Chloride in serum is determined by titration with Hg(NO3)2; 2Cl\u2212 + Hg2+ HgCl2. The Hg(NO3)2 is standardized by titrating 2.00 mL of a 0.0108 M NaCl solution, requiring 1.12 mL to reach the diphenylcarbazone end point. A 0.500-mL serum sample is treated with 3.50 mL water, 0.50 mL 10% sodium tungstate solution, and 0.50 mL of 0.33 M H2SO4 solution to precipitate proteins. After the proteins are precipitated, the sample is filtered through a dry filter into a dry flask. A 2.00-mL aliquot of the filtrate is titrated with the Hg(NO3)2 solution, requiring 1.23 mL. Calculate the mg\/L chloride in the serum. (Note: mercury is rarely used today due to its toxicity. The problem is illustrative.)<\/p>\n\n\n\n
The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n To calculate the chloride concentration in the serum sample, we need to follow several steps, using the information provided in the problem. Let’s break it down into logical steps:<\/p>\n\n\n\n We start by determining the concentration of the Hg(NO3)2 solution used to titrate the chloride ions in the serum.<\/p>\n\n\n\n Next, we calculate the chloride concentration in the serum.<\/p>\n\n\n\n Now we need to calculate the chloride concentration in the entire serum sample, which was diluted before titration.<\/p>\n\n\n\n The chloride concentration in the serum sample is 843.7 mg\/L<\/strong>.<\/p>\n\n\n\n In this problem, the key steps involved standardizing the titrant (Hg(NO3)2) using a known chloride solution (NaCl) and then using the standardized titrant to determine the chloride concentration in a diluted serum sample. By using stoichiometry based on the titration reactions, we calculated the chloride concentration in the original serum sample and converted it to mg\/L.<\/p>\n","protected":false},"excerpt":{"rendered":" Chloride in serum is determined by titration with Hg(NO3)2; 2Cl\u2212 + Hg2+ HgCl2. The Hg(NO3)2 is standardized by titrating 2.00 mL of a 0.0108 M NaCl solution, requiring 1.12 mL to reach the diphenylcarbazone end point. A 0.500-mL serum sample is treated with 3.50 mL water, 0.50 mL 10% sodium tungstate solution, and 0.50 mL […]<\/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-188369","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/188369","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=188369"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/188369\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=188369"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=188369"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=188369"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Step 1: Standardization of the Hg(NO3)2 Solution<\/h3>\n\n\n\n
\n
\n
\n
[
\\text{moles of Cl}^{-} = \\text{molarity} \\times \\text{volume (in L)}
]
[
\\text{moles of Cl}^{-} = 0.0108 \\, \\text{mol\/L} \\times 2.00 \\, \\text{mL} \\times \\left( \\frac{1}{1000} \\right) \\, \\text{L\/mL} = 0.0000216 \\, \\text{mol}
]<\/li>\n\n\n\n
According to the reaction, 1 mole of Hg2+ reacts with 2 moles of Cl-. Therefore, the moles of Hg2+ are half the moles of chloride:
[
\\text{moles of Hg}^{2+} = \\frac{0.0000216}{2} = 0.0000108 \\, \\text{mol}
]<\/li>\n\n\n\n
The volume of Hg(NO3)2 used is 1.12 mL = 0.00112 L. So, the molarity of Hg(NO3)2 is:
[
M_{\\text{Hg(NO3)2}} = \\frac{\\text{moles of Hg}^{2+}}{\\text{volume of Hg(NO3)2 in L}}
]
[
M_{\\text{Hg(NO3)2}} = \\frac{0.0000108 \\, \\text{mol}}{0.00112 \\, \\text{L}} = 0.00964 \\, \\text{M}
]<\/li>\n<\/ol>\n\n\n\nStep 2: Titration of the Serum Sample<\/h3>\n\n\n\n
\n
\n
\n
[
\\text{moles of Hg}^{2+} = M_{\\text{Hg(NO3)2}} \\times \\text{volume (in L)}
]
[
\\text{moles of Hg}^{2+} = 0.00964 \\, \\text{mol\/L} \\times 1.23 \\, \\text{mL} \\times \\left( \\frac{1}{1000} \\right) \\, \\text{L\/mL} = 0.0000119 \\, \\text{mol}
]<\/li>\n\n\n\n
Since 1 mole of Hg2+ reacts with 2 moles of Cl-, the moles of chloride are double the moles of Hg2+:
[
\\text{moles of Cl}^{-} = 2 \\times 0.0000119 \\, \\text{mol} = 0.0000238 \\, \\text{mol}
]<\/li>\n\n\n\n
[
\\text{molarity of Cl}^{-} = \\frac{\\text{moles of Cl}^{-}}{\\text{volume of aliquot in L}}
]
[
\\text{molarity of Cl}^{-} = \\frac{0.0000238 \\, \\text{mol}}{2.00 \\, \\text{mL} \\times \\left( \\frac{1}{1000} \\right)} = 0.0119 \\, \\text{M}
]<\/li>\n<\/ol>\n\n\n\nStep 3: Determining the Chloride Concentration in the Serum<\/h3>\n\n\n\n
\n
[
\\text{dilution factor} = \\frac{\\text{total volume after dilution}}{\\text{aliquot volume}} = \\frac{0.500 + 3.50}{2.00} = \\frac{4.00}{2.00} = 2
]<\/li>\n\n\n\n
The chloride concentration in the 2.00 mL aliquot is 0.0119 M. Therefore, the chloride concentration in the original 0.500 mL serum sample is:
[
\\text{Concentration in serum} = 0.0119 \\, \\text{M} \\times 2 = 0.0238 \\, \\text{M}
]<\/li>\n\n\n\n
The molar mass of chloride (Cl-) is 35.45 g\/mol. To convert molarity to mg\/L:
[
\\text{Concentration in mg\/L} = 0.0238 \\, \\text{mol\/L} \\times 35.45 \\, \\text{g\/mol} \\times 1000 \\, \\text{mg\/g} = 843.7 \\, \\text{mg\/L}
]<\/li>\n<\/ol>\n\n\n\nFinal Answer:<\/h3>\n\n\n\n
Explanation:<\/h3>\n\n\n\n