The solubility of silver chloride (AgCl) is 1.9 * 10 ^ – 3 * g \/ L Calculate kop – Molar mass for silver chloride is 143.5 g\/mol. Select one: a. 3.61 * 10 ^ – 9 * mol \/ L b. 4.68 * 10 ^ – 9 moll L c. 1.75 * 10 ^ – 10 mollL 0.6 * 0.34 * 10 ^ – 4 * mol \/ L<\/p>\n\n\n\n
The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n To calculate the solubility product constant (KspK_{sp}) of silver chloride (AgCl), we need to follow several steps.<\/p>\n\n\n\n The given solubility is in grams per liter, but we need it in moles per liter to calculate KspK_{sp}.<\/p>\n\n\n\n The molar mass of AgCl is 143.5 g\/mol. To convert the solubility into moles per liter: Solubility (mol\/L)=Solubility (g\/L)Molar mass (g\/mol)\\text{Solubility (mol\/L)} = \\frac{\\text{Solubility (g\/L)}}{\\text{Molar mass (g\/mol)}} Solubility (mol\/L)=1.9\u00d710\u22123\u2009g\/L143.5\u2009g\/mol\\text{Solubility (mol\/L)} = \\frac{1.9 \\times 10^{-3} \\, \\text{g\/L}}{143.5 \\, \\text{g\/mol}} Solubility (mol\/L)=1.32\u00d710\u22125\u2009mol\/L\\text{Solubility (mol\/L)} = 1.32 \\times 10^{-5} \\, \\text{mol\/L}<\/p>\n\n\n\n When silver chloride dissociates in water, it dissociates according to the following equation: AgCl (s)\u21ccAg+(aq)+Cl\u2212(aq)\\text{AgCl (s)} \\rightleftharpoons \\text{Ag}^+ (aq) + \\text{Cl}^- (aq)<\/p>\n\n\n\n For every mole of AgCl that dissolves, it produces 1 mole of Ag+\\text{Ag}^+ and 1 mole of Cl\u2212\\text{Cl}^-.<\/p>\n\n\n\n The solubility product constant KspK_{sp} is given by the product of the concentrations of the ions at equilibrium: Ksp=[Ag+][Cl\u2212]K_{sp} = [\\text{Ag}^+][\\text{Cl}^-]<\/p>\n\n\n\n Since the concentrations of Ag+\\text{Ag}^+ and Cl\u2212\\text{Cl}^- are both equal to the solubility of AgCl (1.32 x 10^-5 mol\/L), we can substitute this value: Ksp=(1.32\u00d710\u22125\u2009mol\/L)\u00d7(1.32\u00d710\u22125\u2009mol\/L)K_{sp} = (1.32 \\times 10^{-5} \\, \\text{mol\/L}) \\times (1.32 \\times 10^{-5} \\, \\text{mol\/L}) Ksp=1.74\u00d710\u221210\u2009mol2\/L2K_{sp} = 1.74 \\times 10^{-10} \\, \\text{mol}^2\/\\text{L}^2<\/p>\n\n\n\n We find that the correct value for KspK_{sp} is approximately 1.75\u00d710\u221210\u2009mol2\/L21.75 \\times 10^{-10} \\, \\text{mol}^2\/\\text{L}^2, which matches option c<\/strong>.<\/p>\n\n\n\n Thus, the correct answer is c. 1.75\u00d710\u221210\u2009mol\/L21.75 \\times 10^{-10} \\, \\text{mol\/L}^2<\/strong>.<\/p>\n\n\n\n The solubility product constant for AgCl is calculated by determining its solubility in moles per liter and then applying the formula for the solubility product. This gives a value of Ksp=1.75\u00d710\u221210K_{sp} = 1.75 \\times 10^{-10}, which corresponds to option c<\/strong>.<\/p>\n","protected":false},"excerpt":{"rendered":" The solubility of silver chloride (AgCl) is 1.9 * 10 ^ – 3 * g \/ L Calculate kop – Molar mass for silver chloride is 143.5 g\/mol. Select one: a. 3.61 * 10 ^ – 9 * mol \/ L b. 4.68 * 10 ^ – 9 moll L c. 1.75 * 10 ^ […]<\/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-186466","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/186466","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=186466"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/186466\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=186466"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=186466"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=186466"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Given:<\/h3>\n\n\n\n
\n
Step 1: Convert the solubility from grams per liter to moles per liter<\/h3>\n\n\n\n
Step 2: Set up the dissociation equation for AgCl<\/h3>\n\n\n\n
Step 3: Write the expression for KspK_{sp}<\/h3>\n\n\n\n
Step 4: Compare with the answer choices<\/h3>\n\n\n\n
Conclusion:<\/h3>\n\n\n\n