{"id":248674,"date":"2025-07-08T15:01:14","date_gmt":"2025-07-08T15:01:14","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=248674"},"modified":"2025-07-08T15:01:18","modified_gmt":"2025-07-08T15:01:18","slug":"the-value-of-ka-for-nitrous-acid-hno2-at-25ac-is-4-5a-10-4","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\/","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. Write the chemical equation for the equilibrium that corresponds to Ka:<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">The correct equation for the dissociation of nitrous acid (HNO2) is:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>2. HNO2(aq) \u21cc H+(aq) + NO2-(aq)<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">This represents the dissociation of nitrous acid into hydrogen ions (H+) and nitrite ions (NO2-), which is the equilibrium described by the acid dissociation constant (Ka).<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">b. Calculate \u0394G\u00b0 for the dissociation of nitrous acid in aqueous solution:<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">The relationship between the acid dissociation constant (Ka) and the standard Gibbs free energy change (\u0394G\u00b0) is given by the equation:\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 = 8.314 J\/mol\u00b7K (universal gas constant),<\/li>\n\n\n\n<li>TTT = 298 K (standard temperature in Kelvin),<\/li>\n\n\n\n<li>KaK_aKa\u200b = 4.5 \u00d7 10\u207b\u2074 (given Ka for nitrous acid).<\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\">Now, calculate \u0394G\u00b0:\u0394G\u2218=\u2212(8.314\u2009J\/mol\\cdotpK)\u00d7(298\u2009K)\u00d7ln\u2061(4.5\u00d710\u22124)\\Delta G^\\circ = &#8211; (8.314 \\, \\text{J\/mol\u00b7K}) \\times (298 \\, \\text{K}) \\times \\ln(4.5 \\times 10^{-4})\u0394G\u2218=\u2212(8.314J\/mol\\cdotpK)\u00d7(298K)\u00d7ln(4.5\u00d710\u22124)<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">First, calculate the natural logarithm of Ka:ln\u2061(4.5\u00d710\u22124)\u2248\u22127.81\\ln(4.5 \\times 10^{-4}) \\approx -7.81ln(4.5\u00d710\u22124)\u2248\u22127.81<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Now calculate \u0394G\u00b0:\u0394G\u2218=\u2212(8.314\u00d7298)\u00d7(\u22127.81)\u224819321.5\u2009J\/mol\u224819.3\u2009kJ\/mol\\Delta G^\\circ = &#8211; (8.314 \\times 298) \\times (-7.81) \\approx 19321.5 \\, \\text{J\/mol} \\approx 19.3 \\, \\text{kJ\/mol}\u0394G\u2218=\u2212(8.314\u00d7298)\u00d7(\u22127.81)\u224819321.5J\/mol\u224819.3kJ\/mol<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">So, the standard Gibbs free energy change (\u0394G\u00b0) for the dissociation of nitrous acid is approximately <strong>+19.3 kJ\/mol<\/strong>.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">c. What is the value of \u0394G at equilibrium?<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">At equilibrium, \u0394G = 0, as the system has reached a state where the forward and reverse reactions occur at the same rate, and no net change in concentrations happens.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">So, the value of <strong>\u0394G at equilibrium is 0<\/strong>.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">d. What is the value of \u0394G when [H+] = 5.9 \u00d7 10\u207b\u00b2 M, [NO2-] = 6.7 \u00d7 10\u207b\u2074 M, and [HNO2] = 0.21 M?<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">To calculate the value of \u0394G under 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>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^+][NO_2^-]}{[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)\u00d7(6.7\u00d710\u22124)0.21Q = \\frac{(5.9 \\times 10^{-2}) \\times (6.7 \\times 10^{-4})}{0.21}Q=0.21(5.9\u00d710\u22122)\u00d7(6.7\u00d710\u22124)\u200bQ\u22483.953\u00d710\u221250.21\u22481.88\u00d710\u22124Q \\approx \\frac{3.953 \\times 10^{-5}}{0.21} \\approx 1.88 \\times 10^{-4}Q\u22480.213.953\u00d710\u22125\u200b\u22481.88\u00d710\u22124<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Now, calculate \u0394G:\u0394G=19.3\u2009kJ\/mol+(8.314\u2009J\/mol\\cdotpK)\u00d7(298\u2009K)\u00d7ln\u2061(1.88\u00d710\u22124)\\Delta G = 19.3 \\, \\text{kJ\/mol} + (8.314 \\, \\text{J\/mol\u00b7K}) \\times (298 \\, \\text{K}) \\times \\ln(1.88 \\times 10^{-4})\u0394G=19.3kJ\/mol+(8.314J\/mol\\cdotpK)\u00d7(298K)\u00d7ln(1.88\u00d710\u22124)<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">First, calculate the natural logarithm of Q:ln\u2061(1.88\u00d710\u22124)\u2248\u22128.97\\ln(1.88 \\times 10^{-4}) \\approx -8.97ln(1.88\u00d710\u22124)\u2248\u22128.97<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Now calculate \u0394G:\u0394G=19321.5\u2009J\/mol+(8.314\u00d7298)\u00d7(\u22128.97)\\Delta G = 19321.5 \\, \\text{J\/mol} + (8.314 \\times 298) \\times (-8.97)\u0394G=19321.5J\/mol+(8.314\u00d7298)\u00d7(\u22128.97)\u0394G=19321.5\u2009J\/mol\u221222140.5\u2009J\/mol\u2248\u22122820\u2009J\/mol\u2248\u22122.82\u2009kJ\/mol\\Delta G = 19321.5 \\, \\text{J\/mol} &#8211; 22140.5 \\, \\text{J\/mol} \\approx -2820 \\, \\text{J\/mol} \\approx -2.82 \\, \\text{kJ\/mol}\u0394G=19321.5J\/mol\u221222140.5J\/mol\u2248\u22122820J\/mol\u2248\u22122.82kJ\/mol<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">So, the value of \u0394G when the concentrations are given is approximately <strong>-2.82 kJ\/mol<\/strong>.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Summary:<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>a.<\/strong> The correct equilibrium equation is <strong>HNO2(aq) \u21cc H+(aq) + NO2-(aq)<\/strong>.<\/li>\n\n\n\n<li><strong>b.<\/strong> \u0394G\u00b0 = <strong>+19.3 kJ\/mol<\/strong>.<\/li>\n\n\n\n<li><strong>c.<\/strong> \u0394G at equilibrium = <strong>0<\/strong>.<\/li>\n\n\n\n<li><strong>d.<\/strong> \u0394G at the given concentrations = <strong>-2.82 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-banner8-290.jpeg\" alt=\"\" class=\"wp-image-248675\"\/><\/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-248674","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/248674","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=248674"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/248674\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=248674"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=248674"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=248674"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}