{"id":187297,"date":"2025-02-04T06:10:08","date_gmt":"2025-02-04T06:10:08","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=187297"},"modified":"2025-02-04T06:10:10","modified_gmt":"2025-02-04T06:10:10","slug":"an-edta-solution-was-prepared-by-dissolving-approximately-4-g-of-the-disodium-salt-in-approximately-1-l-of-water","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/02\/04\/an-edta-solution-was-prepared-by-dissolving-approximately-4-g-of-the-disodium-salt-in-approximately-1-l-of-water\/","title":{"rendered":"An EDTA solution was prepared by dissolving approximately 4 g of the disodium salt in approximately 1 L of water"},"content":{"rendered":"\n<p>An EDTA solution was prepared by dissolving approximately 4 g of the disodium salt in approximately 1 L of water. An average of 42.35 mL of this solution was required to titrate 50.00-mL aliquots of a standard that contained 0.7682 g of MgCO3 per liter. Titration of a 25.00-mL sample of mineral water at pH 10 required 18.81 mL of the EDTA solution. A 50.00-mL aliquot of the mineral water was rendered strongly alkaline to precipitate the magnesium at Mg(OH)2. Titration with a calcium-specific indicator required 31.54 mL of the EDTA solution. Calculate<\/p>\n\n\n\n<p>(a) the molar concentration of the EDTA solution.<\/p>\n\n\n\n<p>(b) the concentration of CaCO3 in the mineral water in ppm.<\/p>\n\n\n\n<p>(c) the concentration of MgCO3 in the mineral water in ppm.<\/p>\n\n\n\n<p><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-6-color\"><strong>The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n<p>To determine the concentrations of calcium carbonate (CaCO\u2083) and magnesium carbonate (MgCO\u2083) in the mineral water sample, we can follow these steps:<\/p>\n\n\n\n<p><strong>(a) Molar Concentration of the EDTA Solution<\/strong><\/p>\n\n\n\n<p>First, we standardize the EDTA solution using a magnesium carbonate (MgCO\u2083) standard.<\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Calculate the molarity of the MgCO\u2083 standard solution:<\/strong><\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Molar mass of MgCO\u2083: 24.305 (Mg) + 12.011 (C) + 3\u00d715.999 (O) = 84.313 g\/mol<\/li>\n\n\n\n<li>Concentration of MgCO\u2083 standard: 0.7682 g\/L<\/li>\n\n\n\n<li>Moles of MgCO\u2083 per liter: 0.7682 g\/L \u00f7 84.313 g\/mol \u2248 0.00911 mol\/L<\/li>\n<\/ul>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Determine the moles of MgCO\u2083 in a 50.00 mL aliquot:<\/strong><\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Volume in liters: 50.00 mL \u00d7 (1 L \/ 1000 mL) = 0.05000 L<\/li>\n\n\n\n<li>Moles of MgCO\u2083: 0.00911 mol\/L \u00d7 0.05000 L \u2248 0.0004555 mol<\/li>\n<\/ul>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Since the reaction between Mg\u00b2\u207a and EDTA is 1:1, moles of EDTA required = moles of MgCO\u2083:<\/strong><\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Moles of EDTA: 0.0004555 mol<\/li>\n<\/ul>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Calculate the molarity of the EDTA solution:<\/strong><\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Volume of EDTA solution used: 42.35 mL \u00d7 (1 L \/ 1000 mL) = 0.04235 L<\/li>\n\n\n\n<li>Molarity of EDTA: 0.0004555 mol \u00f7 0.04235 L \u2248 0.01075 M<\/li>\n<\/ul>\n\n\n\n<p><strong>(b) Concentration of CaCO\u2083 in the Mineral Water in ppm<\/strong><\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Titrate a 25.00 mL sample of mineral water at pH 10:<\/strong><\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Volume of EDTA used: 18.81 mL<\/li>\n<\/ul>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Calculate moles of EDTA used:<\/strong><\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Volume in liters: 18.81 mL \u00d7 (1 L \/ 1000 mL) = 0.01881 L<\/li>\n\n\n\n<li>Moles of EDTA: 0.01075 M \u00d7 0.01881 L \u2248 0.0002022 mol<\/li>\n<\/ul>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Assuming all hardness is due to Ca\u00b2\u207a (since Mg\u00b2\u207a is precipitated at pH 10), moles of Ca\u00b2\u207a = moles of EDTA:<\/strong><\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Moles of Ca\u00b2\u207a: 0.0002022 mol<\/li>\n<\/ul>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Calculate the mass of CaCO\u2083 equivalent:<\/strong><\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Molar mass of CaCO\u2083: 40.078 (Ca) + 12.011 (C) + 3\u00d715.999 (O) = 100.086 g\/mol<\/li>\n\n\n\n<li>Mass of CaCO\u2083: 0.0002022 mol \u00d7 100.086 g\/mol \u2248 0.02023 g<\/li>\n<\/ul>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Convert to ppm (mg\/L):<\/strong><\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Since 25.00 mL of sample was used, to find the concentration in ppm:<\/li>\n\n\n\n<li>ppm CaCO\u2083: 0.02023 g \u00d7 (1000 mg \/ 1 g) \u00f7 0.025 L \u2248 809.2 ppm<\/li>\n<\/ul>\n\n\n\n<p><strong>(c) Concentration of MgCO\u2083 in the Mineral Water in ppm<\/strong><\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Titrate a 50.00 mL aliquot of mineral water after precipitating Mg\u00b2\u207a as Mg(OH)\u2082:<\/strong><\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Volume of EDTA used: 31.54 mL<\/li>\n<\/ul>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Calculate moles of EDTA used:<\/strong><\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Volume in liters: 31.54 mL \u00d7 (1 L \/ 1000 mL) = 0.03154 L<\/li>\n\n\n\n<li>Moles of EDTA: 0.01075 M \u00d7 0.03154 L \u2248 0.0003392 mol<\/li>\n<\/ul>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Assuming the titration now measures only Ca\u00b2\u207a (since Mg\u00b2\u207a was precipitated), moles of Ca\u00b2\u207a = moles of EDTA:<\/strong><\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Moles of Ca\u00b2\u207a: 0.0003392 mol<\/li>\n<\/ul>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Calculate the mass of CaCO\u2083 equivalent:<\/strong><\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Mass of CaCO\u2083: 0.0003392 mol \u00d7 100.086 g\/mol \u2248 0.03394 g<\/li>\n<\/ul>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Convert to ppm (mg\/L):<\/strong><\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Since 50.00 mL of sample was used:<\/li>\n\n\n\n<li>ppm CaCO\u2083: 0.03394 g \u00d7 (1000 mg \/ 1 g) \u00f7 0.050 L \u2248 678.8 ppm<\/li>\n<\/ul>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Determine the MgCO\u2083 concentration:<\/strong><\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Total hardness (from part b) is due to both Ca\u00b2\u207a and Mg\u00b2\u207a.<\/li>\n\n\n\n<li>Hardness due to Mg\u00b2\u207a: Total hardness &#8211; Ca\u00b2\u207a hardness<\/li>\n\n\n\n<li>Mg\u00b2\u207a hardness in ppm CaCO\u2083: 809.2 ppm &#8211; 678.8 ppm \u2248 130.4 ppm<\/li>\n<\/ul>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Convert ppm CaCO\u2083 to ppm MgCO\u2083:<\/strong><\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Molar mass ratio: MgCO\u2083 \/ CaCO\u2083 = 84.313 g\/mol \/ 100.086 g\/mol \u2248 0.8425<\/li>\n\n\n\n<li>ppm MgCO\u2083: 130.4 ppm \u00d7 0.8425 \u2248 109.9 ppm<\/li>\n<\/ul>\n\n\n\n<p><strong>Summary:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Molarity of EDTA solution:<\/strong> \u2248 0.01075 M<\/li>\n\n\n\n<li><strong>Concentration of CaCO\u2083 in mineral water:<\/strong> \u2248 809.2 ppm<\/li>\n\n\n\n<li><strong>Concentration of MgCO\u2083 in mineral water:<\/strong> \u2248 109.9 ppm<\/li>\n<\/ul>\n\n\n\n<p>These calculations are based on the assumption that at pH 10, Mg\u00b2\u207a is precipitated as Mg(OH)\u2082, allowing for the selective titration of Ca\u00b2\u207a. The total hardness is first determined, followed by the specific hardness due to Ca\u00b2\u207a, and the difference is attributed to Mg\u00b2\u207a. Converting these values to their respective carbonate forms provides the concentrations in ppm.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>An EDTA solution was prepared by dissolving approximately 4 g of the disodium salt in approximately 1 L of water. An average of 42.35 mL of this solution was required to titrate 50.00-mL aliquots of a standard that contained 0.7682 g of MgCO3 per liter. Titration of a 25.00-mL sample of mineral water at pH [&hellip;]<\/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-187297","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/187297","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=187297"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/187297\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=187297"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=187297"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=187297"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}