{"id":242002,"date":"2025-07-04T05:07:22","date_gmt":"2025-07-04T05:07:22","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=242002"},"modified":"2025-07-04T05:07:24","modified_gmt":"2025-07-04T05:07:24","slug":"draw-a-balanced-chemical-equation-for-the-reaction-of-sulfuric-acid-and-sodium-hydroxide","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/04\/draw-a-balanced-chemical-equation-for-the-reaction-of-sulfuric-acid-and-sodium-hydroxide\/","title":{"rendered":"Draw a balanced chemical equation for the reaction of sulfuric acid and sodium hydroxide"},"content":{"rendered":"\n

. Draw a balanced chemical equation for the reaction of sulfuric acid and sodium hydroxide. Based on that equation, if 9.5 mL of 0.406 M sulfuric acid is required to neutralized 26.0 mL of a sodium hydroxide solution, what is the concentration of the sodium hydroxide solution? (5 Points)<\/p>\n\n\n\n

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

The reaction between sulfuric acid (H\u2082SO\u2084) and sodium hydroxide (NaOH) can be represented by the following balanced chemical equation:<\/p>\n\n\n\n

H\u2082SO\u2084 (aq) + 2 NaOH (aq) \u2192 Na\u2082SO\u2084 (aq) + 2 H\u2082O (l)<\/strong><\/p>\n\n\n\n

This equation indicates that one mole of sulfuric acid reacts with two moles of sodium hydroxide to form sodium sulfate and water.<\/p>\n\n\n\n

Given:<\/h3>\n\n\n\n
    \n
  • Volume of sulfuric acid (H\u2082SO\u2084) = 9.5 mL = 0.0095 L<\/li>\n\n\n\n
  • Molarity of sulfuric acid (H\u2082SO\u2084) = 0.406 M<\/li>\n\n\n\n
  • Volume of sodium hydroxide (NaOH) = 26.0 mL = 0.026 L<\/li>\n\n\n\n
  • Molarity of sodium hydroxide (NaOH) = ? (This is what we need to find)<\/li>\n<\/ul>\n\n\n\n

    Steps to solve:<\/h3>\n\n\n\n
      \n
    1. Calculate the moles of sulfuric acid<\/strong>:
      The formula to calculate moles from molarity is: Moles\u00a0of\u00a0H\u2082SO\u2084=Molarity\u00a0of\u00a0H\u2082SO\u2084\u00d7Volume\u00a0of\u00a0H\u2082SO\u2084\u00a0(in\u00a0L)\\text{Moles of H\u2082SO\u2084} = \\text{Molarity of H\u2082SO\u2084} \\times \\text{Volume of H\u2082SO\u2084 (in L)}Moles\u00a0of\u00a0H\u2082SO\u2084=Molarity\u00a0of\u00a0H\u2082SO\u2084\u00d7Volume\u00a0of\u00a0H\u2082SO\u2084\u00a0(in\u00a0L) Moles\u00a0of\u00a0H\u2082SO\u2084=0.406\u2009mol\/L\u00d70.0095\u2009L=0.003857\u2009mol\\text{Moles of H\u2082SO\u2084} = 0.406 \\, \\text{mol\/L} \\times 0.0095 \\, \\text{L} = 0.003857 \\, \\text{mol}Moles\u00a0of\u00a0H\u2082SO\u2084=0.406mol\/L\u00d70.0095L=0.003857mol<\/li>\n\n\n\n
    2. Use the stoichiometry of the balanced equation<\/strong>:
      According to the equation, 1 mole of H\u2082SO\u2084 reacts with 2 moles of NaOH. Therefore, the moles of NaOH required to react with 0.003857 moles of H\u2082SO\u2084 are: Moles\u00a0of\u00a0NaOH=2\u00d7Moles\u00a0of\u00a0H\u2082SO\u2084\\text{Moles of NaOH} = 2 \\times \\text{Moles of H\u2082SO\u2084}Moles\u00a0of\u00a0NaOH=2\u00d7Moles\u00a0of\u00a0H\u2082SO\u2084 Moles\u00a0of\u00a0NaOH=2\u00d70.003857\u2009mol=0.007714\u2009mol\\text{Moles of NaOH} = 2 \\times 0.003857 \\, \\text{mol} = 0.007714 \\, \\text{mol}Moles\u00a0of\u00a0NaOH=2\u00d70.003857mol=0.007714mol<\/li>\n\n\n\n
    3. Calculate the molarity of NaOH<\/strong>:
      The formula to calculate molarity is: Molarity\u00a0of\u00a0NaOH=Moles\u00a0of\u00a0NaOHVolume\u00a0of\u00a0NaOH\u00a0(in\u00a0L)\\text{Molarity of NaOH} = \\frac{\\text{Moles of NaOH}}{\\text{Volume of NaOH (in L)}}Molarity\u00a0of\u00a0NaOH=Volume\u00a0of\u00a0NaOH\u00a0(in\u00a0L)Moles\u00a0of\u00a0NaOH\u200b Molarity\u00a0of\u00a0NaOH=0.007714\u2009mol0.0260\u2009L=0.296\u2009M\\text{Molarity of NaOH} = \\frac{0.007714 \\, \\text{mol}}{0.0260 \\, \\text{L}} = 0.296 \\, \\text{M}Molarity\u00a0of\u00a0NaOH=0.0260L0.007714mol\u200b=0.296M<\/li>\n<\/ol>\n\n\n\n

      Final Answer:<\/h3>\n\n\n\n

      The concentration of the sodium hydroxide solution is 0.296 M<\/strong>.<\/p>\n\n\n\n

      Explanation:<\/h3>\n\n\n\n

      This problem involves stoichiometry, where we use the balanced chemical equation to relate the moles of sulfuric acid to the moles of sodium hydroxide. By first calculating the moles of sulfuric acid, then using the mole ratio from the balanced equation to determine the moles of sodium hydroxide, we can finally find the concentration (molarity) of the sodium hydroxide solution.<\/p>\n\n\n\n

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

      . Draw a balanced chemical equation for the reaction of sulfuric acid and sodium hydroxide. Based on that equation, if 9.5 mL of 0.406 M sulfuric acid is required to neutralized 26.0 mL of a sodium hydroxide solution, what is the concentration of the sodium hydroxide solution? (5 Points) The Correct Answer and Explanation is: […]<\/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-242002","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/242002","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=242002"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/242002\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=242002"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=242002"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=242002"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}