{"id":231935,"date":"2025-06-11T14:54:21","date_gmt":"2025-06-11T14:54:21","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=231935"},"modified":"2025-06-11T14:54:24","modified_gmt":"2025-06-11T14:54:24","slug":"ammonium-chloride-nh4cl-is-an-acidic-salt","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/11\/ammonium-chloride-nh4cl-is-an-acidic-salt\/","title":{"rendered":"Ammonium chloride (NH4Cl) is an acidic salt."},"content":{"rendered":"\n

Ammonium chloride (NH4Cl) is an acidic salt. When ammonium chloride is dissolved in water, it dissociates into its component ions. This reaction goes to completion, as indicated by the one-way arrow in the following equation: NH4Cl(s) ==========> NH4+(aq) + Cl-(aq) The Cl- ion, being the conjugate base of a strong acid (HCl), is too weak to react with water. However, the NH4+, being the conjugate acid of a weak base (NH3), is strong enough to react slightly with water. The equation for this equilibrium reaction is: NH4+(aq) + H2O(l) <———-> H3O+(aq) + NH3(aq) The equilibrium constant for the above reaction (Ka for NH4+) is normally not published in tables because it can be calculated from two other values that ARE published: the ionization constants for H2O and NH3. 2H2O(l) <———-> H3O+(aq) + OH-(aq) Kw = 1.0 x 10^-14 NH3(aq) + H2O(l) <———-> NH4+(aq) + OH-(aq) Kb = 1.8 x 10^-5 By combining these two equations in the appropriate way, it is possible to obtain the desired reaction, and hence, its equilibrium constant. What is the pOH of a 3.25 M NH4Cl solution? answers: 2.33 4.37 6.28 7.52 9.63 Can someone please explain how to do this step by step? Thank you.<\/p>\n\n\n\n

The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n

Step 1: Understand the Chemistry<\/strong><\/h3>\n\n\n\n

Ammonium chloride (NH\u2084Cl) is a salt<\/strong> formed from a weak base (NH\u2083) and a strong acid (HCl). When it dissolves in water:NH\u2084Cl (s)\u2192NH\u2084\u207a (aq)+Cl\u207b (aq)\\text{NH\u2084Cl (s)} \\rightarrow \\text{NH\u2084\u207a (aq)} + \\text{Cl\u207b (aq)}NH\u2084Cl (s)\u2192NH\u2084\u207a (aq)+Cl\u207b (aq)<\/p>\n\n\n\n

    \n
  • The Cl\u207b<\/strong> ion is the conjugate base of a strong acid<\/strong> (HCl), so it does not hydrolyze<\/strong>.<\/li>\n\n\n\n
  • The NH\u2084\u207a<\/strong> ion is the conjugate acid of a weak base<\/strong> (NH\u2083), so it does hydrolyze<\/strong> and contributes to the acidity of the solution.<\/li>\n<\/ul>\n\n\n\n

    NH\u2084\u207a + H\u2082O\u2194H\u2083O\u207a + NH\u2083\\text{NH\u2084\u207a + H\u2082O} \\leftrightarrow \\text{H\u2083O\u207a + NH\u2083}NH\u2084\u207a + H\u2082O\u2194H\u2083O\u207a + NH\u2083<\/p>\n\n\n\n

    \n\n\n\n

    Step 2: Determine KaK_aKa\u200b for NH\u2084\u207a<\/strong><\/h3>\n\n\n\n

    We use the relationship:Ka\u00d7Kb=KwK_a \\times K_b = K_wKa\u200b\u00d7Kb\u200b=Kw\u200b<\/p>\n\n\n\n

    Given:<\/p>\n\n\n\n

      \n
    • KbK_bKb\u200b for NH\u2083 = 1.8\u00d710\u221251.8 \\times 10^{-5}1.8\u00d710\u22125<\/li>\n\n\n\n
    • KwK_wKw\u200b = 1.0\u00d710\u2212141.0 \\times 10^{-14}1.0\u00d710\u221214<\/li>\n<\/ul>\n\n\n\n

      Ka=KwKb=1.0\u00d710\u2212141.8\u00d710\u22125=5.56\u00d710\u221210K_a = \\frac{K_w}{K_b} = \\frac{1.0 \\times 10^{-14}}{1.8 \\times 10^{-5}} = 5.56 \\times 10^{-10}Ka\u200b=Kb\u200bKw\u200b\u200b=1.8\u00d710\u221251.0\u00d710\u221214\u200b=5.56\u00d710\u221210<\/p>\n\n\n\n

      \n\n\n\n

      Step 3: Set Up ICE Table<\/strong><\/h3>\n\n\n\n

      Initial concentration of NH\u2084\u207a = 3.25 M<\/p>\n\n\n\n

      Let xxx be the concentration of H\u2083O\u207a formed:NH\u2084\u207a+H\u2082O\u2194H\u2083O\u207a+NH\u2083Initial (M):3.2500Change (M):\u2212x+x+xEquilibrium (M):3.25\u2212xxx\\begin{align*} \\text{NH\u2084\u207a} + \\text{H\u2082O} &\\leftrightarrow \\text{H\u2083O\u207a} + \\text{NH\u2083} \\\\ \\text{Initial (M)}: &\\quad 3.25 \\quad 0 \\quad 0 \\\\ \\text{Change (M)}: &\\quad -x \\quad +x \\quad +x \\\\ \\text{Equilibrium (M)}: &\\quad 3.25 – x \\quad x \\quad x \\\\ \\end{align*}NH\u2084\u207a+H\u2082OInitial (M):Change (M):Equilibrium (M):\u200b\u2194H\u2083O\u207a+NH\u20833.2500\u2212x+x+x3.25\u2212xxx\u200b<\/p>\n\n\n\n

      Now apply KaK_aKa\u200b:Ka=x23.25\u2212x\u2248x23.25K_a = \\frac{x^2}{3.25 – x} \\approx \\frac{x^2}{3.25}Ka\u200b=3.25\u2212xx2\u200b\u22483.25×2\u200b5.56\u00d710\u221210=x23.25\u21d2x2=1.807\u00d710\u22129\u21d2x=1.807\u00d710\u22129=4.25\u00d710\u221255.56 \\times 10^{-10} = \\frac{x^2}{3.25} \\Rightarrow x^2 = 1.807 \\times 10^{-9} \\Rightarrow x = \\sqrt{1.807 \\times 10^{-9}} = 4.25 \\times 10^{-5}5.56\u00d710\u221210=3.25×2\u200b\u21d2x2=1.807\u00d710\u22129\u21d2x=1.807\u00d710\u22129\u200b=4.25\u00d710\u22125<\/p>\n\n\n\n

      \n\n\n\n

      Step 4: Calculate pH and pOH<\/strong><\/h3>\n\n\n\n

      [H\u2083O\u207a]=x=4.25\u00d710\u22125\u21d2pH=\u2212log\u2061(4.25\u00d710\u22125)=4.37\u21d2pOH=14\u22124.37=9.63[\\text{H\u2083O\u207a}] = x = 4.25 \\times 10^{-5} \\Rightarrow \\text{pH} = -\\log(4.25 \\times 10^{-5}) = 4.37 \\Rightarrow \\text{pOH} = 14 – 4.37 = 9.63[H\u2083O\u207a]=x=4.25\u00d710\u22125\u21d2pH=\u2212log(4.25\u00d710\u22125)=4.37\u21d2pOH=14\u22124.37=9.63<\/p>\n\n\n\n

      \n\n\n\n

      Final Answer: 9.63\\boxed{9.63}9.63\u200b<\/strong><\/h3>\n\n\n\n
      \n\n\n\n

      Explanation<\/strong><\/h3>\n\n\n\n

      When ammonium chloride (NH\u2084Cl) dissolves in water, it fully dissociates into ammonium ions (NH\u2084\u207a) and chloride ions (Cl\u207b). The chloride ion, as the conjugate base of a strong acid (HCl), is essentially inert in water and does not affect the pH. The ammonium ion, however, is the conjugate acid of the weak base ammonia (NH\u2083), and thus it undergoes a slight hydrolysis with water to produce hydronium ions (H\u2083O\u207a) and ammonia.<\/p>\n\n\n\n

      The acidity from NH\u2084\u207a hydrolysis is determined by its acid dissociation constant, KaK_aKa\u200b, which is not typically found in tables. However, since NH\u2084\u207a and NH\u2083 are conjugate acid-base pairs, KaK_aKa\u200b for NH\u2084\u207a can be calculated using the relationship Ka=KwKbK_a = \\frac{K_w}{K_b}Ka\u200b=Kb\u200bKw\u200b\u200b, where KbK_bKb\u200b is the base constant for ammonia and KwK_wKw\u200b is the ionization constant for water.<\/p>\n\n\n\n

      After calculating KaK_aKa\u200b, we set up an ICE table to determine the concentration of H\u2083O\u207a formed at equilibrium. This concentration allows us to find the pH, and subsequently the pOH using the relationship pH+pOH=14\\text{pH} + \\text{pOH} = 14pH+pOH=14. In a 3.25 M NH\u2084Cl solution, the final pOH comes out to 9.63, confirming that the solution is weakly acidic, as expected from a salt of a weak base and strong acid.<\/p>\n\n\n\n

      \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

      Ammonium chloride (NH4Cl) is an acidic salt. When ammonium chloride is dissolved in water, it dissociates into its component ions. This reaction goes to completion, as indicated by the one-way arrow in the following equation: NH4Cl(s) ==========> NH4+(aq) + Cl-(aq) The Cl- ion, being the conjugate base of a strong acid (HCl), is too weak […]<\/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-231935","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/231935","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=231935"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/231935\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=231935"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=231935"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=231935"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}