{"id":43215,"date":"2025-06-29T15:13:45","date_gmt":"2025-06-29T15:13:45","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=43215"},"modified":"2025-06-29T15:13:47","modified_gmt":"2025-06-29T15:13:47","slug":"choose-the-correct-net-ionic-equation-of-acetic-acid-being-neutralized-with-sodium-hydroxide","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/choose-the-correct-net-ionic-equation-of-acetic-acid-being-neutralized-with-sodium-hydroxide\/","title":{"rendered":"Choose the correct net ionic equation of acetic acid being neutralized with sodium hydroxide"},"content":{"rendered":"\n

Choose the correct net ionic equation of acetic acid being neutralized with sodium hydroxide: H+ (aq) + OH- (aq) \u00e2\u2020\u2019 H2O (aq) CH3COOH (aq) + OH- (aq) \u00e2\u2020\u2019 H2O (l) CH3COOH (aq) + H+ (aq) + OH- (aq) \u00e2\u2020\u2019 CH3COOH (aq) + H2O (l) CH3COOH (aq) + Na+ (aq) + OH- (aq) \u00e2\u2020\u2019 CH3COONa (aq) + H2O (l)<\/p>\n\n\n\n

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

The correct net ionic equation for the neutralization of acetic acid (CH\u2083COOH) with sodium hydroxide (NaOH) is:<\/p>\n\n\n\n

CH\u2083COOH (aq) + OH\u207b (aq) \u2192 CH\u2083COO\u207b (aq) + H\u2082O (l)<\/strong><\/p>\n\n\n\n

Explanation:<\/h3>\n\n\n\n

In this neutralization reaction, acetic acid, which is a weak acid, reacts with sodium hydroxide, a strong base. To derive the net ionic equation, let’s break down the process step by step:<\/p>\n\n\n\n

    \n
  1. Acetic Acid Dissociation:<\/strong>
    Acetic acid (CH\u2083COOH) is a weak acid, meaning it does not dissociate completely in water. However, for the purpose of neutralization, we consider its dissociation in an aqueous solution to some extent: CH\u2083COOH\u00a0(aq)\u21ccCH\u2083COO\u207b\u00a0(aq)+H\u207a\u00a0(aq)\\text{CH\u2083COOH (aq)} \\rightleftharpoons \\text{CH\u2083COO\u207b (aq)} + \\text{H\u207a (aq)}CH\u2083COOH\u00a0(aq)\u21ccCH\u2083COO\u207b\u00a0(aq)+H\u207a\u00a0(aq) Here, the acetic acid molecule dissociates into acetate ions (CH\u2083COO\u207b) and hydrogen ions (H\u207a).<\/li>\n\n\n\n
  2. Sodium Hydroxide Dissociation:<\/strong>
    Sodium hydroxide (NaOH), a strong base, dissociates completely in water: NaOH\u00a0(aq)\u2192Na\u207a\u00a0(aq)+OH\u207b\u00a0(aq)\\text{NaOH (aq)} \\rightarrow \\text{Na\u207a (aq)} + \\text{OH\u207b (aq)}NaOH\u00a0(aq)\u2192Na\u207a\u00a0(aq)+OH\u207b\u00a0(aq) The hydroxide ion (OH\u207b) is responsible for neutralizing the hydrogen ion (H\u207a).<\/li>\n\n\n\n
  3. Neutralization Reaction:<\/strong>
    When the hydrogen ion (H\u207a) from acetic acid reacts with the hydroxide ion (OH\u207b) from sodium hydroxide, they form water (H\u2082O): H\u207a\u00a0(aq)+OH\u207b\u00a0(aq)\u2192H\u2082O\u00a0(l)\\text{H\u207a (aq)} + \\text{OH\u207b (aq)} \\rightarrow \\text{H\u2082O (l)}H\u207a\u00a0(aq)+OH\u207b\u00a0(aq)\u2192H\u2082O\u00a0(l)<\/li>\n\n\n\n
  4. Final Ionic Equation:<\/strong>
    The sodium ion (Na\u207a) does not participate in the reaction because it is a spectator ion, meaning it remains unchanged throughout the reaction. Therefore, the net ionic equation only includes the components that undergo a chemical change: CH\u2083COOH\u00a0(aq)+OH\u207b\u00a0(aq)\u2192CH\u2083COO\u207b\u00a0(aq)+H\u2082O\u00a0(l)\\text{CH\u2083COOH (aq)} + \\text{OH\u207b (aq)} \\rightarrow \\text{CH\u2083COO\u207b (aq)} + \\text{H\u2082O (l)}CH\u2083COOH\u00a0(aq)+OH\u207b\u00a0(aq)\u2192CH\u2083COO\u207b\u00a0(aq)+H\u2082O\u00a0(l) This shows that acetic acid reacts with the hydroxide ion to produce acetate ions and water.<\/li>\n<\/ol>\n\n\n\n

    Thus, the correct net ionic equation is:CH\u2083COOH (aq)+OH\u207b (aq)\u2192CH\u2083COO\u207b (aq)+H\u2082O (l)\\text{CH\u2083COOH (aq)} + \\text{OH\u207b (aq)} \\rightarrow \\text{CH\u2083COO\u207b (aq)} + \\text{H\u2082O (l)}CH\u2083COOH (aq)+OH\u207b (aq)\u2192CH\u2083COO\u207b (aq)+H\u2082O (l)<\/p>\n\n\n\n

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

    Choose the correct net ionic equation of acetic acid being neutralized with sodium hydroxide: H+ (aq) + OH- (aq) \u00e2\u2020\u2019 H2O (aq) CH3COOH (aq) + OH- (aq) \u00e2\u2020\u2019 H2O (l) CH3COOH (aq) + H+ (aq) + OH- (aq) \u00e2\u2020\u2019 CH3COOH (aq) + H2O (l) CH3COOH (aq) + Na+ (aq) + OH- (aq) \u00e2\u2020\u2019 CH3COONa (aq) […]<\/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-43215","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/43215","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=43215"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/43215\/revisions"}],"predecessor-version":[{"id":43217,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/43215\/revisions\/43217"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=43215"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=43215"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=43215"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}