To calculate the acid dissociation constant (KaK_a) for a monoprotic acid, we can follow these steps:<\/p>\n\n\n\n
\n\n\n\n
Given:<\/strong><\/h3>\n\n\n\n\n- Concentration of acid [HA]=0.0192\u2009M[HA] = 0.0192 \\, \\text{M}<\/li>\n\n\n\n
- pH = 2.53<\/li>\n<\/ul>\n\n\n\n
\n\n\n\nStep 1: Find [H+][H^+]<\/strong><\/h3>\n\n\n\n[H+]=10\u2212pH=10\u22122.53\u22482.95\u00d710\u22123\u2009M[H^+] = 10^{-\\text{pH}} = 10^{-2.53} \\approx 2.95 \\times 10^{-3} \\, \\text{M}<\/p>\n\n\n\n
\n\n\n\nStep 2: Set up the equilibrium expression<\/strong><\/h3>\n\n\n\nFor a monoprotic weak acid HAHA: HA\u21ccH++A\u2212HA \\rightleftharpoons H^+ + A^-<\/p>\n\n\n\n
Initial concentration:<\/p>\n\n\n\n
\n- [HA]=0.0192[HA] = 0.0192 M<\/li>\n\n\n\n
- [H+]=0[H^+] = 0, [A\u2212]=0[A^-] = 0<\/li>\n<\/ul>\n\n\n\n
Change:<\/p>\n\n\n\n
\n- [HA]=0.0192\u2212x[HA] = 0.0192 – x, [H+]=x[H^+] = x, [A\u2212]=x[A^-] = x<\/li>\n<\/ul>\n\n\n\n
From Step 1, x=[H+]=2.95\u00d710\u22123x = [H^+] = 2.95 \\times 10^{-3}<\/p>\n\n\n\n
\n\n\n\nStep 3: Use the Ka expression<\/strong><\/h3>\n\n\n\nKa=[H+][A\u2212][HA]=(2.95\u00d710\u22123)20.0192\u22122.95\u00d710\u22123K_a = \\frac{[H^+][A^-]}{[HA]} = \\frac{(2.95 \\times 10^{-3})^2}{0.0192 – 2.95 \\times 10^{-3}} Ka=8.70\u00d710\u221260.01625\u22485.36\u00d710\u22124K_a = \\frac{8.70 \\times 10^{-6}}{0.01625} \\approx 5.36 \\times 10^{-4}<\/p>\n\n\n\n
\n\n\n\n\u2705 Final Answer:<\/strong><\/h3>\n\n\n\nKa\u22485.36\u00d710\u22124K_a \\approx 5.36 \\times 10^{-4}<\/p>\n\n\n\n
\n\n\n\nExplanation<\/strong><\/h3>\n\n\n\nIn aqueous solutions, weak acids like monoprotic acids only partially dissociate. The strength of a weak acid is quantified using its acid dissociation constant (KaK_a). A higher KaK_a means a stronger acid that ionizes more in solution. In this problem, we are given the initial concentration of the monoprotic acid (HAHA) and the pH of the solution. From the pH, we can determine the concentration of hydrogen ions ([H+][H^+]), which directly results from the dissociation of the acid.<\/p>\n\n\n\n
Since the acid is monoprotic (releases only one proton per molecule), each molecule of HAHA that dissociates produces one hydrogen ion and one conjugate base (A\u2212A^-). Thus, the [H+][H^+] is equal to the amount of acid dissociated, represented as xx in the equilibrium expression.<\/p>\n\n\n\n
Once we have [H+][H^+], we subtract it from the initial acid concentration to get the equilibrium concentration of undissociated HAHA. Plugging these values into the KaK_a expression: Ka=[H+][A\u2212][HA]K_a = \\frac{[H^+][A^-]}{[HA]}<\/p>\n\n\n\n
we find the numerical value of KaK_a.<\/p>\n\n\n\n
The calculation assumes the solution behaves ideally and no significant secondary equilibria are present. The result, Ka\u22485.36\u00d710\u22124K_a \\approx 5.36 \\times 10^{-4}, indicates a weak acid (since strong acids have much larger KaK_a values). This approach is fundamental in analytical chemistry, especially when studying buffer systems, titrations, and the behavior of weak acids in biological or environmental systems.<\/p>\n\n\n\n![\"\"](\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/05\/learnexams-banner5-136.jpeg\")
<\/figure>\n","protected":false},"excerpt":{"rendered":"Enough of a monoprotic acid is dissolved in water to produce a 0.0192 M solution. The pH of the resulting solution is 2.53. Calculate the Ka for the acid. The Correct Answer and Explanation is: To calculate the acid dissociation constant (KaK_a) for a monoprotic acid, we can follow these steps: Given: Step 1: Find […]<\/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-221682","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/221682","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=221682"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/221682\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=221682"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=221682"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=221682"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\n\n\n\n
Step 1: Find [H+][H^+]<\/strong><\/h3>\n\n\n\n[H+]=10\u2212pH=10\u22122.53\u22482.95\u00d710\u22123\u2009M[H^+] = 10^{-\\text{pH}} = 10^{-2.53} \\approx 2.95 \\times 10^{-3} \\, \\text{M}<\/p>\n\n\n\n
\n\n\n\nStep 2: Set up the equilibrium expression<\/strong><\/h3>\n\n\n\nFor a monoprotic weak acid HAHA: HA\u21ccH++A\u2212HA \\rightleftharpoons H^+ + A^-<\/p>\n\n\n\n
Initial concentration:<\/p>\n\n\n\n
\n- [HA]=0.0192[HA] = 0.0192 M<\/li>\n\n\n\n
- [H+]=0[H^+] = 0, [A\u2212]=0[A^-] = 0<\/li>\n<\/ul>\n\n\n\n
Change:<\/p>\n\n\n\n
\n- [HA]=0.0192\u2212x[HA] = 0.0192 – x, [H+]=x[H^+] = x, [A\u2212]=x[A^-] = x<\/li>\n<\/ul>\n\n\n\n
From Step 1, x=[H+]=2.95\u00d710\u22123x = [H^+] = 2.95 \\times 10^{-3}<\/p>\n\n\n\n
\n\n\n\nStep 3: Use the Ka expression<\/strong><\/h3>\n\n\n\nKa=[H+][A\u2212][HA]=(2.95\u00d710\u22123)20.0192\u22122.95\u00d710\u22123K_a = \\frac{[H^+][A^-]}{[HA]} = \\frac{(2.95 \\times 10^{-3})^2}{0.0192 – 2.95 \\times 10^{-3}} Ka=8.70\u00d710\u221260.01625\u22485.36\u00d710\u22124K_a = \\frac{8.70 \\times 10^{-6}}{0.01625} \\approx 5.36 \\times 10^{-4}<\/p>\n\n\n\n
\n\n\n\n\u2705 Final Answer:<\/strong><\/h3>\n\n\n\nKa\u22485.36\u00d710\u22124K_a \\approx 5.36 \\times 10^{-4}<\/p>\n\n\n\n
\n\n\n\nExplanation<\/strong><\/h3>\n\n\n\nIn aqueous solutions, weak acids like monoprotic acids only partially dissociate. The strength of a weak acid is quantified using its acid dissociation constant (KaK_a). A higher KaK_a means a stronger acid that ionizes more in solution. In this problem, we are given the initial concentration of the monoprotic acid (HAHA) and the pH of the solution. From the pH, we can determine the concentration of hydrogen ions ([H+][H^+]), which directly results from the dissociation of the acid.<\/p>\n\n\n\n
Since the acid is monoprotic (releases only one proton per molecule), each molecule of HAHA that dissociates produces one hydrogen ion and one conjugate base (A\u2212A^-). Thus, the [H+][H^+] is equal to the amount of acid dissociated, represented as xx in the equilibrium expression.<\/p>\n\n\n\n
Once we have [H+][H^+], we subtract it from the initial acid concentration to get the equilibrium concentration of undissociated HAHA. Plugging these values into the KaK_a expression: Ka=[H+][A\u2212][HA]K_a = \\frac{[H^+][A^-]}{[HA]}<\/p>\n\n\n\n
we find the numerical value of KaK_a.<\/p>\n\n\n\n
The calculation assumes the solution behaves ideally and no significant secondary equilibria are present. The result, Ka\u22485.36\u00d710\u22124K_a \\approx 5.36 \\times 10^{-4}, indicates a weak acid (since strong acids have much larger KaK_a values). This approach is fundamental in analytical chemistry, especially when studying buffer systems, titrations, and the behavior of weak acids in biological or environmental systems.<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"Enough of a monoprotic acid is dissolved in water to produce a 0.0192 M solution. The pH of the resulting solution is 2.53. Calculate the Ka for the acid. The Correct Answer and Explanation is: To calculate the acid dissociation constant (KaK_a) for a monoprotic acid, we can follow these steps: Given: Step 1: Find […]<\/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-221682","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/221682","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=221682"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/221682\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=221682"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=221682"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=221682"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
For a monoprotic weak acid HAHA: HA\u21ccH++A\u2212HA \\rightleftharpoons H^+ + A^-<\/p>\n\n\n\n
Initial concentration:<\/p>\n\n\n\n
- \n
- [HA]=0.0192[HA] = 0.0192 M<\/li>\n\n\n\n
- [H+]=0[H^+] = 0, [A\u2212]=0[A^-] = 0<\/li>\n<\/ul>\n\n\n\n
Change:<\/p>\n\n\n\n
- \n
- [HA]=0.0192\u2212x[HA] = 0.0192 – x, [H+]=x[H^+] = x, [A\u2212]=x[A^-] = x<\/li>\n<\/ul>\n\n\n\n
From Step 1, x=[H+]=2.95\u00d710\u22123x = [H^+] = 2.95 \\times 10^{-3}<\/p>\n\n\n\n
\n\n\n\nStep 3: Use the Ka expression<\/strong><\/h3>\n\n\n\n
Ka=[H+][A\u2212][HA]=(2.95\u00d710\u22123)20.0192\u22122.95\u00d710\u22123K_a = \\frac{[H^+][A^-]}{[HA]} = \\frac{(2.95 \\times 10^{-3})^2}{0.0192 – 2.95 \\times 10^{-3}} Ka=8.70\u00d710\u221260.01625\u22485.36\u00d710\u22124K_a = \\frac{8.70 \\times 10^{-6}}{0.01625} \\approx 5.36 \\times 10^{-4}<\/p>\n\n\n\n
\n\n\n\n\u2705 Final Answer:<\/strong><\/h3>\n\n\n\n
Ka\u22485.36\u00d710\u22124K_a \\approx 5.36 \\times 10^{-4}<\/p>\n\n\n\n
\n\n\n\nExplanation<\/strong><\/h3>\n\n\n\n
In aqueous solutions, weak acids like monoprotic acids only partially dissociate. The strength of a weak acid is quantified using its acid dissociation constant (KaK_a). A higher KaK_a means a stronger acid that ionizes more in solution. In this problem, we are given the initial concentration of the monoprotic acid (HAHA) and the pH of the solution. From the pH, we can determine the concentration of hydrogen ions ([H+][H^+]), which directly results from the dissociation of the acid.<\/p>\n\n\n\n
Since the acid is monoprotic (releases only one proton per molecule), each molecule of HAHA that dissociates produces one hydrogen ion and one conjugate base (A\u2212A^-). Thus, the [H+][H^+] is equal to the amount of acid dissociated, represented as xx in the equilibrium expression.<\/p>\n\n\n\n
Once we have [H+][H^+], we subtract it from the initial acid concentration to get the equilibrium concentration of undissociated HAHA. Plugging these values into the KaK_a expression: Ka=[H+][A\u2212][HA]K_a = \\frac{[H^+][A^-]}{[HA]}<\/p>\n\n\n\n
we find the numerical value of KaK_a.<\/p>\n\n\n\n
The calculation assumes the solution behaves ideally and no significant secondary equilibria are present. The result, Ka\u22485.36\u00d710\u22124K_a \\approx 5.36 \\times 10^{-4}, indicates a weak acid (since strong acids have much larger KaK_a values). This approach is fundamental in analytical chemistry, especially when studying buffer systems, titrations, and the behavior of weak acids in biological or environmental systems.<\/p>\n\n\n\n
<\/figure>\n","protected":false},"excerpt":{"rendered":"Enough of a monoprotic acid is dissolved in water to produce a 0.0192 M solution. The pH of the resulting solution is 2.53. Calculate the Ka for the acid. The Correct Answer and Explanation is: To calculate the acid dissociation constant (KaK_a) for a monoprotic acid, we can follow these steps: Given: Step 1: Find […]<\/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-221682","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/221682","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=221682"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/221682\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=221682"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=221682"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=221682"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
- [HA]=0.0192\u2212x[HA] = 0.0192 – x, [H+]=x[H^+] = x, [A\u2212]=x[A^-] = x<\/li>\n<\/ul>\n\n\n\n