{"id":221552,"date":"2025-05-29T19:14:01","date_gmt":"2025-05-29T19:14:01","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=221552"},"modified":"2025-05-29T19:14:03","modified_gmt":"2025-05-29T19:14:03","slug":"a-10-00-ml-sample-of-sulfuric-acid-from-an-automobile-battery-requires-35-08-ml-of-2-12-m-sodium-hydroxide-solution-for-complete-neutralization","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/29\/a-10-00-ml-sample-of-sulfuric-acid-from-an-automobile-battery-requires-35-08-ml-of-2-12-m-sodium-hydroxide-solution-for-complete-neutralization\/","title":{"rendered":"A 10.00-mL sample of sulfuric acid from an automobile battery requires 35.08 mL of 2.12 M sodium hydroxide solution for complete neutralization"},"content":{"rendered":"\n<p>A 10.00-mL sample of sulfuric acid from an automobile battery requires 35.08 mL of 2.12 M sodium hydroxide solution for complete neutralization. What is the molarity of the sulfuric acid?<\/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 find the molarity of the sulfuric acid (H\u2082SO\u2084), we must first understand the neutralization reaction between sulfuric acid and sodium hydroxide (NaOH):<\/p>\n\n\n\n<p><strong>Balanced chemical equation:<\/strong> H2SO4(aq)+2NaOH(aq)\u2192Na2SO4(aq)+2H2O(l)\\text{H}_2\\text{SO}_4(aq) + 2\\text{NaOH}(aq) \\rightarrow \\text{Na}_2\\text{SO}_4(aq) + 2\\text{H}_2\\text{O}(l)<\/p>\n\n\n\n<p>This equation tells us that <strong>1 mole of sulfuric acid reacts with 2 moles of sodium hydroxide<\/strong>.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: Calculate moles of NaOH added<\/h3>\n\n\n\n<p>Given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Volume of NaOH = 35.08 mL = 0.03508 L<\/li>\n\n\n\n<li>Molarity of NaOH = 2.12 M<\/li>\n<\/ul>\n\n\n\n<p>Moles&nbsp;of&nbsp;NaOH=M\u00d7V=2.12\u2009mol\/L\u00d70.03508\u2009L=0.07438\u2009mol\\text{Moles of NaOH} = M \\times V = 2.12 \\, \\text{mol\/L} \\times 0.03508 \\, \\text{L} = 0.07438 \\, \\text{mol}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 2: Use the mole ratio to find moles of H\u2082SO\u2084<\/h3>\n\n\n\n<p>From the balanced equation, the mole ratio is: 1\u2009mol&nbsp;H2SO42\u2009mol&nbsp;NaOH=x0.07438\u21d2x=0.074382=0.03719\u2009mol&nbsp;H2SO4\\frac{1 \\, \\text{mol H}_2\\text{SO}_4}{2 \\, \\text{mol NaOH}} = \\frac{x}{0.07438} \\Rightarrow x = \\frac{0.07438}{2} = 0.03719 \\, \\text{mol H}_2\\text{SO}_4<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 3: Calculate molarity of H\u2082SO\u2084<\/h3>\n\n\n\n<p>Given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Volume of H\u2082SO\u2084 = 10.00 mL = 0.01000 L<\/li>\n<\/ul>\n\n\n\n<p>Molarity&nbsp;of&nbsp;H2SO4=molesvolume=0.037190.01000=3.72\u2009M\\text{Molarity of H}_2\\text{SO}_4 = \\frac{\\text{moles}}{\\text{volume}} = \\frac{0.03719}{0.01000} = \\boxed{3.72 \\, \\text{M}}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Conclusion<\/h3>\n\n\n\n<p>To determine the concentration of a sulfuric acid solution from an automobile battery, a titration with a standard sodium hydroxide solution was used. The balanced chemical reaction between sulfuric acid (H\u2082SO\u2084) and sodium hydroxide (NaOH) shows that one mole of sulfuric acid reacts with two moles of sodium hydroxide. In this titration, 10.00 mL of the acid required 35.08 mL of 2.12 M sodium hydroxide for complete neutralization.<\/p>\n\n\n\n<p>First, we calculated the number of moles of NaOH used in the reaction: multiplying the volume in liters (0.03508 L) by the molarity (2.12 M) gave 0.07438 moles of NaOH. Since two moles of NaOH react with one mole of H\u2082SO\u2084, we divide the moles of NaOH by 2 to find the number of moles of H\u2082SO\u2084, which is 0.03719 mol.<\/p>\n\n\n\n<p>Next, we determined the molarity of the sulfuric acid by dividing the number of moles of acid by its volume in liters (0.01000 L), resulting in a molarity of <strong>3.72 M<\/strong>.<\/p>\n\n\n\n<p>This titration illustrates a practical application of stoichiometry and molarity in real-world chemical analysis. Knowing the concentration of battery acid is crucial for evaluating battery health and preventing damage due to overly concentrated or diluted solutions. This method is widely used in analytical chemistry to determine unknown concentrations of acids or bases accurately.To find the molarity of the sulfuric acid (H\u2082SO\u2084), we must first understand the neutralization reaction between sulfuric acid and sodium hydroxide (NaOH):<\/p>\n\n\n\n<p><strong>Balanced chemical equation:<\/strong> H2SO4(aq)+2NaOH(aq)\u2192Na2SO4(aq)+2H2O(l)\\text{H}_2\\text{SO}_4(aq) + 2\\text{NaOH}(aq) \\rightarrow \\text{Na}_2\\text{SO}_4(aq) + 2\\text{H}_2\\text{O}(l)<\/p>\n\n\n\n<p>This equation tells us that <strong>1 mole of sulfuric acid reacts with 2 moles of sodium hydroxide<\/strong>.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: Calculate moles of NaOH added<\/h3>\n\n\n\n<p>Given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Volume of NaOH = 35.08 mL = 0.03508 L<\/li>\n\n\n\n<li>Molarity of NaOH = 2.12 M<\/li>\n<\/ul>\n\n\n\n<p>Moles&nbsp;of&nbsp;NaOH=M\u00d7V=2.12\u2009mol\/L\u00d70.03508\u2009L=0.07438\u2009mol\\text{Moles of NaOH} = M \\times V = 2.12 \\, \\text{mol\/L} \\times 0.03508 \\, \\text{L} = 0.07438 \\, \\text{mol}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 2: Use the mole ratio to find moles of H\u2082SO\u2084<\/h3>\n\n\n\n<p>From the balanced equation, the mole ratio is: 1\u2009mol&nbsp;H2SO42\u2009mol&nbsp;NaOH=x0.07438\u21d2x=0.074382=0.03719\u2009mol&nbsp;H2SO4\\frac{1 \\, \\text{mol H}_2\\text{SO}_4}{2 \\, \\text{mol NaOH}} = \\frac{x}{0.07438} \\Rightarrow x = \\frac{0.07438}{2} = 0.03719 \\, \\text{mol H}_2\\text{SO}_4<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 3: Calculate molarity of H\u2082SO\u2084<\/h3>\n\n\n\n<p>Given:<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: Calculate moles of NaOH added<\/h3>\n\n\n\n<p>Given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Volume of NaOH = 35.08 mL = 0.03508 L<\/li>\n\n\n\n<li>Molarity of NaOH = 2.12 M<\/li>\n<\/ul>\n\n\n\n<p>Moles&nbsp;of&nbsp;NaOH=M\u00d7V=2.12\u2009mol\/L\u00d70.03508\u2009L=0.07438\u2009mol\\text{Moles of NaOH} = M \\times V = 2.12 \\, \\text{mol\/L} \\times 0.03508 \\, \\text{L} = 0.07438 \\, \\text{mol}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 2: Use the mole ratio to find moles of H\u2082SO\u2084<\/h3>\n\n\n\n<p>From the balanced equation, the mole ratio is: 1\u2009mol&nbsp;H2SO42\u2009mol&nbsp;NaOH=x0.07438\u21d2x=0.074382=0.03719\u2009mol&nbsp;H2SO4\\frac{1 \\, \\text{mol H}_2\\text{SO}_4}{2 \\, \\text{mol NaOH}} = \\frac{x}{0.07438} \\Rightarrow x = \\frac{0.07438}{2} = 0.03719 \\, \\text{mol H}_2\\text{SO}_4<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 3: Calculate molarity of H\u2082SO\u2084<\/h3>\n\n\n\n<p>Given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Volume of H\u2082SO\u2084 = 10.00 mL = 0.01000 L<\/li>\n<\/ul>\n\n\n\n<p>Molarity&nbsp;of&nbsp;H2SO4=molesvolume=0.037190.01000=3.72\u2009M\\text{Molarity of H}_2\\text{SO}_4 = \\frac{\\text{moles}}{\\text{volume}} = \\frac{0.03719}{0.01000} = \\boxed{3.72 \\, \\text{M}}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><\/h3>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/05\/learnexams-banner9-80.jpeg\" alt=\"\" class=\"wp-image-221553\"\/><\/figure>\n\n\n\n<p>.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>A 10.00-mL sample of sulfuric acid from an automobile battery requires 35.08 mL of 2.12 M sodium hydroxide solution for complete neutralization. What is the molarity of the sulfuric acid? The Correct Answer and Explanation is: To find the molarity of the sulfuric acid (H\u2082SO\u2084), we must first understand the neutralization reaction between sulfuric acid [&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-221552","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/221552","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=221552"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/221552\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=221552"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=221552"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=221552"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}