Some K2SO3 and KHSO3 are dissolved in 150.0 mL of solution and the resulting pH is 7.40. Which is greater in this buffer solution, the concentration of SO32 or the concentration of HSO3? Ka (HSO3) = 1.0 \u00d7 10-7 Oconcentration of SO32- concentration of HSO3 Oboth are equal If [SO32] [HSO3] = = 0.60 M in this solution, calculate the concentration of HSO3 M<\/p>\n\n\n\n
The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n To solve this, we need to evaluate the relationship between the concentrations of ( \\text{SO}_3^{2-} ) and ( \\text{HSO}_3^- ) in a buffer solution, given the pH and the dissociation constant ( K_a ) of ( \\text{HSO}_3^- ).<\/p>\n\n\n\n The dissociation of ( \\text{HSO}_3^- ) in water is:<\/p>\n\n\n\n [ The acid dissociation constant ( K_a ) for ( \\text{HSO}_3^- ) is given as ( 1.0 \\times 10^{-7} ).<\/p>\n\n\n\n [ The pH of the solution is given as 7.40. To find the concentration of hydrogen ions, we use the equation:<\/p>\n\n\n\n [ Rearranging to solve for ( [\\text{H}^+] ):<\/p>\n\n\n\n [ We are given that the concentrations of ( \\text{SO}_3^{2-} ) and ( \\text{HSO}_3^- ) are equal, both at 0.60 M, initially.<\/p>\n\n\n\n At equilibrium, using the equation for ( K_a ), we have:<\/p>\n\n\n\n [ Substituting the known values:<\/p>\n\n\n\n [ Solving for ( [\\text{HSO}_3^-] ):<\/p>\n\n\n\n [ From the calculation, the concentration of ( \\text{SO}_3^{2-} ) (0.60 M) is greater than the concentration of ( \\text{HSO}_3^- ) (2.4 \u00d7 10\u207b\u2078 M). Therefore, the concentration of ( \\text{SO}_3^{2-} ) is greater than the concentration of ( \\text{HSO}_3^- ).<\/p>\n\n\n\n The concentration of ( \\text{SO}_3^{2-} ) is greater in this buffer solution.<\/p>\n\n\n\n Explanation<\/strong>:<\/p>\n\n\n\n Some K2SO3 and KHSO3 are dissolved in 150.0 mL of solution and the resulting pH is 7.40. Which is greater in this buffer solution, the concentration of SO32 or the concentration of HSO3? Ka (HSO3) = 1.0 \u00d7 10-7 Oconcentration of SO32- concentration of HSO3 Oboth are equal If [SO32] [HSO3] = = 0.60 M […]<\/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-186652","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/186652","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=186652"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/186652\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=186652"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=186652"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=186652"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Step 1: Write the equilibrium expression<\/h3>\n\n\n\n
\\text{HSO}_3^- \\rightleftharpoons \\text{H}^+ + \\text{SO}_3^{2-}
]<\/p>\n\n\n\n
K_a = \\frac{[\\text{H}^+][\\text{SO}_3^{2-}]}{[\\text{HSO}_3^-]}
]<\/p>\n\n\n\nStep 2: Use the pH to find ([H^+])<\/h3>\n\n\n\n
\\text{pH} = -\\log[\\text{H}^+]
]<\/p>\n\n\n\n
[\\text{H}^+] = 10^{-\\text{pH}} = 10^{-7.40} \\approx 4.0 \\times 10^{-8} \\, \\text{M}
]<\/p>\n\n\n\nStep 3: Set up the equilibrium expression using the concentrations<\/h3>\n\n\n\n
K_a = \\frac{[\\text{H}^+][\\text{SO}_3^{2-}]}{[\\text{HSO}_3^-]}
]<\/p>\n\n\n\n
1.0 \\times 10^{-7} = \\frac{(4.0 \\times 10^{-8})(0.60)}{[\\text{HSO}_3^-]}
]<\/p>\n\n\n\n
[\\text{HSO}_3^-] = \\frac{(4.0 \\times 10^{-8})(0.60)}{1.0 \\times 10^{-7}} = 2.4 \\times 10^{-8} \\, \\text{M}
]<\/p>\n\n\n\nStep 4: Conclusion<\/h3>\n\n\n\n
Final Answer:<\/h3>\n\n\n\n
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