{"id":37814,"date":"2025-06-26T06:27:21","date_gmt":"2025-06-26T06:27:21","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=37814"},"modified":"2025-06-26T07:34:19","modified_gmt":"2025-06-26T07:34:19","slug":"oxalic-acid-h2c2o4-lewis-structure-shown-is-a-diprotic-acid-with-the-following-ka-values","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/oxalic-acid-h2c2o4-lewis-structure-shown-is-a-diprotic-acid-with-the-following-ka-values\/","title":{"rendered":"Oxalic acid (H2C2O4, Lewis structure shown) is a diprotic acid with the following\u00a0Ka\u00a0values"},"content":{"rendered":"\n<p>Oxalic acid (H2C2O4, Lewis structure shown) is a diprotic acid with the following\u00a0Ka\u00a0values<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"1024\" height=\"391\" src=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/image-355.png\" alt=\"\" class=\"wp-image-37817\" srcset=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/image-355.png 1024w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/image-355-300x115.png 300w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/image-355-768x293.png 768w\" sizes=\"auto, (max-width: 1024px) 100vw, 1024px\" \/><\/figure>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-vivid-cyan-blue-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>The correct answer is <strong>A. A solution of NaHC\u2082O\u2084 is acidic because Ka(HC\u2082O\u2084\u207b) &gt; Kb(HC\u2082O\u2084\u207b).<\/strong><\/p>\n\n\n\n<p>To understand this, we must analyze the nature of the salt NaHC\u2082O\u2084. Oxalic acid (H\u2082C\u2082O\u2084) is a diprotic acid, meaning it donates two protons sequentially. The first dissociation yields HC\u2082O\u2084\u207b, and the second produces C\u2082O\u2084\u00b2\u207b. When NaHC\u2082O\u2084 dissolves, it forms Na\u207a and HC\u2082O\u2084\u207b ions. Sodium ions do not affect the solution\u2019s pH, so the key lies in evaluating the behavior of the HC\u2082O\u2084\u207b ion.<\/p>\n\n\n\n<p>HC\u2082O\u2084\u207b is amphiprotic. It can either donate a proton (acting as an acid) or accept a proton (acting as a base). To determine whether it acts more strongly as an acid or a base, we compare its acid dissociation constant (Ka) to its base ionization constant (Kb).<\/p>\n\n\n\n<p>Given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Ka for HC\u2082O\u2084\u207b = 1.5 \u00d7 10\u207b\u2074<\/li>\n\n\n\n<li>To find Kb for HC\u2082O\u2084\u207b, we use the expression Kb = Kw \/ Ka, where Kw = 1.0 \u00d7 10\u207b\u00b9\u2074<\/li>\n\n\n\n<li>Kb = (1.0 \u00d7 10\u207b\u00b9\u2074) \/ (1.5 \u00d7 10\u207b\u2074) = 6.67 \u00d7 10\u207b\u00b9\u00b9<\/li>\n<\/ul>\n\n\n\n<p>Since Ka for HC\u2082O\u2084\u207b is several orders of magnitude larger than its Kb, the ion acts predominantly as an acid in solution. As a result, when NaHC\u2082O\u2084 dissolves in water, the HC\u2082O\u2084\u207b ions donate protons to water molecules, forming H\u2083O\u207a and lowering the pH below 7.<\/p>\n\n\n\n<p>Therefore, the solution is acidic, and this confirms that <strong>option A is correct<\/strong>. The remaining choices either misrepresent the species involved or incorrectly compare the strength of acidic and basic behavior.<\/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-banner8-979.jpeg\" alt=\"\" class=\"wp-image-37818\" srcset=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-979.jpeg 852w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-979-250x300.jpeg 250w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-979-768x923.jpeg 768w\" sizes=\"auto, (max-width: 852px) 100vw, 852px\" \/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Oxalic acid (H2C2O4, Lewis structure shown) is a diprotic acid with the following\u00a0Ka\u00a0values The Correct Answer and Explanation is: The correct answer is A. A solution of NaHC\u2082O\u2084 is acidic because Ka(HC\u2082O\u2084\u207b) &gt; Kb(HC\u2082O\u2084\u207b). To understand this, we must analyze the nature of the salt NaHC\u2082O\u2084. Oxalic acid (H\u2082C\u2082O\u2084) is a diprotic acid, meaning it [&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-37814","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/37814","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=37814"}],"version-history":[{"count":2,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/37814\/revisions"}],"predecessor-version":[{"id":38020,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/37814\/revisions\/38020"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=37814"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=37814"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=37814"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}