{"id":229345,"date":"2025-06-08T06:25:25","date_gmt":"2025-06-08T06:25:25","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=229345"},"modified":"2025-06-08T06:25:27","modified_gmt":"2025-06-08T06:25:27","slug":"report-sheet-molar-solubility-common-ion-effect-name-desk-no-date-lab-sec-molar-solubility-and-solubility-product-of-calcium-hydroxide-trial-1-trial-2-trial-3-volume-of-solid","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/08\/report-sheet-molar-solubility-common-ion-effect-name-desk-no-date-lab-sec-molar-solubility-and-solubility-product-of-calcium-hydroxide-trial-1-trial-2-trial-3-volume-of-solid\/","title":{"rendered":"Report Sheet Molar Solubility; Common-Ion Effect Name: Desk No: Date: Lab Sec: Molar Solubility and Solubility Product of Calcium Hydroxide Trial 1 Trial 2 Trial 3 Volume of solid"},"content":{"rendered":"\n<p>Report Sheet Molar Solubility; Common-Ion Effect Name: Desk No: Date: Lab Sec: Molar Solubility and Solubility Product of Calcium Hydroxide Trial 1 Trial 2 Trial 3 Volume of solid Ca(OH)2 solution (mL): 25.0 25.0 30 Concentration of standardized HCl solution (mol\/L): JyY Buret reading: initial (mL): Buret reading: final (mL): Volume of HCl added (mL): 10.40 10.45 10.35 Moles of HCl added (mol): Moles of OH- in Ca(OH)2 solution (mol): [OH-] at equilibrium (mol\/L): 0.1080 * 10.40 \/ 25.00 [Ca2+] at equilibrium (mol\/L): Molar solubility of Ca(OH)2 (mol\/L): Average molar solubility of Ca(OH)2 (mol\/L): Ksp of Ca(OH)2: Average Ksp: Standard deviation of Ksp: Relative standard deviation of Ksp (SRSD): Calculations for Trial:<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-1-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>To complete your <strong>Experiment 22 Report Sheet<\/strong> on the <strong>Molar Solubility and Solubility Product (Ksp) of Calcium Hydroxide<\/strong>, we need to walk through calculations step-by-step for <strong>Trial 1<\/strong>, assuming:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Volume of Ca(OH)\u2082 solution = 25.0 mL = 0.02500 L<\/li>\n\n\n\n<li>Volume of HCl added = 10.40 mL = 0.01040 L<\/li>\n\n\n\n<li><strong>Concentration of HCl (standardized)<\/strong> = 0.1080 mol\/L<\/li>\n<\/ul>\n\n\n\n<p>We will assume this concentration applies for now since it&#8217;s partly shown.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Trial 1 Calculations:<\/strong><\/h3>\n\n\n\n<h4 class=\"wp-block-heading\">1. <strong>Moles of HCl added<\/strong><\/h4>\n\n\n\n<p>mol&nbsp;HCl=MHCl\u00d7VHCl=0.1080\u2009mol\/L\u00d70.01040\u2009L=1.1232\u00d710\u22123\u2009mol\\text{mol HCl} = M_{\\text{HCl}} \\times V_{\\text{HCl}} = 0.1080 \\, \\text{mol\/L} \\times 0.01040 \\, \\text{L} = 1.1232 \\times 10^{-3} \\, \\text{mol}mol&nbsp;HCl=MHCl\u200b\u00d7VHCl\u200b=0.1080mol\/L\u00d70.01040L=1.1232\u00d710\u22123mol<\/p>\n\n\n\n<h4 class=\"wp-block-heading\">2. <strong>Moles of OH\u207b in Ca(OH)\u2082 solution<\/strong><\/h4>\n\n\n\n<p>Ca(OH)\u2082 reacts with HCl as:Ca(OH)2\u21ccCa2++2OH\u2212\\text{Ca(OH)}_2 \\rightleftharpoons \\text{Ca}^{2+} + 2\\text{OH}^-Ca(OH)2\u200b\u21ccCa2++2OH\u2212HCl+OH\u2212\u2192H2O\\text{HCl} + \\text{OH}^- \\rightarrow \\text{H}_2\\text{O}HCl+OH\u2212\u2192H2\u200bO<\/p>\n\n\n\n<p>Since 1 mol HCl reacts with 1 mol OH\u207b:mol&nbsp;OH\u2212=1.1232\u00d710\u22123\u2009mol\\text{mol OH}^- = 1.1232 \\times 10^{-3} \\, \\text{mol}mol&nbsp;OH\u2212=1.1232\u00d710\u22123mol<\/p>\n\n\n\n<h4 class=\"wp-block-heading\">3. <strong>[OH\u207b] at equilibrium<\/strong><\/h4>\n\n\n\n<p>[OH\u2212]=mol&nbsp;OH\u2212volume&nbsp;of&nbsp;Ca(OH)2&nbsp;solution=1.1232\u00d710\u221230.02500=0.04493\u2009mol\/L[\\text{OH}^-] = \\frac{\\text{mol OH}^-}{\\text{volume of Ca(OH)}_2 \\text{ solution}} = \\frac{1.1232 \\times 10^{-3}}{0.02500} = 0.04493 \\, \\text{mol\/L}[OH\u2212]=volume&nbsp;of&nbsp;Ca(OH)2\u200b&nbsp;solutionmol&nbsp;OH\u2212\u200b=0.025001.1232\u00d710\u22123\u200b=0.04493mol\/L<\/p>\n\n\n\n<h4 class=\"wp-block-heading\">4. <strong>[Ca\u00b2\u207a] at equilibrium<\/strong><\/h4>\n\n\n\n<p>From stoichiometry:<br>Each 1 mol of Ca(OH)\u2082 releases <strong>2 mol OH\u207b<\/strong> and <strong>1 mol Ca\u00b2\u207a<\/strong>[Ca2+]=[OH\u2212]2=0.044932=0.02247\u2009mol\/L[\\text{Ca}^{2+}] = \\frac{[\\text{OH}^-]}{2} = \\frac{0.04493}{2} = 0.02247 \\, \\text{mol\/L}[Ca2+]=2[OH\u2212]\u200b=20.04493\u200b=0.02247mol\/L<\/p>\n\n\n\n<h4 class=\"wp-block-heading\">5. <strong>Molar Solubility of Ca(OH)\u2082<\/strong><\/h4>\n\n\n\n<p>Molar&nbsp;solubility=[Ca2+]=0.02247\u2009mol\/L\\text{Molar solubility} = [\\text{Ca}^{2+}] = 0.02247 \\, \\text{mol\/L}Molar&nbsp;solubility=[Ca2+]=0.02247mol\/L<\/p>\n\n\n\n<h4 class=\"wp-block-heading\">6. <strong>Ksp of Ca(OH)\u2082<\/strong><\/h4>\n\n\n\n<p>Ksp=[Ca2+][OH\u2212]2=(0.02247)(0.04493)2=(0.02247)(0.002018)=4.534\u00d710\u22125K_{sp} = [\\text{Ca}^{2+}][\\text{OH}^-]^2 = (0.02247)(0.04493)^2 = (0.02247)(0.002018) = 4.534 \\times 10^{-5}Ksp\u200b=[Ca2+][OH\u2212]2=(0.02247)(0.04493)2=(0.02247)(0.002018)=4.534\u00d710\u22125<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Explanation<\/strong><\/h3>\n\n\n\n<p>This experiment determines the <strong>molar solubility<\/strong> and <strong>solubility product constant (Ksp)<\/strong> of calcium hydroxide, Ca(OH)\u2082, using a titration approach with hydrochloric acid (HCl). When Ca(OH)\u2082 dissolves in water, it dissociates into calcium (Ca\u00b2\u207a) and hydroxide (OH\u207b) ions. The solubility equilibrium is expressed as:Ca(OH)2(s)\u21ccCa2+(aq)+2OH\u2212(aq)\\text{Ca(OH)}_2 (s) \\rightleftharpoons \\text{Ca}^{2+} (aq) + 2\\text{OH}^- (aq)Ca(OH)2\u200b(s)\u21ccCa2+(aq)+2OH\u2212(aq)<\/p>\n\n\n\n<p>A known volume of the saturated Ca(OH)\u2082 solution is titrated with a standardized HCl solution. Because each mole of HCl neutralizes one mole of OH\u207b, the amount of acid used tells us the total number of moles of OH\u207b in the solution.<\/p>\n\n\n\n<p>In Trial 1, we added 10.40 mL of 0.1080 M HCl, introducing 1.1232 \u00d7 10\u207b\u00b3 moles of HCl, and thus neutralizing the same amount of OH\u207b. Dividing by the volume of the Ca(OH)\u2082 solution gives the [OH\u207b] concentration. The [Ca\u00b2\u207a] is then calculated by dividing [OH\u207b] by 2, because each Ca\u00b2\u207a corresponds to 2 OH\u207b ions.<\/p>\n\n\n\n<p>This [Ca\u00b2\u207a] is taken as the molar solubility, since it represents how much Ca(OH)\u2082 dissociated. Finally, the Ksp is calculated using the equilibrium concentrations in the formula:Ksp=[Ca2+][OH\u2212]2K_{sp} = [\\text{Ca}^{2+}][\\text{OH}^-]^2Ksp\u200b=[Ca2+][OH\u2212]2<\/p>\n\n\n\n<p>This approach allows for determining the <strong>solubility equilibrium<\/strong> of sparingly soluble salts. By performing multiple trials and averaging the Ksp, the reliability of the result increases. Calculating the <strong>standard deviation<\/strong> and <strong>relative standard deviation (SRSD)<\/strong> further evaluates the precision of the experiment.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner10-17.jpeg\" alt=\"\" class=\"wp-image-229346\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Report Sheet Molar Solubility; Common-Ion Effect Name: Desk No: Date: Lab Sec: Molar Solubility and Solubility Product of Calcium Hydroxide Trial 1 Trial 2 Trial 3 Volume of solid Ca(OH)2 solution (mL): 25.0 25.0 30 Concentration of standardized HCl solution (mol\/L): JyY Buret reading: initial (mL): Buret reading: final (mL): Volume of HCl added (mL): [&hellip;]<\/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-229345","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/229345","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=229345"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/229345\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=229345"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=229345"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=229345"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}