{"id":248974,"date":"2025-07-08T20:54:52","date_gmt":"2025-07-08T20:54:52","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=248974"},"modified":"2025-07-08T20:54:54","modified_gmt":"2025-07-08T20:54:54","slug":"nh3-is-a-weak-base-kb1-8x10%e2%88%925-and-so-the-salt-nh4cl-acts-as-a-weak-acid","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/08\/nh3-is-a-weak-base-kb1-8x10%e2%88%925-and-so-the-salt-nh4cl-acts-as-a-weak-acid\/","title":{"rendered":"NH3\u00a0is a weak base (Kb=1.8\u00d710\u22125) and so the salt NH4Cl acts as a weak acid"},"content":{"rendered":"\n<pre id=\"preorder-ask-header-text\" class=\"wp-block-preformatted\">NH3&nbsp;is a weak base (Kb=1.8\u00d710\u22125) and so the salt NH4Cl acts as a weak acid<\/pre>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/07\/image-331.png\" alt=\"\" class=\"wp-image-248975\"\/><\/figure>\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><strong>5.26<\/strong><\/p>\n\n\n\n<p>To determine the pH of the 0.054 M ammonium chloride (NH\u2084Cl) solution, we first need to understand the chemical reaction occurring in the water.<\/p>\n\n\n\n<p>Ammonium chloride is a salt formed from a weak base, ammonia (NH\u2083), and a strong acid, hydrochloric acid (HCl). When dissolved in water, it dissociates completely into its ions:<br>NH\u2084Cl(aq) \u2192 NH\u2084\u207a(aq) + Cl\u207b(aq)<\/p>\n\n\n\n<p>The chloride ion (Cl\u207b) is the conjugate base of a strong acid, so it is a spectator ion and does not affect the pH. The ammonium ion (NH\u2084\u207a), however, is the conjugate acid of the weak base ammonia. It will react with water in a process called hydrolysis, which makes the solution acidic. The equilibrium reaction is:<br>NH\u2084\u207a(aq) + H\u2082O(l) \u21cc NH\u2083(aq) + H\u2083O\u207a(aq)<\/p>\n\n\n\n<p>To find the pH, we need the concentration of hydronium ions [H\u2083O\u207a] at equilibrium. This requires the acid dissociation constant, Ka, for the ammonium ion. We can calculate Ka from the given base dissociation constant, Kb, for ammonia (1.8 \u00d7 10\u207b\u2075) and the ion product constant for water, Kw (1.0 \u00d7 10\u207b\u00b9\u2074 at 25 \u00b0C), using the relationship:<br>Ka \u00d7 Kb = Kw<br>Ka = Kw \/ Kb = (1.0 \u00d7 10\u207b\u00b9\u2074) \/ (1.8 \u00d7 10\u207b\u2075) = 5.56 \u00d7 10\u207b\u00b9\u2070<\/p>\n\n\n\n<p>Next, we use an ICE (Initial, Change, Equilibrium) table to find the equilibrium concentrations. Let &#8216;x&#8217; be the change in concentration.<br>Initial: [NH\u2084\u207a] = 0.054 M; [NH\u2083] = 0; [H\u2083O\u207a] = 0<br>Change: [NH\u2084\u207a] = -x; [NH\u2083] = +x; [H\u2083O\u207a] = +x<br>Equilibrium: [NH\u2084\u207a] = 0.054 &#8211; x; [NH\u2083] = x; [H\u2083O\u207a] = x<\/p>\n\n\n\n<p>The Ka expression is:<br>Ka = ([NH\u2083][H\u2083O\u207a]) \/ [NH\u2084\u207a]<br>5.56 \u00d7 10\u207b\u00b9\u2070 = (x)(x) \/ (0.054 &#8211; x)<\/p>\n\n\n\n<p>Since Ka is very small, we can assume x is negligible compared to 0.054, simplifying the equation to:<br>5.56 \u00d7 10\u207b\u00b9\u2070 \u2248 x\u00b2 \/ 0.054<br>x\u00b2 = (5.56 \u00d7 10\u207b\u00b9\u2070) * 0.054 = 3.00 \u00d7 10\u207b\u00b9\u00b9<br>x = \u221a(3.00 \u00d7 10\u207b\u00b9\u00b9) = 5.48 \u00d7 10\u207b\u2076 M<\/p>\n\n\n\n<p>So, the equilibrium concentration of H\u2083O\u207a is 5.48 \u00d7 10\u207b\u2076 M. Finally, we calculate the pH:<br>pH = -log[H\u2083O\u207a] = -log(5.48 \u00d7 10\u207b\u2076) = 5.26<\/p>\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-banner8-361.jpeg\" alt=\"\" class=\"wp-image-248976\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>NH3&nbsp;is a weak base (Kb=1.8\u00d710\u22125) and so the salt NH4Cl acts as a weak acid The Correct Answer and Explanation is: 5.26 To determine the pH of the 0.054 M ammonium chloride (NH\u2084Cl) solution, we first need to understand the chemical reaction occurring in the water. Ammonium chloride is a salt formed from a weak [&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-248974","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/248974","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=248974"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/248974\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=248974"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=248974"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=248974"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}