{"id":248720,"date":"2025-07-08T17:31:41","date_gmt":"2025-07-08T17:31:41","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=248720"},"modified":"2025-07-08T17:31:43","modified_gmt":"2025-07-08T17:31:43","slug":"the-value-of-ka-for-nitrous-acid-hno2-at-25ac-is-4-5a-10-4-3","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-10-4-3\/","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 dissociation of nitrous acid (HNO\u2082) in water is represented by the equation:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>HNO\u2082(aq) \u21cc H\u207a(aq) + NO\u2082\u207b(aq)<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">This is the correct equilibrium equation because Ka represents the dissociation constant for acids, and the equation must show the acid dissociating into hydrogen ions (H\u207a) and the conjugate base (NO\u2082\u207b). Therefore, option <strong>2<\/strong> is the correct choice.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">b. Calculation of \u0394G\u00b0<\/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 at 25\u00b0C, we can use the following formula: \u0394G\u00b0=\u2212RTln\u2061Ka\\Delta G\u00b0 = -RT \\ln K_a\u0394G\u00b0=\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 = 8.314 J\/mol\u00b7K (universal gas constant)<\/li>\n\n\n\n<li>TTT = 298 K (temperature at 25\u00b0C)<\/li>\n\n\n\n<li>KaK_aKa\u200b = 4.5\u00d710\u221244.5 \\times 10^{-4}4.5\u00d710\u22124 (given)<\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\">Substitute the values into the equation: \u0394G\u00b0=\u2212(8.314\u2009J\/mol\\cdotpK)\u00d7(298\u2009K)\u00d7ln\u2061(4.5\u00d710\u22124)\\Delta G\u00b0 = &#8211; (8.314 \\, \\text{J\/mol\u00b7K}) \\times (298 \\, \\text{K}) \\times \\ln(4.5 \\times 10^{-4})\u0394G\u00b0=\u2212(8.314J\/mol\\cdotpK)\u00d7(298K)\u00d7ln(4.5\u00d710\u22124)<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Now calculate: \u0394G\u00b0=\u22128.314\u00d7298\u00d7ln\u2061(4.5\u00d710\u22124)\\Delta G\u00b0 = &#8211; 8.314 \\times 298 \\times \\ln(4.5 \\times 10^{-4})\u0394G\u00b0=\u22128.314\u00d7298\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)\u2248\u22127.803\\ln(4.5 \\times 10^{-4}) \u2248 -7.803ln(4.5\u00d710\u22124)\u2248\u22127.803<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Now, calculate: \u0394G\u00b0=\u22128.314\u00d7298\u00d7(\u22127.803)\\Delta G\u00b0 = &#8211; 8.314 \\times 298 \\times (-7.803)\u0394G\u00b0=\u22128.314\u00d7298\u00d7(\u22127.803) \u0394G\u00b0\u224819378.6\u2009J\/mol=19.38\u2009kJ\/mol\\Delta G\u00b0 \u2248 19378.6 \\, \\text{J\/mol} = 19.38 \\, \\text{kJ\/mol}\u0394G\u00b0\u224819378.6J\/mol=19.38kJ\/mol<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">So, <strong>\u0394G\u00b0 \u2248 19.38 kJ\/mol<\/strong>.<\/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 change in Gibbs free energy (\u0394G) is <strong>zero<\/strong> because the system is in a state of no net change: \u0394G=0\\Delta G = 0\u0394G=0<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Therefore, the value of \u0394G at equilibrium is <strong>0 J\/mol<\/strong>.<\/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 find the value of \u0394G under non-standard conditions, we use the following equation: \u0394G=\u0394G\u00b0+RTln\u2061Q\\Delta G = \\Delta G\u00b0 + RT \\ln Q\u0394G=\u0394G\u00b0+RTlnQ<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>QQQ is the reaction quotient, given by:<\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\">Q=[H+][NO2\u2212][HNO2]Q = \\frac{[H\u207a][NO\u2082\u207b]}{[HNO\u2082]}Q=[HNO2\u200b][H+][NO2\u2212\u200b]\u200b<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Substitute the concentrations: Q=(5.9\u00d710\u22122)(6.7\u00d710\u22124)0.21\u22481.87\u00d710\u22124Q = \\frac{(5.9 \\times 10^{-2})(6.7 \\times 10^{-4})}{0.21} \u2248 1.87 \\times 10^{-4}Q=0.21(5.9\u00d710\u22122)(6.7\u00d710\u22124)\u200b\u22481.87\u00d710\u22124<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Now, substitute the known values into the \u0394G equation: \u0394G=19.38\u2009kJ\/mol+(8.314\u2009J\/mol\\cdotpK)\u00d7(298\u2009K)\u00d7ln\u2061(1.87\u00d710\u22124)\\Delta G = 19.38 \\, \\text{kJ\/mol} + (8.314 \\, \\text{J\/mol\u00b7K}) \\times (298 \\, \\text{K}) \\times \\ln(1.87 \\times 10^{-4})\u0394G=19.38kJ\/mol+(8.314J\/mol\\cdotpK)\u00d7(298K)\u00d7ln(1.87\u00d710\u22124)<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Calculate the natural logarithm: ln\u2061(1.87\u00d710\u22124)\u2248\u22128.591\\ln(1.87 \\times 10^{-4}) \u2248 -8.591ln(1.87\u00d710\u22124)\u2248\u22128.591<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Now, calculate: \u0394G=19380\u2009J\/mol+(8.314\u00d7298)\u00d7(\u22128.591)\\Delta G = 19380 \\, \\text{J\/mol} + (8.314 \\times 298) \\times (-8.591)\u0394G=19380J\/mol+(8.314\u00d7298)\u00d7(\u22128.591) \u0394G=19380+(\u221221163.4)\\Delta G = 19380 + (-21163.4)\u0394G=19380+(\u221221163.4) \u0394G\u2248\u221217783.4\u2009J\/mol=\u221217.78\u2009kJ\/mol\\Delta G \u2248 -17783.4 \\, \\text{J\/mol} = -17.78 \\, \\text{kJ\/mol}\u0394G\u2248\u221217783.4J\/mol=\u221217.78kJ\/mol<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">So, <strong>\u0394G \u2248 -17.78 kJ\/mol<\/strong>.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Summary of Answers:<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>a.<\/strong> The correct equilibrium equation is: <strong>HNO\u2082(aq) \u21cc H\u207a(aq) + NO\u2082\u207b(aq)<\/strong><\/li>\n\n\n\n<li><strong>b.<\/strong> \u0394G\u00b0 \u2248 19.38 kJ\/mol<\/li>\n\n\n\n<li><strong>c.<\/strong> \u0394G at equilibrium = 0 J\/mol<\/li>\n\n\n\n<li><strong>d.<\/strong> \u0394G \u2248 -17.78 kJ\/mol at the given concentrations.<\/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-banner8-301.jpeg\" alt=\"\" class=\"wp-image-248721\"\/><\/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-248720","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/248720","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=248720"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/248720\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=248720"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=248720"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=248720"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}