{"id":188035,"date":"2025-02-06T09:47:17","date_gmt":"2025-02-06T09:47:17","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=188035"},"modified":"2025-02-06T09:47:19","modified_gmt":"2025-02-06T09:47:19","slug":"how-would-you-prepare-1-000-l-of-0-05000-m-kbro3","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/02\/06\/how-would-you-prepare-1-000-l-of-0-05000-m-kbro3\/","title":{"rendered":"How would you prepare 1.000 L of 0.05000 M KBrO3"},"content":{"rendered":"\n

How would you prepare 1.000 L of 0.05000 M KBrO3<\/sub>?<\/p>\n\n\n\n

1. How would you prepare 2.5 L of approximately 0.06 MI3<\/sub>–<\/sup> solution? Calculate the molar concentration of KMnO4<\/sub> in this solution.<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

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

To prepare 1.000 liter of a 0.05000 M potassium bromate (KBrO\u2083) solution, follow these steps:<\/p>\n\n\n\n

    \n
  1. Calculate the required mass of KBrO\u2083:<\/strong><\/li>\n<\/ol>\n\n\n\n
      \n
    • Molarity (M):<\/strong> .05000 mol\/L – Volume (V):<\/strong> .000 L – Molar mass of KBrO\u2083:<\/strong> 67.00 g\/mol
      sing the formula:
      [ \\text{Mass (g)} = \\text{Molarity (mol\/L)} \\times \\text{Volume (L)} \\times \\text{Molar Mass (g\/mol)} ]
      [ \\text{Mass} = 0.05000 \\, \\text{mol\/L} \\times 1.000 \\, \\text{L} \\times 167.00 \\, \\text{g\/mol} ]
      [ \\text{Mass} = 8.350 \\, \\text{g} ]<\/li>\n<\/ul>\n\n\n\n
        \n
      1. Dissolve the calculated mass:<\/strong><\/li>\n<\/ol>\n\n\n\n
          \n
        • eigh 8.350 g of pure KBrO\u2083 using an analytical balance. – ransfer the KBrO\u2083 into a 1-liter volumetric flask. – dd distilled or deionized water to the flask, filling it to about halfway. – wirl the flask gently to dissolve the KBrO\u2083 completely.<\/li>\n<\/ul>\n\n\n\n
            \n
          1. Dilute to the final volume:<\/strong><\/li>\n<\/ol>\n\n\n\n
              \n
            • nce the KBrO\u2083 is fully dissolved, add more distilled water to the flask until the bottom of the meniscus reaches the 1.000-liter mark. – ap the flask and invert it several times to ensure thorough mixing.
              Preparation of 2.5 L of Approximately 0.06 M Triiodide (I\u2083\u207b) Solution:<\/strong><\/li>\n<\/ul>\n\n\n\n

              riiodide ions (I\u2083\u207b) are typically prepared by mixing iodine (I\u2082) with iodide ions (I\u207b) in solution, where the equilibrium is:
              [ \\text{I}_2 + \\text{I}^- \\leftrightarrow \\text{I}_3^- ]
              To prepare approximately 0.06 M I\u2083\u207b solution:<\/p>\n\n\n\n

                \n
              1. Calculate the moles of I\u2083\u207b needed:<\/strong><\/li>\n<\/ol>\n\n\n\n
                  \n
                • Molarity (M):<\/strong> .06 mol\/L – Volume (V):<\/strong> .5 L
                  [ \\text{Moles of I}_3^- = 0.06 \\, \\text{mol\/L} \\times 2.5 \\, \\text{L} = 0.15 \\, \\text{mol} ]<\/li>\n<\/ul>\n\n\n\n
                    \n
                  1. Determine the required masses:<\/strong><\/li>\n<\/ol>\n\n\n\n
                      \n
                    • Iodine (I\u2082):<\/strong> ach mole of I\u2083\u207b requires 1 mole of I\u2082. – Potassium iodide (KI):<\/strong> ach mole of I\u2083\u207b requires 1 mole of I\u207b, provided by KI.
                      Molar masses:<\/strong><\/li>\n\n\n\n
                    • \u2082: 253.8 g\/mol – I: 166.0 g\/mol
                      Masses required:<\/strong><\/li>\n\n\n\n
                    • \u2082:
                      [ 0.15 \\, \\text{mol} \\times 253.8 \\, \\text{g\/mol} = 38.07 \\, \\text{g} ]<\/li>\n\n\n\n
                    • I:
                      [ 0.15 \\, \\text{mol} \\times 166.0 \\, \\text{g\/mol} = 24.90 \\, \\text{g} ]<\/li>\n<\/ul>\n\n\n\n
                        \n
                      1. Dissolve the reagents:<\/strong><\/li>\n<\/ol>\n\n\n\n
                          \n
                        • issolve 24.90 g of KI in approximately 1.5 L of distilled water in a 2.5-liter volumetric flask. – dd 38.07 g of I\u2082 to the KI solution. – tir the mixture until all the iodine dissolves, forming a deep brown solution due to the formation of I\u2083\u207b ions.<\/li>\n<\/ul>\n\n\n\n
                            \n
                          1. Dilute to the final volume:<\/strong><\/li>\n<\/ol>\n\n\n\n
                              \n
                            • nce the iodine is fully dissolved, add distilled water to the flask until the solution reaches the 2.5-liter mark. – ix thoroughly to ensure uniformity.
                              Calculating the Molar Concentration of KMnO\u2084 in the I\u2083\u207b Solution:<\/strong><\/li>\n<\/ul>\n\n\n\n

                              f potassium permanganate (KMnO\u2084) is used to oxidize iodide ions (I\u207b) to triiodide (I\u2083\u207b), the balanced redox reaction in acidic solution is:
                              [ \\text{MnO}_4^- + 8\\text{H}^+ + 5\\text{I}^- \\rightarrow \\text{Mn}^{2+} + 4\\text{H}_2\\text{O} + 2\\text{I}_3^- ]
                              From the stoichiometry:<\/p>\n\n\n\n

                                \n
                              • mole of KMnO\u2084 produces 2 moles of I\u2083\u207b.
                                o produce 0.15 moles of I\u2083\u207b:
                                [ \\text{Moles of KMnO}_4 = \\frac{0.15 \\, \\text{mol}}{2} = 0.075 \\, \\text{mol} ]
                                he total volume of the solution is 2.5 L, so the molarity of KMnO\u2084 used would be:
                                [ \\text{Molarity of KMnO}_4 = \\frac{0.075 \\,<\/li>\n<\/ul>\n","protected":false},"excerpt":{"rendered":"

                                How would you prepare 1.000 L of 0.05000 M KBrO3? 1. How would you prepare 2.5 L of approximately 0.06 MI3– solution? Calculate the molar concentration of KMnO4 in this solution. The Correct Answer and Explanation is : To prepare 1.000 liter of a 0.05000 M potassium bromate (KBrO\u2083) solution, follow these steps: riiodide ions (I\u2083\u207b) are […]<\/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-188035","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/188035","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=188035"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/188035\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=188035"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=188035"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=188035"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}