{"id":244386,"date":"2025-07-05T05:17:43","date_gmt":"2025-07-05T05:17:43","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=244386"},"modified":"2025-07-05T05:17:45","modified_gmt":"2025-07-05T05:17:45","slug":"m1v1-n1-m2v2-n2-m1-molarity-of-edta-0-01-m-m2-molarity-of-bottled-water-2","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-2\/","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. This is how I did it, and I don&#8217;t know if it&#8217;s right or wrong: Express the result in the form of Ca2+ ppm and CaCO3 ppm: 0.01 x 0.0034 = M x 0.04 M = (0.01 x 0.0034) \/ 0.04 M = 8.5 x 10^-4 Ca2+ Moles per liter: 40 x 8.5 x 10^-4 = 0.034 g per liter Ppm = mg per liter 34 mg per 1 liter 34 ppm Ca2+ Weight CaCO3 = 100 grams Weight Ca2+ = 40 34 x (100\/40) = 85 ppm CaCO3<\/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 in terms of Ca\u00b2\u207a ppm and CaCO\u2083 ppm, you have the right approach but there are a few minor adjustments to make. Let\u2019s break it down step by step:<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Given:<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>M\u2081 = 0.01 M (molarity of EDTA)<\/li>\n\n\n\n<li>V\u2081 = 3.46 mL (volume of EDTA used)<\/li>\n\n\n\n<li>V\u2082 = 40 mL (volume of bottled water used)<\/li>\n\n\n\n<li>N\u2081 = 1 (mole equivalence of EDTA)<\/li>\n\n\n\n<li>N\u2082 = 1 (mole equivalence of Ca\u00b2\u207a)<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: Use the equation to find the molarity of Ca\u00b2\u207a in the sample<\/h3>\n\n\n\n<p>From the equation M\u2081V\u2081 \/ N\u2081 = M\u2082V\u2082 \/ N\u2082, you can solve for M\u2082 (the molarity of Ca\u00b2\u207a in the water). M2=M1V1V2=0.01\u00d73.4640=8.65\u00d710\u22124\u2009MM\u2082 = \\frac{M\u2081V\u2081}{V\u2082} = \\frac{0.01 \\times 3.46}{40} = 8.65 \\times 10^{-4} \\, MM2\u200b=V2\u200bM1\u200bV1\u200b\u200b=400.01\u00d73.46\u200b=8.65\u00d710\u22124M<\/p>\n\n\n\n<p>So, the molarity of Ca\u00b2\u207a in the water is <strong>8.65 \u00d7 10\u207b\u2074 M<\/strong>.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 2: Calculate the mass of Ca\u00b2\u207a in 1 liter of water (in grams)<\/h3>\n\n\n\n<p>Now, to convert the moles of Ca\u00b2\u207a to grams, you multiply by the molar mass of Ca\u00b2\u207a. The molar mass of Ca\u00b2\u207a is <strong>40.08 g\/mol<\/strong>. Mass&nbsp;of&nbsp;Ca\u00b2\u207a=8.65\u00d710\u22124\u2009M\u00d740.08\u2009g\/mol=0.0347\u2009g\/L\\text{Mass of Ca\u00b2\u207a} = 8.65 \\times 10^{-4} \\, M \\times 40.08 \\, \\text{g\/mol} = 0.0347 \\, \\text{g\/L}Mass&nbsp;of&nbsp;Ca\u00b2\u207a=8.65\u00d710\u22124M\u00d740.08g\/mol=0.0347g\/L<\/p>\n\n\n\n<p>This is the <strong>mass of Ca\u00b2\u207a in 1 liter of water<\/strong>.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 3: Convert to ppm (Ca\u00b2\u207a)<\/h3>\n\n\n\n<p>Since 1 ppm = 1 mg\/L, you have: 0.0347\u2009g\/L=34.7\u2009mg\/L=34.7\u2009ppm&nbsp;of&nbsp;Ca\u00b2\u207a0.0347 \\, \\text{g\/L} = 34.7 \\, \\text{mg\/L} = 34.7 \\, \\text{ppm of Ca\u00b2\u207a}0.0347g\/L=34.7mg\/L=34.7ppm&nbsp;of&nbsp;Ca\u00b2\u207a<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 4: Convert to CaCO\u2083 ppm<\/h3>\n\n\n\n<p>To convert the Ca\u00b2\u207a concentration to the equivalent amount of CaCO\u2083, use the molar mass ratio. The molar mass of CaCO\u2083 is 100.09 g\/mol, while the molar mass of Ca\u00b2\u207a is 40.08 g\/mol. So: ppm&nbsp;of&nbsp;CaCO\u2083=34.7\u00d7100.0940.08=86.6\u2009ppm&nbsp;of&nbsp;CaCO\u2083\\text{ppm of CaCO\u2083} = 34.7 \\times \\frac{100.09}{40.08} = 86.6 \\, \\text{ppm of CaCO\u2083}ppm&nbsp;of&nbsp;CaCO\u2083=34.7\u00d740.08100.09\u200b=86.6ppm&nbsp;of&nbsp;CaCO\u2083<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Final Answer:<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Hardness in Ca\u00b2\u207a ppm<\/strong> = 34.7 ppm<\/li>\n\n\n\n<li><strong>Hardness in CaCO\u2083 ppm<\/strong> = 86.6 ppm<\/li>\n<\/ul>\n\n\n\n<p>Your calculation is very close but a little off due to rounding. The final results for Ca\u00b2\u207a are around <strong>34.7 ppm<\/strong> and for <strong>CaCO\u2083<\/strong> around <strong>86.6 ppm<\/strong>.<\/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-banner5-444.jpeg\" alt=\"\" class=\"wp-image-244387\"\/><\/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-244386","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/244386","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=244386"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/244386\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=244386"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=244386"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=244386"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}