Find the pH of 0.2 M a weak acid (HA) with Ka = 8.80 x 10-6.<\/p>\n\n\n\n
The correct answer and explanation is :<\/strong><\/mark><\/p>\n\n\n\n To find the pH of a 0.2 M solution of a weak acid (HA) with a given acid dissociation constant ( K_a = 8.80 \\times 10^{-6} ), we follow these steps:<\/p>\n\n\n\n The weak acid dissociates in water as follows:<\/p>\n\n\n\n [ Let\u2019s define the initial concentration of HA as 0.2 M and assume that the concentration of ( H^+ ) and ( A^- ) starts at 0.<\/p>\n\n\n\n [ Substituting the values:<\/p>\n\n\n\n [ Since ( K_a ) is small, we assume ( x ) is much smaller than 0.2, so ( 0.2 – x \\approx 0.2 ):<\/p>\n\n\n\n [ [ [ [ [ Since ( x = [H^+] ), we find the pH:<\/p>\n\n\n\n [ [ This calculation follows a systematic approach to finding the pH of a weak acid. By using the ICE table and assuming ( x ) is small compared to the initial concentration, we simplify the quadratic equation, making it easier to solve. The small ( K_a ) confirms that HA only partially dissociates, leading to a pH greater than that of a strong acid of the same concentration.<\/p>\n","protected":false},"excerpt":{"rendered":" Find the pH of 0.2 M a weak acid (HA) with Ka = 8.80 x 10-6. The correct answer and explanation is : To find the pH of a 0.2 M solution of a weak acid (HA) with a given acid dissociation constant ( K_a = 8.80 \\times 10^{-6} ), we follow these steps: Step […]<\/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-204852","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/204852","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=204852"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/204852\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=204852"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=204852"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=204852"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Step 1: Write the Ionization Equation<\/h3>\n\n\n\n
HA \\rightleftharpoons H^+ + A^-
]<\/p>\n\n\n\nStep 2: Set Up the ICE Table<\/h3>\n\n\n\n
Species<\/th> Initial (M)<\/th> Change (M)<\/th> Equilibrium (M)<\/th><\/tr><\/thead> HA<\/td> 0.2<\/td> -x<\/td> 0.2 – x<\/td><\/tr> ( H^+ )<\/td> 0<\/td> +x<\/td> x<\/td><\/tr> ( A^- )<\/td> 0<\/td> +x<\/td> x<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n Step 3: Apply the Ka Expression<\/h3>\n\n\n\n
K_a = \\frac{[H^+][A^-]}{[HA]}
]<\/p>\n\n\n\n
8.80 \\times 10^{-6} = \\frac{x^2}{0.2 – x}
]<\/p>\n\n\n\n
8.80 \\times 10^{-6} = \\frac{x^2}{0.2}
]<\/p>\n\n\n\nStep 4: Solve for ( x ) (Hydrogen Ion Concentration)<\/h3>\n\n\n\n
x^2 = (8.80 \\times 10^{-6}) \\times 0.2
]<\/p>\n\n\n\n
x^2 = 1.76 \\times 10^{-6}
]<\/p>\n\n\n\n
x = \\sqrt{1.76 \\times 10^{-6}}
]<\/p>\n\n\n\n
x = 1.33 \\times 10^{-3} \\text{ M}
]<\/p>\n\n\n\n
\\text{pH} = -\\log(1.33 \\times 10^{-3})
]<\/p>\n\n\n\n
\\text{pH} \\approx 2.88
]<\/p>\n\n\n\nExplanation:<\/h3>\n\n\n\n