{"id":44020,"date":"2025-06-30T12:50:22","date_gmt":"2025-06-30T12:50:22","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=44020"},"modified":"2025-06-30T12:50:23","modified_gmt":"2025-06-30T12:50:23","slug":"formic-acid-dissociates-in-water-according-to-the-equation","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/formic-acid-dissociates-in-water-according-to-the-equation\/","title":{"rendered":"Formic acid dissociates in water according to the equation"},"content":{"rendered":"\n<p>Formic acid dissociates in water according to the equation: HCOOH(aq) \u00e2\u2020\u2019 H+(aq) + HCOO-(aq) Given the following equilibrium concentrations, calculate the equilibrium constant: [HCOO-] = 0.550 M, [H+] = 0.060 M, [HCOOH] = 0.060 M.<\/p>\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>To calculate the equilibrium constant KaK_aKa\u200b for formic acid dissociation, we can use the equation for the dissociation of formic acid:HCOOH&nbsp;(aq)\u21ccH+(aq)+HCOO\u2212(aq)\\text{HCOOH (aq)} \\rightleftharpoons \\text{H}^+ (aq) + \\text{HCOO}^- (aq)HCOOH&nbsp;(aq)\u21ccH+(aq)+HCOO\u2212(aq)<\/p>\n\n\n\n<p>The equilibrium expression for this reaction is:Ka=[H+][HCOO\u2212][HCOOH]K_a = \\frac{[\\text{H}^+][\\text{HCOO}^-]}{[\\text{HCOOH}]}Ka\u200b=[HCOOH][H+][HCOO\u2212]\u200b<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: Plug the given concentrations into the equation<\/h3>\n\n\n\n<p>From the problem, the following equilibrium concentrations are given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>[HCOO\u2212]=0.550\u2009M[ \\text{HCOO}^- ] = 0.550 \\, \\text{M}[HCOO\u2212]=0.550M<\/li>\n\n\n\n<li>[H+]=0.060\u2009M[ \\text{H}^+ ] = 0.060 \\, \\text{M}[H+]=0.060M<\/li>\n\n\n\n<li>[HCOOH]=0.060\u2009M[ \\text{HCOOH} ] = 0.060 \\, \\text{M}[HCOOH]=0.060M<\/li>\n<\/ul>\n\n\n\n<p>Now, substitute these values into the equilibrium constant equation:Ka=(0.060)(0.550)0.060K_a = \\frac{(0.060)(0.550)}{0.060}Ka\u200b=0.060(0.060)(0.550)\u200b<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 2: Simplify the equation<\/h3>\n\n\n\n<p>Ka=0.0330.060K_a = \\frac{0.033}{0.060}Ka\u200b=0.0600.033\u200b<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 3: Perform the division<\/h3>\n\n\n\n<p>Ka=0.55K_a = 0.55Ka\u200b=0.55<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Conclusion<\/h3>\n\n\n\n<p>The equilibrium constant KaK_aKa\u200b for the dissociation of formic acid is 0.550.550.55.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation<\/h3>\n\n\n\n<p>Formic acid is a weak acid, meaning it does not fully dissociate in water. The equilibrium constant KaK_aKa\u200b provides a measure of the extent of dissociation. The expression for KaK_aKa\u200b takes into account the concentrations of the products (H+ and HCOO-) and the undissociated reactant (HCOOH) at equilibrium. The calculated value of Ka=0.55K_a = 0.55Ka\u200b=0.55 suggests that formic acid is a weak acid, since it does not dissociate completely, and its KaK_aKa\u200b value is moderate, indicating partial dissociation. The lower the value of KaK_aKa\u200b, the weaker the acid, and the higher the value, the stronger the acid.<\/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-banner8-1458.jpeg\" alt=\"\" class=\"wp-image-44036\" srcset=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-1458.jpeg 852w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-1458-250x300.jpeg 250w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-1458-768x923.jpeg 768w\" sizes=\"auto, (max-width: 852px) 100vw, 852px\" \/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Formic acid dissociates in water according to the equation: HCOOH(aq) \u00e2\u2020\u2019 H+(aq) + HCOO-(aq) Given the following equilibrium concentrations, calculate the equilibrium constant: [HCOO-] = 0.550 M, [H+] = 0.060 M, [HCOOH] = 0.060 M. The Correct Answer and Explanation is: To calculate the equilibrium constant KaK_aKa\u200b for formic acid dissociation, we can use 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-44020","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/44020","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=44020"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/44020\/revisions"}],"predecessor-version":[{"id":44042,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/44020\/revisions\/44042"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=44020"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=44020"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=44020"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}