{"id":237850,"date":"2025-06-17T17:18:09","date_gmt":"2025-06-17T17:18:09","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=237850"},"modified":"2025-06-17T17:18:12","modified_gmt":"2025-06-17T17:18:12","slug":"what-is-the-original-molarity-of-a-solution-of-formic-acid-hcooh-whose-ph-is-3-26-at-equilibrium","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/17\/what-is-the-original-molarity-of-a-solution-of-formic-acid-hcooh-whose-ph-is-3-26-at-equilibrium\/","title":{"rendered":"What IS the original molarity of a solution of formic acid (HCOOH) whose pH is 3.26 at equilibrium"},"content":{"rendered":"\n

What IS the original molarity of a solution of formic acid (HCOOH) whose pH is 3.26 at equilibrium?<\/p>\n\n\n\n

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

To find the original molarity of a solution of formic acid (HCOOH) with a pH of 3.26 at equilibrium, we need to use the acid dissociation constant (Ka) for formic acid and apply the ICE (Initial, Change, Equilibrium) method.<\/p>\n\n\n\n

Step 1: Write the dissociation equation<\/h3>\n\n\n\n

Formic acid is a weak acid and partially ionizes:
HCOOH \u21cc H\u207a + HCOO\u207b<\/strong><\/p>\n\n\n\n

Step 2: Use the pH to find H+H\u207aH+<\/h3>\n\n\n\n

pH = 3.26
H+H\u207aH+ = 10^(\u2013pH) = 10^(\u20133.26) \u2248 5.50 \u00d7 10\u207b\u2074 M<\/strong><\/p>\n\n\n\n

Step 3: Use the Ka of formic acid<\/h3>\n\n\n\n

Ka of formic acid = 1.8 \u00d7 10\u207b\u2074<\/strong><\/p>\n\n\n\n

Let the initial concentration of HCOOH be C<\/strong>
At equilibrium:<\/p>\n\n\n\n

    \n
  • H+H\u207aH+ = HCOO\u2212HCOO\u207bHCOO\u2212 = 5.50 \u00d7 10\u207b\u2074 M<\/strong><\/li>\n\n\n\n
  • HCOOHHCOOHHCOOH = C \u2013 5.50 \u00d7 10\u207b\u2074<\/li>\n<\/ul>\n\n\n\n

    Using the Ka expression:
    Ka = H+H\u207aH+HCOO\u2212HCOO\u207bHCOO\u2212 \/ HCOOHHCOOHHCOOH
    1.8 \u00d7 10\u207b\u2074 = (5.50 \u00d7 10\u207b\u2074)\u00b2 \/ (C \u2013 5.50 \u00d7 10\u207b\u2074)<\/p>\n\n\n\n

    Solve for C:
    1.8 \u00d7 10\u207b\u2074 = (3.03 \u00d7 10\u207b\u2077) \/ (C \u2013 5.50 \u00d7 10\u207b\u2074)
    Multiply both sides by (C \u2013 5.50 \u00d7 10\u207b\u2074):
    1.8 \u00d7 10\u207b\u2074(C \u2013 5.50 \u00d7 10\u207b\u2074) = 3.03 \u00d7 10\u207b\u2077
    Expand the left side:
    1.8 \u00d7 10\u207b\u2074C \u2013 9.9 \u00d7 10\u207b\u2078 = 3.03 \u00d7 10\u207b\u2077
    Add 9.9 \u00d7 10\u207b\u2078 to both sides:
    1.8 \u00d7 10\u207b\u2074C = 3.03 \u00d7 10\u207b\u2077 + 9.9 \u00d7 10\u207b\u2078 = 3.999 \u00d7 10\u207b\u2077
    Now divide both sides:
    C = (3.999 \u00d7 10\u207b\u2077) \/ (1.8 \u00d7 10\u207b\u2074) \u2248 2.22 \u00d7 10\u207b\u00b3 M<\/strong><\/p>\n\n\n\n

    \n\n\n\n

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

    The original molarity of formic acid is 0.00222 M<\/strong><\/p>\n\n\n\n

    \n\n\n\n

    Explanation (300 words):<\/h3>\n\n\n\n

    To determine the original molarity of a formic acid solution with a pH of 3.26, we must understand the relationship between pH, hydrogen ion concentration, and the dissociation behavior of weak acids. The pH value indicates how acidic a solution is and can be converted into the hydrogen ion concentration by using the formula H+H\u207aH+ = 10^(\u2013pH). For a pH of 3.26, this gives approximately 5.50 \u00d7 10\u207b\u2074 M.<\/p>\n\n\n\n

    Formic acid is a weak acid that does not completely ionize in water. It partially dissociates into hydrogen ions and formate ions. The acid dissociation constant (Ka) quantifies how much of the acid dissociates at equilibrium. For formic acid, Ka is 1.8 \u00d7 10\u207b\u2074. Using the ICE method, we let the initial concentration of the acid be C. At equilibrium, the amount that has dissociated is equal to the concentration of hydrogen ions, which is also equal to the concentration of formate ions.<\/p>\n\n\n\n

    We then plug these values into the Ka expression, solving for C, the initial concentration of the acid. After doing algebraic manipulation, we find that C is equal to 0.00222 M. This means that to produce a pH of 3.26 at equilibrium, the solution must have originally contained formic acid at a concentration of about 2.22 millimoles per liter. This demonstrates how the behavior of weak acids in solution can be quantified through equilibrium chemistry and pH calculations.<\/p>\n\n\n\n

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

    What IS the original molarity of a solution of formic acid (HCOOH) whose pH is 3.26 at equilibrium? The Correct Answer and Explanation is: To find the original molarity of a solution of formic acid (HCOOH) with a pH of 3.26 at equilibrium, we need to use the acid dissociation constant (Ka) for formic acid […]<\/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-237850","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/237850","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=237850"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/237850\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=237850"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=237850"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=237850"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}