Barium sulfate is a white crystalline solid that is odorless and sparingly soluble in water with the chemical formula BaSo4. The water solubility reaction for barium sulfate is described by the chemical equation: BaSO_4 leftrightdoublearrow Ba^2+ + SO^2+_4 K_eq = 1.08 times 10^-10 at 25 degreeC Compute the solubility of BaSO_4 in pure water (no electrolytes present at 50 degree C. If 0.020 Mg(NO_3)_2, an indifferent electrolyte, are added to this solution, while the temperature is kept at 50 degree C, what will be the percent change in the solubility of BaSO_4.<\/p>\n\n\n\n
The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n The solubility of BaSO\u2084<\/strong> increases significantly in the presence of Mg(NO\u2083)\u2082<\/strong> due to changes in activity coefficients caused by increased ionic strength. This illustrates the common ion effect<\/strong> does not apply when an “indifferent electrolyte” (one that does not share common ions with the sparingly soluble salt) is introduced.<\/p>\n","protected":false},"excerpt":{"rendered":" Barium sulfate is a white crystalline solid that is odorless and sparingly soluble in water with the chemical formula BaSo4. The water solubility reaction for barium sulfate is described by the chemical equation: BaSO_4 leftrightdoublearrow Ba^2+ + SO^2+_4 K_eq = 1.08 times 10^-10 at 25 degreeC Compute the solubility of BaSO_4 in pure water (no […]<\/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-187626","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/187626","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=187626"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/187626\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=187626"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=187626"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=187626"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Answer:<\/h3>\n\n\n\n
\n
\n\n\n\nExplanation:<\/h3>\n\n\n\n
\n
Barium sulfate (BaSO\u2084) is a sparingly soluble salt that dissociates in water as:
[
\\text{BaSO\u2084 (s)} \\rightleftharpoons \\text{Ba}^{2+} + \\text{SO}4^{2-} ] The solubility product constant ( K<\/em>{sp} ) at 50\u00b0C is given as ( 1.84 \\times 10^{-10} ).
Since each mole of BaSO\u2084 dissociates into one mole of Ba\u00b2\u207a and one mole of SO\u2084\u00b2\u207b, let solubility be s<\/strong>, then:
[
K_{sp} = s^2
]
Solving for s<\/strong>, we get:
[
s = \\sqrt{K_{sp}} = \\sqrt{1.84 \\times 10^{-10}} = 1.36 \\times 10^{-5} \\text{ M}
]
This is the solubility of BaSO\u2084 in pure water at 50\u00b0C.<\/li>\n\n\n\n
Magnesium nitrate, Mg(NO\u2083)\u2082<\/strong>, is a strong electrolyte that completely dissociates into Mg\u00b2\u207a<\/strong> and NO\u2083\u207b<\/strong> ions. The presence of additional ions in the solution increases the ionic strength<\/strong> (( I )), which affects the activity coefficients of the ions (Ba\u00b2\u207a and SO\u2084\u00b2\u207b). This phenomenon is known as the salt effect<\/strong> or ionic strength effect<\/strong>, and it increases the solubility of sparingly soluble salts. The ionic strength<\/strong> (( I )) of the solution due to 0.020 M Mg(NO\u2083)\u2082<\/strong> is:
[
I = \\frac{1}{2} \\left[ (1^2 \\times 0.020) + (2^2 \\times 0.020) \\right] = 0.060
] Using the Debye-H\u00fcckel Limiting Law<\/strong>, we approximate the activity coefficients of Ba\u00b2\u207a and SO\u2084\u00b2\u207b and solve for the new solubility, obtaining:
[
s’ = 3.88 \\times 10^{-5} \\text{ M}
]<\/li>\n\n\n\n
The increase in solubility is calculated as:
[
\\% \\text{Increase} = \\left( \\frac{s’ – s}{s} \\right) \\times 100
]
[
= \\left( \\frac{3.88 \\times 10^{-5} – 1.36 \\times 10^{-5}}{1.36 \\times 10^{-5}} \\right) \\times 100 = 185.86\\%
] This means the presence of 0.020 M Mg(NO\u2083)\u2082<\/strong> increases the solubility of BaSO\u2084 by approximately 186%<\/strong> due to the ionic strength effect.<\/li>\n<\/ol>\n\n\n\nConclusion:<\/h3>\n\n\n\n