{"id":193852,"date":"2025-02-21T05:27:31","date_gmt":"2025-02-21T05:27:31","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=193852"},"modified":"2025-02-21T05:27:33","modified_gmt":"2025-02-21T05:27:33","slug":"how-to-calculate-the-normality-of-sodium-thiosulphate-in-this-titration","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/02\/21\/how-to-calculate-the-normality-of-sodium-thiosulphate-in-this-titration\/","title":{"rendered":"How to calculate the normality of sodium thiosulphate in this titration"},"content":{"rendered":"\n

How to calculate the normality of sodium thiosulphate in this titration. The volume of sodium thiosulphate used at the endpoint is 24ml.<\/p>\n\n\n\n

The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n

To calculate the normality of sodium thiosulfate (Na\u2082S\u2082O\u2083) in a titration, you will need to know the amount of the substance with which it reacts, usually expressed through its equivalent weight, and the volume of Na\u2082S\u2082O\u2083 solution used to reach the endpoint. Here\u2019s how you can calculate it:<\/p>\n\n\n\n

    \n
  1. Volume of Sodium Thiosulfate Used<\/strong>: Given as 24 mL or 0.024 liters.<\/li>\n\n\n\n
  2. Reaction Information<\/strong>: The reaction commonly involves iodine (I\u2082) being titrated with sodium thiosulfate. The balanced chemical equation for this reaction is: I2+2Na2S2O3\u21922NaI+Na2S4O6I\u2082 + 2 Na\u2082S\u2082O\u2083 \u2192 2 NaI + Na\u2082S\u2084O\u2086 This equation tells us that 1 mole of iodine reacts with 2 moles of sodium thiosulfate.<\/li>\n\n\n\n
  3. Mole Ratio<\/strong>: Since the reaction ratio between I\u2082 and Na\u2082S\u2082O\u2083 is 1:2, the equivalent weight of Na\u2082S\u2082O\u2083 is half its molar mass (because 1 mole of I\u2082 requires 2 moles of Na\u2082S\u2082O\u2083).<\/li>\n\n\n\n
  4. Calculation of Normality<\/strong>: Normality (N) = Number\u00a0of\u00a0equivalents\u00a0of\u00a0soluteVolume\u00a0of\u00a0solution\u00a0in\u00a0liters\\frac{\\text{Number of equivalents of solute}}{\\text{Volume of solution in liters}} An equivalent of Na\u2082S\u2082O\u2083 in this reaction corresponds to its ability to react with 1 mole of iodine (since 2 moles of Na\u2082S\u2082O\u2083 react with 1 mole of I\u2082). Assuming you know the molarity of iodine (or can calculate it based on the amount used), you can calculate the normality of Na\u2082S\u2082O\u2083 by multiplying the molarity of the iodine solution by 2 (because each mole of iodine reacts with 2 equivalents of Na\u2082S\u2082O\u2083). For example, if 0.1 M of I\u2082 is used: Normality\u00a0of\u00a0Na\u2082S\u2082O\u2083=2\u00d7Molarity\u00a0of\u00a0I\u2082\\text{Normality of Na\u2082S\u2082O\u2083} = 2 \\times \\text{Molarity of I\u2082} Normality\u00a0of\u00a0Na\u2082S\u2082O\u2083=2\u00d70.1=0.2\u00a0N\\text{Normality of Na\u2082S\u2082O\u2083} = 2 \\times 0.1 = 0.2 \\text{ N} However, without the molarity of the iodine solution used in your titration, we can’t compute the exact normality of Na\u2082S\u2082O\u2083. It\u2019s essential to either know or be able to calculate the molarity of the iodine for a precise calculation.<\/li>\n<\/ol>\n\n\n\n

    Conclusion<\/strong>: Calculating the normality in a titration requires understanding the stoichiometry of the reaction involved. For sodium thiosulfate titrations involving iodine, remember that the normality is double the molarity of the iodine solution due to the 1:2 stoichiometric reaction ratio. This calculation method assumes accurate measurement of reaction components and complete reaction according to the stoichiometry.<\/p>\n\n\n\n

    I’ll now create an image that visually represents this calculation process for better understanding.<\/p>\n\n\n\n

    Here is the visual representation of a chemistry lab setup for a titration experiment, showing the process of adding sodium thiosulfate to an iodine solution. This image illustrates the careful measurement and observation involved in determining the endpoint of the titration, which is critical for accurate calculation of the normality.<\/p>\n\n\n\n

    \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

    How to calculate the normality of sodium thiosulphate in this titration. The volume of sodium thiosulphate used at the endpoint is 24ml. The Correct Answer and Explanation is : To calculate the normality of sodium thiosulfate (Na\u2082S\u2082O\u2083) in a titration, you will need to know the amount of the substance with which it reacts, usually […]<\/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-193852","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/193852","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=193852"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/193852\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=193852"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=193852"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=193852"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}