What is the pH of a 0.250 M solution of NH4Br, given that the Ka of NH4+ is 5.6 x 10^-10?<\/p>\n\n\n\n
The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n To calculate the pH of a 0.250 M solution of ammonium bromide (NH\u2084Br), we first need to understand the dissociation and behavior of the ammonium ion (NH\u2084\u207a) in water. Since ammonium bromide is a salt formed from the weak base ammonia (NH\u2083) and a strong acid (HBr), it dissociates in water to produce NH\u2084\u207a and Br\u207b ions. The Br\u207b ion, being the conjugate base of a strong acid, does not affect the pH, while NH\u2084\u207a is acidic and will affect the pH of the solution.<\/p>\n\n\n\n Ammonium ion (NH\u2084\u207a) is a weak acid and partially dissociates in water:<\/p>\n\n\n\n [ The dissociation of NH\u2084\u207a can be represented by the acid dissociation constant (Ka) expression:<\/p>\n\n\n\n [ We are given that the Ka for NH\u2084\u207a is (5.6 \\times 10^{-10}), and the initial concentration of NH\u2084\u207a is 0.250 M.<\/p>\n\n\n\n Let (x) represent the concentration of H\u207a ions that dissociate from NH\u2084\u207a:<\/p>\n\n\n\n Substitute the equilibrium concentrations into the Ka expression:<\/p>\n\n\n\n [ Since Ka is very small, we can assume that (x) is much smaller than 0.250, so we approximate (0.250 – x \\approx 0.250). Thus, the equation becomes:<\/p>\n\n\n\n [ Solve for (x):<\/p>\n\n\n\n [ Since (x) represents the concentration of H\u207a ions, the pH is:<\/p>\n\n\n\n [ The pH of the 0.250 M NH\u2084Br solution is approximately 4.93<\/strong>.<\/p>\n\n\n\n Ammonium bromide (NH\u2084Br) dissociates in water to produce NH\u2084\u207a, which is a weak acid. The Ka value for NH\u2084\u207a indicates that it dissociates to a small extent, producing H\u207a ions in the solution, which results in a slightly acidic pH. Using the provided Ka and the initial concentration, we used an ICE table and approximations to calculate the H\u207a concentration, and from that, the pH of the solution.<\/p>\n","protected":false},"excerpt":{"rendered":" What is the pH of a 0.250 M solution of NH4Br, given that the Ka of NH4+ is 5.6 x 10^-10? The Correct Answer and Explanation is : To calculate the pH of a 0.250 M solution of ammonium bromide (NH\u2084Br), we first need to understand the dissociation and behavior of the ammonium ion (NH\u2084\u207a) […]<\/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-169202","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/169202","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=169202"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/169202\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=169202"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=169202"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=169202"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Step 1: Write the dissociation equation for NH\u2084\u207a<\/h3>\n\n\n\n
NH\u2084\u207a \\rightleftharpoons NH\u2083 + H\u207a
]<\/p>\n\n\n\nStep 2: Set up the equilibrium expression<\/h3>\n\n\n\n
K_a = \\frac{[NH\u2083][H\u207a]}{[NH\u2084\u207a]}
]<\/p>\n\n\n\nStep 3: Set up an ICE table<\/h3>\n\n\n\n
Species<\/th> Initial (M)<\/th> Change (M)<\/th> Equilibrium (M)<\/th><\/tr><\/thead> NH\u2084\u207a<\/td> 0.250<\/td> -x<\/td> 0.250 – x<\/td><\/tr> NH\u2083<\/td> 0<\/td> +x<\/td> x<\/td><\/tr> H\u207a<\/td> 0<\/td> +x<\/td> x<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n Step 4: Substitute into the Ka expression<\/h3>\n\n\n\n
5.6 \\times 10^{-10} = \\frac{x \\cdot x}{0.250 – x}
]<\/p>\n\n\n\n
5.6 \\times 10^{-10} = \\frac{x^2}{0.250}
]<\/p>\n\n\n\n
x^2 = (5.6 \\times 10^{-10}) \\times 0.250
]
[
x^2 = 1.4 \\times 10^{-10}
]
[
x = \\sqrt{1.4 \\times 10^{-10}} = 1.18 \\times 10^{-5} \\, \\text{M}
]<\/p>\n\n\n\nStep 5: Calculate pH<\/h3>\n\n\n\n
\\text{pH} = -\\log[H\u207a] = -\\log(1.18 \\times 10^{-5}) \\approx 4.93
]<\/p>\n\n\n\nFinal Answer:<\/h3>\n\n\n\n
Explanation:<\/h3>\n\n\n\n