{"id":244382,"date":"2025-07-05T05:15:23","date_gmt":"2025-07-05T05:15:23","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=244382"},"modified":"2025-07-05T05:15:25","modified_gmt":"2025-07-05T05:15:25","slug":"m1v1-n1-m2v2-n2-m1-molarity-of-edta-0-01-m-m2-molarity-of-bottled-water","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/05\/m1v1-n1-m2v2-n2-m1-molarity-of-edta-0-01-m-m2-molarity-of-bottled-water\/","title":{"rendered":"M1V1\/n1 = M2V2\/n2 M1 = molarity of EDTA = 0.01 M M2 = molarity of bottled water"},"content":{"rendered":"\n<p>M1V1\/n1 = M2V2\/n2 M1 = molarity of EDTA = 0.01 M M2 = molarity of bottled water = ? V1 = volume of titre used = 3.46 mL V2 = volume of bottled water used = 40 mL N1 = mole equivalence = 1 N2 = mole equivalence = 1 Calculate the hardness of water. Express the result in the form of Ca2+ ppm and CaCO3 ppm. Hardness is expressed as the equivalent amount of Ca2+ in molarity from titration.<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-1-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>To calculate the hardness of water based on the given formula, let&#8217;s break it down:<\/p>\n\n\n\n<p>We are given the following information:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>M1 = 0.01 M (molarity of EDTA)<\/li>\n\n\n\n<li>V1 = 3.46 mL (volume of EDTA used for titration)<\/li>\n\n\n\n<li>V2 = 40 mL (volume of bottled water used)<\/li>\n\n\n\n<li>N1 = 1 (mole equivalence of EDTA)<\/li>\n\n\n\n<li>N2 = 1 (mole equivalence of bottled water)<\/li>\n<\/ul>\n\n\n\n<p>The formula for the calculation of molarity (M) is:M1\u00d7V1N1=M2\u00d7V2N2\\frac{M1 \\times V1}{N1} = \\frac{M2 \\times V2}{N2}N1M1\u00d7V1\u200b=N2M2\u00d7V2\u200b<\/p>\n\n\n\n<p>We can rearrange the equation to solve for M2M2M2 (the molarity of bottled water):M2=M1\u00d7V1\u00d7N2V2\u00d7N1M2 = \\frac{M1 \\times V1 \\times N2}{V2 \\times N1}M2=V2\u00d7N1M1\u00d7V1\u00d7N2\u200b<\/p>\n\n\n\n<p>Now, substitute the given values:M2=0.01\u2009M\u00d73.46\u2009mL\u00d7140\u2009mL\u00d71M2 = \\frac{0.01 \\, \\text{M} \\times 3.46 \\, \\text{mL} \\times 1}{40 \\, \\text{mL} \\times 1}M2=40mL\u00d710.01M\u00d73.46mL\u00d71\u200bM2=0.01\u00d73.4640M2 = \\frac{0.01 \\times 3.46}{40}M2=400.01\u00d73.46\u200bM2=0.034640M2 = \\frac{0.0346}{40}M2=400.0346\u200bM2=0.000865\u2009MM2 = 0.000865 \\, \\text{M}M2=0.000865M<\/p>\n\n\n\n<p>So, the molarity of bottled water (M2) is <strong>0.000865 M<\/strong>.<\/p>\n\n\n\n<p>Next, to calculate the hardness of water in terms of Ca2+\\text{Ca}^{2+}Ca2+ in ppm (parts per million) and CaCO3\\text{CaCO}_3CaCO3\u200b in ppm, we need to relate the molarity of Ca2+\\text{Ca}^{2+}Ca2+ to the equivalent concentration of CaCO3\\text{CaCO}_3CaCO3\u200b.<\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Hardness as Ca\u00b2\u207a (ppm):<\/strong><br>The molarity of Ca2+\\text{Ca}^{2+}Ca2+ is equivalent to the molarity of EDTA used in the titration. Since 1 mole of EDTA reacts with 1 mole of Ca2+\\text{Ca}^{2+}Ca2+, we can convert this molarity directly to ppm.<\/li>\n<\/ol>\n\n\n\n<p>ppm&nbsp;Ca2+=M2\u00d740,000\\text{ppm Ca}^{2+} = M2 \\times 40,000ppm&nbsp;Ca2+=M2\u00d740,000ppm&nbsp;Ca2+=0.000865\u2009M\u00d740,000=34.6\u2009ppm&nbsp;Ca2+\\text{ppm Ca}^{2+} = 0.000865 \\, \\text{M} \\times 40,000 = 34.6 \\, \\text{ppm Ca}^{2+}ppm&nbsp;Ca2+=0.000865M\u00d740,000=34.6ppm&nbsp;Ca2+<\/p>\n\n\n\n<ol start=\"2\" class=\"wp-block-list\">\n<li><strong>Hardness as CaCO\u2083 (ppm):<\/strong><br>The molecular weight of CaCO3\\text{CaCO}_3CaCO3\u200b is approximately 100 g\/mol, and the equivalent weight for calcium in CaCO3\\text{CaCO}_3CaCO3\u200b is 40 g\/mol. Thus, we use a conversion factor of 2.5 to convert the Ca2+\\text{Ca}^{2+}Ca2+ hardness to CaCO3\\text{CaCO}_3CaCO3\u200b hardness:<\/li>\n<\/ol>\n\n\n\n<p>ppm&nbsp;CaCO3=ppm&nbsp;Ca2+\u00d72.5\\text{ppm CaCO}_3 = \\text{ppm Ca}^{2+} \\times 2.5ppm&nbsp;CaCO3\u200b=ppm&nbsp;Ca2+\u00d72.5ppm&nbsp;CaCO3=34.6\u00d72.5=86.5\u2009ppm&nbsp;CaCO3\\text{ppm CaCO}_3 = 34.6 \\times 2.5 = 86.5 \\, \\text{ppm CaCO}_3ppm&nbsp;CaCO3\u200b=34.6\u00d72.5=86.5ppm&nbsp;CaCO3\u200b<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Final Result:<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Hardness of water in Ca\u00b2\u207a<\/strong>: 34.6 ppm<\/li>\n\n\n\n<li><strong>Hardness of water in CaCO\u2083<\/strong>: 86.5 ppm<\/li>\n<\/ul>\n\n\n\n<p>This means that the hardness of the water sample is equivalent to 34.6 ppm of Ca2+\\text{Ca}^{2+}Ca2+ or 86.5 ppm of CaCO3\\text{CaCO}_3CaCO3\u200b.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner5-443.jpeg\" alt=\"\" class=\"wp-image-244383\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>M1V1\/n1 = M2V2\/n2 M1 = molarity of EDTA = 0.01 M M2 = molarity of bottled water = ? V1 = volume of titre used = 3.46 mL V2 = volume of bottled water used = 40 mL N1 = mole equivalence = 1 N2 = mole equivalence = 1 Calculate the hardness of water. [&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-244382","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/244382","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=244382"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/244382\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=244382"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=244382"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=244382"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}