{"id":258686,"date":"2025-07-18T08:52:06","date_gmt":"2025-07-18T08:52:06","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=258686"},"modified":"2025-07-18T08:52:08","modified_gmt":"2025-07-18T08:52:08","slug":"what-concentration-of-clo3-results-when-765-ml-of-0-733-m-agclo2-is-mixed-with-789-ml-of-0-337-m-mnclo32-clo3-1-0514","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/18\/what-concentration-of-clo3-results-when-765-ml-of-0-733-m-agclo2-is-mixed-with-789-ml-of-0-337-m-mnclo32-clo3-1-0514\/","title":{"rendered":"What concentration of ClO3 results when 765 mL of 0.733 M AgClO2 is mixed with 789 mL of 0.337 M Mn(ClO3)2? [ClO3] 1.0514"},"content":{"rendered":"\n

What concentration of ClO3 results when 765 mL of 0.733 M AgClO2 is mixed with 789 mL of 0.337 M Mn(ClO3)2? [ClO3] 1.0514<\/p>\n\n\n\n

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

To find the concentration of ClO3\u2212\\text{ClO}_3^-ClO3\u2212\u200b when 765 mL of 0.733 M AgClO2\\text{AgClO}_2AgClO2\u200b is mixed with 789 mL of 0.337 M Mn(ClO3)2\\text{Mn(ClO}_3)_2Mn(ClO3\u200b)2\u200b, we can follow a systematic approach.<\/p>\n\n\n\n

Step 1: Identify the relevant reactions<\/h3>\n\n\n\n
    \n
  • AgClO2\\text{AgClO}_2AgClO2\u200b contains the ClO2\u2212\\text{ClO}_2^-ClO2\u2212\u200b ion, which can react to form ClO3\u2212\\text{ClO}_3^-ClO3\u2212\u200b.<\/li>\n\n\n\n
  • Mn(ClO3)2\\text{Mn(ClO}_3)_2Mn(ClO3\u200b)2\u200b contains ClO3\u2212\\text{ClO}_3^-ClO3\u2212\u200b, which is already in its final form.<\/li>\n<\/ul>\n\n\n\n

    Step 2: Write the balanced reactions (if necessary)<\/h3>\n\n\n\n

    The silver chlorite (AgClO2\\text{AgClO}_2AgClO2\u200b) will undergo oxidation to form ClO3\u2212\\text{ClO}_3^-ClO3\u2212\u200b, but for simplicity, we assume the reaction goes to completion, and the ClO2\u2212\\text{ClO}_2^-ClO2\u2212\u200b ion is oxidized to ClO3\u2212\\text{ClO}_3^-ClO3\u2212\u200b.<\/p>\n\n\n\n

    So, the primary change is the production of ClO3\u2212\\text{ClO}_3^-ClO3\u2212\u200b from ClO2\u2212\\text{ClO}_2^-ClO2\u2212\u200b.<\/p>\n\n\n\n

    Step 3: Calculate the moles of ClO2\u2212\\text{ClO}_2^-ClO2\u2212\u200b and ClO3\u2212\\text{ClO}_3^-ClO3\u2212\u200b from each solution<\/h3>\n\n\n\n
      \n
    • For AgClO2\\text{AgClO}_2AgClO2\u200b, the number of moles of ClO2\u2212\\text{ClO}_2^-ClO2\u2212\u200b can be calculated as:<\/li>\n<\/ul>\n\n\n\n

      moles of ClO2\u2212=0.733\u2009M\u00d70.765\u2009L=0.561\u2009mol\\text{moles of } \\text{ClO}_2^- = 0.733 \\, \\text{M} \\times 0.765 \\, \\text{L} = 0.561 \\, \\text{mol}moles of ClO2\u2212\u200b=0.733M\u00d70.765L=0.561mol<\/p>\n\n\n\n

        \n
      • For Mn(ClO3)2\\text{Mn(ClO}_3)_2Mn(ClO3\u200b)2\u200b, the number of moles of ClO3\u2212\\text{ClO}_3^-ClO3\u2212\u200b is:<\/li>\n<\/ul>\n\n\n\n

        moles of ClO3\u2212=0.337\u2009M\u00d70.789\u2009L=0.266\u2009mol\\text{moles of } \\text{ClO}_3^- = 0.337 \\, \\text{M} \\times 0.789 \\, \\text{L} = 0.266 \\, \\text{mol}moles of ClO3\u2212\u200b=0.337M\u00d70.789L=0.266mol<\/p>\n\n\n\n

        Step 4: Add the moles of ClO3\u2212\\text{ClO}_3^-ClO3\u2212\u200b<\/h3>\n\n\n\n
          \n
        • From AgClO2\\text{AgClO}_2AgClO2\u200b, 0.561 moles of ClO2\u2212\\text{ClO}_2^-ClO2\u2212\u200b will be converted into ClO3\u2212\\text{ClO}_3^-ClO3\u2212\u200b (assuming complete oxidation).<\/li>\n\n\n\n
        • From Mn(ClO3)2\\text{Mn(ClO}_3)_2Mn(ClO3\u200b)2\u200b, 0.266 moles of ClO3\u2212\\text{ClO}_3^-ClO3\u2212\u200b are already present.<\/li>\n<\/ul>\n\n\n\n

          Thus, the total moles of ClO3\u2212\\text{ClO}_3^-ClO3\u2212\u200b is:Total moles of ClO3\u2212=0.561\u2009mol+0.266\u2009mol=0.827\u2009mol\\text{Total moles of } \\text{ClO}_3^- = 0.561 \\, \\text{mol} + 0.266 \\, \\text{mol} = 0.827 \\, \\text{mol}Total moles of ClO3\u2212\u200b=0.561mol+0.266mol=0.827mol<\/p>\n\n\n\n

          Step 5: Calculate the final volume of the solution<\/h3>\n\n\n\n

          The total volume is the sum of the individual volumes:Total volume=765\u2009mL+789\u2009mL=1554\u2009mL=1.554\u2009L\\text{Total volume} = 765 \\, \\text{mL} + 789 \\, \\text{mL} = 1554 \\, \\text{mL} = 1.554 \\, \\text{L}Total volume=765mL+789mL=1554mL=1.554L<\/p>\n\n\n\n

          Step 6: Calculate the final concentration of ClO3\u2212\\text{ClO}_3^-ClO3\u2212\u200b<\/h3>\n\n\n\n

          The concentration of ClO3\u2212\\text{ClO}_3^-ClO3\u2212\u200b is the total moles of ClO3\u2212\\text{ClO}_3^-ClO3\u2212\u200b divided by the total volume:[ClO3\u2212]=0.827\u2009mol1.554\u2009L=0.532\u2009M[\\text{ClO}_3^-] = \\frac{0.827 \\, \\text{mol}}{1.554 \\, \\text{L}} = 0.532 \\, \\text{M}[ClO3\u2212\u200b]=1.554L0.827mol\u200b=0.532M<\/p>\n\n\n\n

          However, it seems you provided an answer of 1.0514 M. To reach that, we need to carefully reconsider the reactions involved. If there was a more efficient conversion of ClO2\u2212\\text{ClO}_2^-ClO2\u2212\u200b to ClO3\u2212\\text{ClO}_3^-ClO3\u2212\u200b or if we assumed additional oxidation steps, the resulting concentration could match closer to 1.0514 M. However, based on the basic stoichiometry, the result is 0.532 M, which seems reasonable given the provided numbers. If more details on oxidation are available, this could be adjusted.<\/p>\n\n\n\n

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

          What concentration of ClO3 results when 765 mL of 0.733 M AgClO2 is mixed with 789 mL of 0.337 M Mn(ClO3)2? [ClO3] 1.0514 The Correct Answer and Explanation is: To find the concentration of ClO3\u2212\\text{ClO}_3^-ClO3\u2212\u200b when 765 mL of 0.733 M AgClO2\\text{AgClO}_2AgClO2\u200b is mixed with 789 mL of 0.337 M Mn(ClO3)2\\text{Mn(ClO}_3)_2Mn(ClO3\u200b)2\u200b, we can follow a […]<\/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-258686","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/258686","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=258686"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/258686\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=258686"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=258686"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=258686"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}