38:48","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=229714"},"modified":"2025-06-08T11:38:50","modified_gmt":"2025-06-08T11:38:50","slug":"advance-study-assignment-properties-of-systems-in-chemical-equilibrium-methyl-orange-hmo-is-a-common-acid-base-indicator","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/08\/advance-study-assignment-properties-of-systems-in-chemical-equilibrium-methyl-orange-hmo-is-a-common-acid-base-indicator\/","title":{"rendered":"Advance Study Assignment: Properties of Systems in Chemical Equilibrium Methyl orange, HMO, is a common acid-base indicator."},"content":{"rendered":"\n<p>Advance Study Assignment: Properties of Systems in Chemical Equilibrium Methyl orange, HMO, is a common acid-base indicator. In solution, it ionizes according to the equation: HMO(aq) \u00e2\u2020\u2019 H+(aq) + MO-(aq) If methyl orange is added to distilled water, the solution turns yellow. If a drop or two of 6 M HCl is added to the yellow solution, it turns red. If a few drops of 6 M NaOH are added to the red solution, the color reverts to yellow. Why does adding 6 M HCl to the yellow solution of methyl orange tend to cause the color to change to red? (Note that in solution, HCl exists as H+ and Cl- ions) Why does adding 6 M NaOH to the red solution tend to make it turn back to yellow? (Note that in solution, NaOH exists as Na+ and OH- ions) How does increasing [OH-] shift Reaction 3 in the discussion section? How would the resulting change in [H+] affect the dissociation reaction of HMO<\/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>When methyl orange (HMO) is added to distilled water, it partially dissociates according to the equilibrium:<\/p>\n\n\n\n<p><strong>HMO(aq) \u21cc H\u207a(aq) + MO\u207b(aq)<\/strong><\/p>\n\n\n\n<p>This is an acid-base equilibrium where HMO is the weak acid and MO\u207b is its conjugate base. The color of the solution depends on the ratio of HMO to MO\u207b:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Red color<\/strong> indicates more <strong>HMO<\/strong> (acidic conditions).<\/li>\n\n\n\n<li><strong>Yellow color<\/strong> indicates more <strong>MO\u207b<\/strong> (basic conditions).<\/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\">1. <strong>Why does adding 6 M HCl turn the yellow solution red?<\/strong><\/h3>\n\n\n\n<p>Adding <strong>HCl<\/strong> introduces a high concentration of <strong>H\u207a<\/strong> ions to the solution. According to <strong>Le Chatelier&#8217;s Principle<\/strong>, increasing [H\u207a] shifts the equilibrium <strong>to the left<\/strong>, favoring the formation of <strong>HMO<\/strong>:<\/p>\n\n\n\n<p><strong>HMO \u21cc H\u207a + MO\u207b \u2190<\/strong> (shift to the left)<\/p>\n\n\n\n<p>As more HMO is formed and MO\u207b decreases, the solution turns <strong>red<\/strong>, the color associated with the undissociated HMO.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">2. <strong>Why does adding 6 M NaOH turn the red solution back to yellow?<\/strong><\/h3>\n\n\n\n<p>Adding <strong>NaOH<\/strong> introduces a high concentration of <strong>OH\u207b<\/strong> ions, which react with H\u207a to form water:<\/p>\n\n\n\n<p><strong>OH\u207b + H\u207a \u2192 H\u2082O<\/strong><\/p>\n\n\n\n<p>This decreases [H\u207a] in solution, disturbing the equilibrium. Again, per <strong>Le Chatelier\u2019s Principle<\/strong>, the system shifts <strong>to the right<\/strong> to produce more H\u207a:<\/p>\n\n\n\n<p><strong>HMO \u21cc H\u207a + MO\u207b \u2192<\/strong> (shift to the right)<\/p>\n\n\n\n<p>This increases [MO\u207b] (the conjugate base), causing the solution to turn <strong>yellow<\/strong>, the color of MO\u207b.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">3. <strong>How does increasing [OH\u207b] shift Reaction 3?<\/strong><\/h3>\n\n\n\n<p>Reaction 3 refers to the dissociation of HMO:<\/p>\n\n\n\n<p><strong>HMO \u21cc H\u207a + MO\u207b<\/strong><\/p>\n\n\n\n<p>When [OH\u207b] increases, [H\u207a] is reduced as OH\u207b neutralizes H\u207a. The equilibrium shifts <strong>to the right<\/strong> to compensate, increasing the dissociation of HMO.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">4. <strong>How does this affect HMO dissociation?<\/strong><\/h3>\n\n\n\n<p>With reduced [H\u207a], the system favors more dissociation of HMO to restore equilibrium. Thus, <strong>[HMO] decreases<\/strong>, <strong>[MO\u207b] increases<\/strong>, and the color changes accordingly\u2014turning yellow.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Summary:<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>HCl adds H\u207a<\/strong>, shifts equilibrium <strong>left<\/strong>, forms more <strong>HMO<\/strong>, color turns <strong>red<\/strong>.<\/li>\n\n\n\n<li><strong>NaOH adds OH\u207b<\/strong>, removes H\u207a, shifts equilibrium <strong>right<\/strong>, forms more <strong>MO\u207b<\/strong>, color turns <strong>yellow<\/strong>.<\/li>\n\n\n\n<li>Increasing [OH\u207b] promotes more <strong>HMO dissociation<\/strong> by lowering [H\u207a], shifting equilibrium to the right.<\/li>\n<\/ul>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner6-151.jpeg\" alt=\"\" class=\"wp-image-229716\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Advance Study Assignment: Properties of Systems in Chemical Equilibrium Methyl orange, HMO, is a common acid-base indicator. In solution, it ionizes according to the equation: HMO(aq) \u00e2\u2020\u2019 H+(aq) + MO-(aq) If methyl orange is added to distilled water, the solution turns yellow. If a drop or two of 6 M HCl is added to the [&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-229714","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/229714","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=229714"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/229714\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=229714"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=229714"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=229714"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}