{"id":248678,"date":"2025-07-08T15:04:58","date_gmt":"2025-07-08T15:04:58","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=248678"},"modified":"2025-07-08T15:05:00","modified_gmt":"2025-07-08T15:05:00","slug":"the-value-of-ka-for-nitrous-acid-hno2-at-25ac-is-4-5a-10-4-2","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/08\/the-value-of-ka-for-nitrous-acid-hno2-at-25ac-is-4-5a-10-4-2\/","title":{"rendered":"The value of Ka for nitrous acid (HNO2) at 25\u00c2\u00b0C is 4.5\u00c3\u201410^(-4)."},"content":{"rendered":"\n<p class=\"wp-block-paragraph\">The value of Ka for nitrous acid (HNO2) at 25\u00c2\u00b0C is 4.5\u00c3\u201410^(-4). a. Write the chemical equation for the equilibrium that corresponds to Ka: 1. H+(aq) + NO2-(aq) \u00e2\u2021\u0152 HNO2(aq) 2. HNO2(aq) \u00e2\u2021\u0152 H+(aq) + NO2-(aq) 3. HNO2(aq) \u00e2\u2021\u0152 H-(aq) + NO2+(aq) 4. HNO2(aq) + H+(aq) \u00e2\u2021\u0152 H2NO2+(aq) 5. HNO2(aq) + H-(aq) \u00e2\u2021\u0152 H2NO2+(aq) b. By using the value of Ka, calculate \u00ce\u201dG\u00c2\u00b0 for the dissociation of nitrous acid in aqueous solution. c. What is the value of \u00ce\u201dG at equilibrium? d. What is the value of \u00ce\u201dG when [H+] = 5.9\u00c3\u201410^(-2) M, [NO2-] = 6.7\u00c3\u201410^(-4) M, and [HNO2] = 0.21 M?<\/p>\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<h3 class=\"wp-block-heading\">a. Chemical Equation for Ka<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">The equilibrium constant KaK_aKa\u200b for nitrous acid (HNO\u2082) represents the dissociation of HNO\u2082 into its ions. The correct equilibrium equation for the dissociation of nitrous acid is:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">HNO2(aq)\u21ccH+(aq)+NO2\u2212(aq)\\text{HNO}_2(aq) \\rightleftharpoons \\text{H}^+(aq) + \\text{NO}_2^-(aq)HNO2\u200b(aq)\u21ccH+(aq)+NO2\u2212\u200b(aq)<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">So, the correct option is:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>2. HNO\u2082(aq) \u21cc H\u207a(aq) + NO\u2082\u207b(aq)<\/strong><\/p>\n\n\n\n<h3 class=\"wp-block-heading\">b. Calculation of \u0394G\u00b0 for the Dissociation of Nitrous Acid<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">To calculate the standard Gibbs free energy change (\u0394G\u00b0) for the dissociation of nitrous acid, we can use the following relationship between KaK_aKa\u200b and \u0394G\u2218\\Delta G^\\circ\u0394G\u2218:\u0394G\u2218=\u2212RTln\u2061Ka\\Delta G^\\circ = -RT \\ln K_a\u0394G\u2218=\u2212RTlnKa\u200b<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>RRR is the gas constant = 8.314 J\/(mol\u00b7K)<\/li>\n\n\n\n<li>TTT is the temperature in Kelvin (given as 25\u00b0C, so T=298\u2009KT = 298 \\, \\text{K}T=298K)<\/li>\n\n\n\n<li>KaK_aKa\u200b is the equilibrium constant for the dissociation of nitrous acid, given as 4.5\u00d710\u221244.5 \\times 10^{-4}4.5\u00d710\u22124<\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\">Substituting the values:\u0394G\u2218=\u2212(8.314)\u00d7(298)\u00d7ln\u2061(4.5\u00d710\u22124)\\Delta G^\\circ = &#8211; (8.314) \\times (298) \\times \\ln(4.5 \\times 10^{-4})\u0394G\u2218=\u2212(8.314)\u00d7(298)\u00d7ln(4.5\u00d710\u22124)<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">First, calculate the natural logarithm:ln\u2061(4.5\u00d710\u22124)=ln\u2061(4.5)+ln\u2061(10\u22124)=1.5041\u22129.2103=\u22127.7062\\ln(4.5 \\times 10^{-4}) = \\ln(4.5) + \\ln(10^{-4}) = 1.5041 &#8211; 9.2103 = -7.7062ln(4.5\u00d710\u22124)=ln(4.5)+ln(10\u22124)=1.5041\u22129.2103=\u22127.7062<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Now calculate \u0394G\u2218\\Delta G^\\circ\u0394G\u2218:\u0394G\u2218=\u2212(8.314)\u00d7(298)\u00d7(\u22127.7062)\\Delta G^\\circ = &#8211; (8.314) \\times (298) \\times (-7.7062)\u0394G\u2218=\u2212(8.314)\u00d7(298)\u00d7(\u22127.7062)\u0394G\u2218=19104.5\u2009J\/mol\u224819.1\u2009kJ\/mol\\Delta G^\\circ = 19104.5 \\, \\text{J\/mol} \\approx 19.1 \\, \\text{kJ\/mol}\u0394G\u2218=19104.5J\/mol\u224819.1kJ\/mol<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">So, \u0394G\u2218\u224819.1\u2009kJ\/mol\\Delta G^\\circ \\approx 19.1 \\, \\text{kJ\/mol}\u0394G\u2218\u224819.1kJ\/mol.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">c. Value of \u0394G at Equilibrium<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">At equilibrium, the Gibbs free energy change \u0394G\\Delta G\u0394G is zero because the system has reached a state of balance between the forward and reverse reactions. Therefore:\u0394G=0\u2009kJ\/mol\\Delta G = 0 \\, \\text{kJ\/mol}\u0394G=0kJ\/mol<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">d. Calculation of \u0394G when [H\u207a] = 5.9 \u00d7 10\u207b\u00b2 M, [NO\u2082\u207b] = 6.7 \u00d7 10\u207b\u2074 M, and [HNO\u2082] = 0.21 M<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">To calculate the Gibbs free energy change at non-standard conditions, we use the equation:\u0394G=\u0394G\u2218+RTln\u2061Q\\Delta G = \\Delta G^\\circ + RT \\ln Q\u0394G=\u0394G\u2218+RTlnQ<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>\u0394G\u2218\\Delta G^\\circ\u0394G\u2218 is the standard Gibbs free energy change (19.1 kJ\/mol)<\/li>\n\n\n\n<li>RRR is the gas constant (8.314 J\/mol\u00b7K)<\/li>\n\n\n\n<li>TTT is the temperature in Kelvin (298 K)<\/li>\n\n\n\n<li>QQQ is the reaction quotient, calculated as:<\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\">Q=[H+][NO2\u2212][HNO2]Q = \\frac{[\\text{H}^+][\\text{NO}_2^-]}{[\\text{HNO}_2]}Q=[HNO2\u200b][H+][NO2\u2212\u200b]\u200b<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Substituting the given concentrations:Q=(5.9\u00d710\u22122)(6.7\u00d710\u22124)0.21=3.953\u00d710\u221250.21=1.88\u00d710\u22124Q = \\frac{(5.9 \\times 10^{-2})(6.7 \\times 10^{-4})}{0.21} = \\frac{3.953 \\times 10^{-5}}{0.21} = 1.88 \\times 10^{-4}Q=0.21(5.9\u00d710\u22122)(6.7\u00d710\u22124)\u200b=0.213.953\u00d710\u22125\u200b=1.88\u00d710\u22124<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Now calculate \u0394G\\Delta G\u0394G:\u0394G=19104.5+(8.314)(298)ln\u2061(1.88\u00d710\u22124)\\Delta G = 19104.5 + (8.314)(298) \\ln(1.88 \\times 10^{-4})\u0394G=19104.5+(8.314)(298)ln(1.88\u00d710\u22124)<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">First, calculate the natural logarithm:ln\u2061(1.88\u00d710\u22124)=ln\u2061(1.88)+ln\u2061(10\u22124)=0.6290\u22129.6439=\u22129.0149\\ln(1.88 \\times 10^{-4}) = \\ln(1.88) + \\ln(10^{-4}) = 0.6290 &#8211; 9.6439 = -9.0149ln(1.88\u00d710\u22124)=ln(1.88)+ln(10\u22124)=0.6290\u22129.6439=\u22129.0149<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Now calculate \u0394G\\Delta G\u0394G:\u0394G=19104.5+(8.314)(298)(\u22129.0149)=19104.5\u221222229.7=\u22123125.2\u2009J\/mol\\Delta G = 19104.5 + (8.314)(298)(-9.0149) = 19104.5 &#8211; 22229.7 = -3125.2 \\, \\text{J\/mol}\u0394G=19104.5+(8.314)(298)(\u22129.0149)=19104.5\u221222229.7=\u22123125.2J\/mol<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">So, \u0394G\u2248\u22123.13\u2009kJ\/mol\\Delta G \\approx -3.13 \\, \\text{kJ\/mol}\u0394G\u2248\u22123.13kJ\/mol.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Final Answers:<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>a. The correct chemical equation is <strong>HNO\u2082(aq) \u21cc H\u207a(aq) + NO\u2082\u207b(aq)<\/strong>.<\/li>\n\n\n\n<li>b. The standard Gibbs free energy change \u0394G\u2218\\Delta G^\\circ\u0394G\u2218 is approximately <strong>19.1 kJ\/mol<\/strong>.<\/li>\n\n\n\n<li>c. The Gibbs free energy at equilibrium is <strong>0 kJ\/mol<\/strong>.<\/li>\n\n\n\n<li>d. The Gibbs free energy at the given concentrations is approximately <strong>-3.13 kJ\/mol<\/strong>.<\/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\/07\/learnexams-banner8-291.jpeg\" alt=\"\" class=\"wp-image-248679\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>The value of Ka for nitrous acid (HNO2) at 25\u00c2\u00b0C is 4.5\u00c3\u201410^(-4). a. Write the chemical equation for the equilibrium that corresponds to Ka: 1. H+(aq) + NO2-(aq) \u00e2\u2021\u0152 HNO2(aq) 2. HNO2(aq) \u00e2\u2021\u0152 H+(aq) + NO2-(aq) 3. HNO2(aq) \u00e2\u2021\u0152 H-(aq) + NO2+(aq) 4. HNO2(aq) + H+(aq) \u00e2\u2021\u0152 H2NO2+(aq) 5. HNO2(aq) + H-(aq) \u00e2\u2021\u0152 H2NO2+(aq) b. [&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-248678","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/248678","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=248678"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/248678\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=248678"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=248678"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=248678"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}