{"id":226387,"date":"2025-06-05T08:26:39","date_gmt":"2025-06-05T08:26:39","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=226387"},"modified":"2025-06-05T08:26:42","modified_gmt":"2025-06-05T08:26:42","slug":"component-ions","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/05\/component-ions\/","title":{"rendered":"Component ions."},"content":{"rendered":"\n

component ions. This reaction goes to completion, as indicated by the one-way arrow in the following equation: NaC2H3O2(s) ==========> Na+(aq) + C2H3O2-(aq) The Na+, being the conjugate acid of a strong base (NaOH), is too weak to react with water. However, the C2H3O2-, being the conjugate base of a weak acid (HC2H3O2), is strong enough to react slightly with water. The equation for this equilibrium reaction is: C2H3O2-(aq) + H2O(l) <———-> HC2H3O2(aq) + OH-(aq) The equilibrium constant for the above reaction (Kb for C2H3O2-) is normally not published in tables because it can be calculated from two other values that ARE published: the ionization constants for H2O and HC2H3O2. 2H2O(l) <———-> H3O+(aq) + OH-(aq) Kw = 1.0 x 10^-14 HC2H3O2(aq) + H2O(l) <———-> H3O+(aq) + C2H3O2-(aq) Ka = 1.8 x 10^-5 By combining these two equations in the appropriate way, it is possible to obtain the desired reaction and hence its equilibrium constant. What is the hydronium ion concentration ([H3O+]) in a 4.75 M NaC2H3O2 solution? answers: 3.8 x 10^-13 mol\/L 4.2 x 10^-12 mol\/L 1.9 x 10^-10 mol\/L 5.1 x 10^-5 mol\/L 2.4 x 10^-3 mol\/L Can someone please explain how to do this step by step? Thank you.
Sodium acetate (NaC2H3O2) is a basic salt. When sodium acetate is dissolved in water, it dissociates into its component ions. This reaction goes to completion, as indicated by the one-way arrow in the following equation: NaC2H3O2(s) ==========> Na+(aq) + C2H3O2-(aq) The Na+, being the conjugate acid of a strong base (NaOH), is too weak to react with water. However, the C2H3O2-, being the conjugate base of a weak acid (HC2H3O2), is strong enough to react slightly with water. The equation for this equilibrium reaction is: C2H3O2-(aq) + H2O(l) <———-> HC2H3O2(aq) + OH-(aq) The equilibrium constant for the above reaction (Kb for C2H3O2-) is normally not published in tables because it can be calculated from two other values that ARE published: the ionization constants for H2O and HC2H3O2. 2H2O(l) <———-> H3O+(aq) + OH-(aq) Kw = 1.0 x 10^-14 HC2H3O2(aq) + H2O(l) <———-> H3O+(aq) + C2H3O2-(aq) Ka = 1.8 x 10^-5 By combining these two equations in the appropriate way, it is possible to obtain the desired reaction and hence its equilibrium constant. What is the hydronium ion concentration ([H3O+]) in a 4.75 M NaC2H3O2 solution? answers: 3.8 x 10^-13 mol\/L 4.2 x 10^-12 mol\/L 1.9 x 10^-10 mol\/L 5.1 x 10^-5 mol\/L 2.4 x 10^-3 mol\/L Can someone please explain how to do this step by step? Thank you.<\/p>\n\n\n\n

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

Correct Answer: 1.9 \u00d7 10\u207b\u00b9\u2070 mol\/L<\/strong><\/p>\n\n\n\n

Step-by-Step Explanation:<\/h3>\n\n\n\n

We are asked to find the [H\u2083O\u207a]<\/strong> in a 4.75 M NaC\u2082H\u2083O\u2082<\/strong> (sodium acetate) solution. Sodium acetate is a salt of a weak acid (acetic acid)<\/strong> and a strong base (NaOH)<\/strong>, so the solution is basic<\/strong>.<\/p>\n\n\n\n

The acetate ion (C\u2082H\u2083O\u2082\u207b) can react with water: C\u2082H\u2083O\u2082\u207b (aq)+H\u2082O (l)\u21ccHC\u2082H\u2083O\u2082 (aq)+OH\u207b (aq)\\text{C\u2082H\u2083O\u2082\u207b (aq)} + \\text{H\u2082O (l)} \\rightleftharpoons \\text{HC\u2082H\u2083O\u2082 (aq)} + \\text{OH\u207b (aq)}<\/p>\n\n\n\n

We are not given the base dissociation constant (K_b<\/strong>) directly, but we can calculate it: Kb=KwKa=1.0\u00d710\u2212141.8\u00d710\u22125=5.56\u00d710\u221210K_b = \\frac{K_w}{K_a} = \\frac{1.0 \\times 10^{-14}}{1.8 \\times 10^{-5}} = 5.56 \\times 10^{-10}<\/p>\n\n\n\n

Now, we use an ICE table<\/strong> (Initial, Change, Equilibrium) for the above equilibrium reaction. Let x<\/strong> be the amount of OH\u207b formed:<\/p>\n\n\n\n

Species<\/th>Initial (M)<\/th>Change (M)<\/th>Equilibrium (M)<\/th><\/tr><\/thead>
C\u2082H\u2083O\u2082\u207b<\/td>4.75<\/td>-x<\/td>4.75 – x \u2248 4.75<\/td><\/tr>
OH\u207b<\/td>0<\/td>+x<\/td>x<\/td><\/tr>
HC\u2082H\u2083O\u2082<\/td>0<\/td>+x<\/td>x<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

Using the expression for Kb: Kb=x2[C2H3O2\u2212]=x24.75K_b = \\frac{x^2}{[C\u2082H\u2083O\u2082\u207b]} = \\frac{x^2}{4.75}<\/p>\n\n\n\n

Substitute and solve: 5.56\u00d710\u221210=x24.75\u21d2x2=(5.56\u00d710\u221210)(4.75)=2.641\u00d710\u22129\u21d2x=2.641\u00d710\u22129\u22485.14\u00d710\u221255.56 \\times 10^{-10} = \\frac{x^2}{4.75} \\Rightarrow x^2 = (5.56 \\times 10^{-10})(4.75) = 2.641 \\times 10^{-9} \\Rightarrow x = \\sqrt{2.641 \\times 10^{-9}} \u2248 5.14 \\times 10^{-5}<\/p>\n\n\n\n

So, [OH\u207b] \u2248 5.14 \u00d7 10\u207b\u2075 M<\/strong><\/p>\n\n\n\n

Now find [H\u2083O\u207a] using: [H3O+]=Kw[OH\u2212]=1.0\u00d710\u2212145.14\u00d710\u22125\u22481.95\u00d710\u221210[H\u2083O\u207a] = \\frac{K_w}{[OH\u207b]} = \\frac{1.0 \\times 10^{-14}}{5.14 \\times 10^{-5}} \u2248 1.95 \\times 10^{-10}<\/p>\n\n\n\n

Rounded to two significant figures: [H3O+]=1.9\u00d710\u221210 mol\/L\\boxed{[H\u2083O\u207a] = 1.9 \\times 10^{-10} \\text{ mol\/L}}<\/p>\n\n\n\n

This value confirms the correct multiple-choice answer is:
1.9 \u00d7 10\u207b\u00b9\u2070 mol\/L<\/strong>.<\/p>\n\n\n\n

This calculation shows how weak bases (conjugate bases of weak acids) slightly increase pH by producing OH\u207b, and thus very low [H\u2083O\u207a].<\/p>\n\n\n\n

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

component ions. This reaction goes to completion, as indicated by the one-way arrow in the following equation: NaC2H3O2(s) ==========> Na+(aq) + C2H3O2-(aq) The Na+, being the conjugate acid of a strong base (NaOH), is too weak to react with water. However, the C2H3O2-, being the conjugate base of a weak acid (HC2H3O2), is strong enough […]<\/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-226387","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/226387","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=226387"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/226387\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=226387"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=226387"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=226387"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}