{"id":248979,"date":"2025-07-08T20:56:54","date_gmt":"2025-07-08T20:56:54","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=248979"},"modified":"2025-07-08T20:56:56","modified_gmt":"2025-07-08T20:56:56","slug":"nh3-is-a-weak-base-kb-1-8-a-10ae5-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-kb-1-8-a-10ae5-and-so-the-salt-nh4cl-acts-as-a-weak-acid\/","title":{"rendered":"NH3 is a weak base (Kb = 1.8 \u00c3\u2014 10^\u00e2\u20ac\u201c5) and so the salt NH4Cl acts as a weak acid."},"content":{"rendered":"\n

NH3 is a weak base (Kb = 1.8 \u00c3\u2014 10^\u00e2\u20ac\u201c5) and so the salt NH4Cl acts as a weak acid. What is the pH of a solution that is 0.038 M in NH4Cl at 25 \u00c2\u00b0C?<\/p>\n\n\n\n

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

To calculate the pH of a 0.038 M solution of NH4Cl, we first need to understand that NH4Cl dissociates in water to form NH4\u207a and Cl\u207b ions. The NH4\u207a ion is a weak acid, and it will hydrolyze in water to produce NH3 (a weak base) and H\u207a ions.<\/p>\n\n\n\n

Step 1: Set up the equilibrium expression for NH4\u207a dissociation.<\/strong>
The dissociation of NH4\u207a in water can be represented by:<\/p>\n\n\n\n

NH4++H2O\u21ccNH3+H3O+NH_4^+ + H_2O \\rightleftharpoons NH_3 + H_3O^+NH4+\u200b+H2\u200bO\u21ccNH3\u200b+H3\u200bO+<\/p>\n\n\n\n

The equilibrium constant for this reaction is the Ka<\/strong> for NH4\u207a. Since NH4Cl is the salt of NH3, we can find the Ka using the relationship between the Ka of NH4\u207a and the Kb of NH3:<\/p>\n\n\n\n

Kw=Ka\u00d7KbK_w = K_a \\times K_bKw\u200b=Ka\u200b\u00d7Kb\u200b<\/p>\n\n\n\n

where K_w = 1\u00d710\u2212141 \\times 10^{-14}1\u00d710\u221214 at 25\u00b0C, and Kb=1.8\u00d710\u22125K_b = 1.8 \\times 10^{-5}Kb\u200b=1.8\u00d710\u22125 for NH3. Rearranging, we can solve for Ka:<\/p>\n\n\n\n

Ka=KwKb=1\u00d710\u2212141.8\u00d710\u22125=5.56\u00d710\u221210K_a = \\frac{K_w}{K_b} = \\frac{1 \\times 10^{-14}}{1.8 \\times 10^{-5}} = 5.56 \\times 10^{-10}Ka\u200b=Kb\u200bKw\u200b\u200b=1.8\u00d710\u221251\u00d710\u221214\u200b=5.56\u00d710\u221210<\/p>\n\n\n\n

Step 2: Set up an ICE table for the dissociation of NH4\u207a.<\/strong>
The initial concentration of NH4Cl is 0.038 M, and since it dissociates 1:1, the initial concentration of NH4\u207a is also 0.038 M.<\/p>\n\n\n\n

Species<\/th>Initial (M)<\/th>Change (M)<\/th>Equilibrium (M)<\/th><\/tr><\/thead>
NH4\u207a<\/td>0.038<\/td>-x<\/td>0.038 – x<\/td><\/tr>
H\u207a (H3O\u207a)<\/td>0<\/td>+x<\/td>x<\/td><\/tr>
NH3<\/td>0<\/td>+x<\/td>x<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

Using the Ka expression for the dissociation of NH4\u207a:<\/p>\n\n\n\n

Ka=[NH3][H3O+][NH4+]=x20.038\u2212xK_a = \\frac{[NH_3][H_3O^+]}{[NH_4^+]} = \\frac{x^2}{0.038 – x}Ka\u200b=[NH4+\u200b][NH3\u200b][H3\u200bO+]\u200b=0.038\u2212xx2\u200b<\/p>\n\n\n\n

Since Ka is very small, we can assume that x will be much smaller than 0.038 M, so 0.038\u2212x\u22480.0380.038 – x \\approx 0.0380.038\u2212x\u22480.038. This simplifies the equation to:<\/p>\n\n\n\n

5.56\u00d710\u221210=x20.0385.56 \\times 10^{-10} = \\frac{x^2}{0.038}5.56\u00d710\u221210=0.038×2\u200b<\/p>\n\n\n\n

Step 3: Solve for x.<\/strong>
Rearranging the equation to solve for x:<\/p>\n\n\n\n

x2=(5.56\u00d710\u221210)\u00d70.038x^2 = (5.56 \\times 10^{-10}) \\times 0.038×2=(5.56\u00d710\u221210)\u00d70.038<\/p>\n\n\n\n

x2=2.11\u00d710\u221211x^2 = 2.11 \\times 10^{-11}x2=2.11\u00d710\u221211<\/p>\n\n\n\n

Taking the square root of both sides:<\/p>\n\n\n\n

x=2.11\u00d710\u221211x = \\sqrt{2.11 \\times 10^{-11}}x=2.11\u00d710\u221211\u200b<\/p>\n\n\n\n

x=1.45\u00d710\u22125x = 1.45 \\times 10^{-5}x=1.45\u00d710\u22125<\/p>\n\n\n\n

This value of x represents the concentration of H\u207a ions in the solution.<\/p>\n\n\n\n

Step 4: Calculate the pH.<\/strong>
The pH is given by:<\/p>\n\n\n\n

pH=\u2212log\u2061[H+]pH = -\\log[H^+]pH=\u2212log[H+]<\/p>\n\n\n\n

Substituting the value of x:<\/p>\n\n\n\n

pH=\u2212log\u2061(1.45\u00d710\u22125)pH = -\\log(1.45 \\times 10^{-5})pH=\u2212log(1.45\u00d710\u22125)<\/p>\n\n\n\n

pH=4.84pH = 4.84pH=4.84<\/p>\n\n\n\n

Therefore, the pH of the 0.038 M NH4Cl solution is 4.84<\/strong>.<\/p>\n\n\n\n

Explanation:<\/h3>\n\n\n\n

The weak acid behavior of NH4\u207a in water leads to the formation of hydronium ions (H\u2083O\u207a), lowering the pH below 7. The pH calculated here reflects the weak acidic nature of the NH4\u207a ion in solution.<\/p>\n\n\n\n

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

NH3 is a weak base (Kb = 1.8 \u00c3\u2014 10^\u00e2\u20ac\u201c5) and so the salt NH4Cl acts as a weak acid. What is the pH of a solution that is 0.038 M in NH4Cl at 25 \u00c2\u00b0C? The Correct Answer and Explanation is: To calculate the pH of a 0.038 M solution of NH4Cl, we first […]<\/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-248979","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/248979","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=248979"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/248979\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=248979"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=248979"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=248979"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}