{"id":29445,"date":"2025-06-21T04:38:14","date_gmt":"2025-06-21T04:38:14","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=29445"},"modified":"2025-06-21T04:38:15","modified_gmt":"2025-06-21T04:38:15","slug":"standardization-of-sodium-hydroxide-solution-balanced-chemical-equation","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/standardization-of-sodium-hydroxide-solution-balanced-chemical-equation\/","title":{"rendered":"Standardization of Sodium Hydroxide Solution Balanced chemical equation"},"content":{"rendered":"\n<p>A. Standardization of Sodium Hydroxide Solution Balanced chemical equation:<br>Data: Trial 1 Trial 2 Trial 3 Mass of oxalic acid 0.1503g 0.1511g 0.1504g Initial burette reading 0.00ml 0.00ml 0.00ml Final burette reading 14.1ml 11.3ml 12.9ml Calculations (show work): Volume NaOH added moles of oxalic acid moles of NaOH added Molarity of NaOH Average molarity<\/p>\n\n\n\n<p>;<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"935\" height=\"1024\" src=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/image-126.png\" alt=\"\" class=\"wp-image-29446\" srcset=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/image-126.png 935w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/image-126-274x300.png 274w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/image-126-768x841.png 768w\" sizes=\"auto, (max-width: 935px) 100vw, 935px\" \/><\/figure>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-vivid-green-cyan-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>A. Standardization of Sodium Hydroxide Solution Balanced chemical equation: Data: Trial 1 Trial 2 Trial 3 Mass of oxalic acid 0.1503g 0.1511g 0.1504g Initial burette reading 0.00ml 0.00ml 0.00ml Final burette reading 14.1ml 11.3ml 12.9ml Calculations (show work): Volume NaOH added moles of oxalic acid moles of NaOH added Molarity of NaOH Average molaritygenerate the correct answer and 300 words explanation without using the em dashes<\/p>\n\n\n\n<h6 class=\"wp-block-heading\">ChatGPT said:<\/h6>\n\n\n\n<p>To calculate the molarity of NaOH using standardization with oxalic acid, we follow these steps for each trial. The balanced chemical equation is:<\/p>\n\n\n\n<p><strong>H\u2082C\u2082O\u2084 + 2 NaOH \u2192 Na\u2082C\u2082O\u2084 + 2 H\u2082O<\/strong><\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: Volume of NaOH added<\/h3>\n\n\n\n<p>This is calculated from the difference in burette readings (Final \u2013 Initial):<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Trial 1:<\/strong> 14.1 mL<\/li>\n\n\n\n<li><strong>Trial 2:<\/strong> 11.3 mL<\/li>\n\n\n\n<li><strong>Trial 3:<\/strong> 12.9 mL<\/li>\n<\/ul>\n\n\n\n<p>Convert to liters (1 mL = 0.001 L):<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Trial 1: 0.0141 L<\/li>\n\n\n\n<li>Trial 2: 0.0113 L<\/li>\n\n\n\n<li>Trial 3: 0.0129 L<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 2: Moles of oxalic acid<\/h3>\n\n\n\n<p>Molar mass of oxalic acid (H\u2082C\u2082O\u2084):<br>= 2(1.008) + 2(12.01) + 4(16.00) = <strong>90.03 g\/mol<\/strong><\/p>\n\n\n\n<p>Moles = mass \/ molar mass:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Trial 1: 0.1503 g \/ 90.03 g\/mol = <strong>0.00167 mol<\/strong><\/li>\n\n\n\n<li>Trial 2: 0.1511 g \/ 90.03 g\/mol = <strong>0.001678 mol<\/strong><\/li>\n\n\n\n<li>Trial 3: 0.1504 g \/ 90.03 g\/mol = <strong>0.001671 mol<\/strong><\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 3: Moles of NaOH added<\/h3>\n\n\n\n<p>From the balanced equation, 1 mol oxalic acid reacts with 2 mol NaOH:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Trial 1: 0.00167 \u00d7 2 = <strong>0.00334 mol<\/strong><\/li>\n\n\n\n<li>Trial 2: 0.001678 \u00d7 2 = <strong>0.003356 mol<\/strong><\/li>\n\n\n\n<li>Trial 3: 0.001671 \u00d7 2 = <strong>0.003342 mol<\/strong><\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 4: Molarity of NaOH<\/h3>\n\n\n\n<p>Molarity = moles \/ volume (in L)<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Trial 1: 0.00334 mol \/ 0.0141 L = <strong>0.237 mol\/L<\/strong><\/li>\n\n\n\n<li>Trial 2: 0.003356 mol \/ 0.0113 L = <strong>0.297 mol\/L<\/strong><\/li>\n\n\n\n<li>Trial 3: 0.003342 mol \/ 0.0129 L = <strong>0.259 mol\/L<\/strong><\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 5: Average molarity<\/h3>\n\n\n\n<p>= (0.237 + 0.297 + 0.259) \/ 3 = <strong>0.264 mol\/L<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Final Answers:<\/h3>\n\n\n\n<figure class=\"wp-block-table\"><table class=\"has-fixed-layout\"><thead><tr><th>Calculation<\/th><th>Trial 1<\/th><th>Trial 2<\/th><th>Trial 3<\/th><\/tr><\/thead><tbody><tr><td>Volume NaOH added (L)<\/td><td>0.0141<\/td><td>0.0113<\/td><td>0.0129<\/td><\/tr><tr><td>Moles of oxalic acid<\/td><td>0.00167<\/td><td>0.001678<\/td><td>0.001671<\/td><\/tr><tr><td>Moles of NaOH added<\/td><td>0.00334<\/td><td>0.003356<\/td><td>0.003342<\/td><\/tr><tr><td>Molarity of NaOH (mol\/L)<\/td><td>0.237<\/td><td>0.297<\/td><td>0.259<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p><strong>Average Molarity: 0.264 mol\/L<\/strong><\/p>\n\n\n\n<p>This process standardizes NaOH solution concentration through stoichiometric relationships, where oxalic acid acts as a primary standard due to its known purity and stability.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"852\" height=\"1024\" src=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-301.jpeg\" alt=\"\" class=\"wp-image-29447\" srcset=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-301.jpeg 852w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-301-250x300.jpeg 250w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-301-768x923.jpeg 768w\" sizes=\"auto, (max-width: 852px) 100vw, 852px\" \/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>A. Standardization of Sodium Hydroxide Solution Balanced chemical equation:Data: Trial 1 Trial 2 Trial 3 Mass of oxalic acid 0.1503g 0.1511g 0.1504g Initial burette reading 0.00ml 0.00ml 0.00ml Final burette reading 14.1ml 11.3ml 12.9ml Calculations (show work): Volume NaOH added moles of oxalic acid moles of NaOH added Molarity of NaOH Average molarity ; The [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-29445","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/29445","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=29445"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/29445\/revisions"}],"predecessor-version":[{"id":29448,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/29445\/revisions\/29448"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=29445"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=29445"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=29445"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}