Write a balanced chemical equation describing the dissolution (dissolving) of strontium fluoride. Determine the molar solubility of SrF2 in a 0.5 mol\/L solution of Sr(NO3)2. (Strontium nitrate is a highly soluble salt)<\/p>\n\n\n\n
The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n Balanced chemical equation for the dissolution of strontium fluoride:<\/strong><\/p>\n\n\n\n Strontium fluoride (SrF\u2082) dissociates in water as follows:<\/p>\n\n\n\n [ This is the chemical equation representing the dissolution of SrF\u2082 into its ions.<\/p>\n\n\n\n [ Let’s denote the molar solubility of SrF\u2082 as “s” mol\/L. This means that for every mole of SrF\u2082 that dissolves, “s” moles of Sr\u00b2\u207a and “2s” moles of F\u207b are released into solution. However, since Sr(NO\u2083)\u2082 is already providing Sr\u00b2\u207a ions (0.5 mol\/L), we must account for this in the equilibrium calculation.<\/p>\n\n\n\n Thus, the Ksp expression becomes:<\/p>\n\n\n\n [ For SrF\u2082, the Ksp is approximately ( 2.6 \\times 10^{-9} ) at room temperature. Since “s” is typically very small compared to 0.5 M, we can approximate the equation as:<\/p>\n\n\n\n [ Solving for “s”:<\/p>\n\n\n\n [ The molar solubility of SrF\u2082 is the amount of SrF\u2082 that dissolves in the solution. In the presence of 0.5 M Sr(NO\u2083)\u2082, the large concentration of Sr\u00b2\u207a from strontium nitrate reduces the solubility of SrF\u2082 due to the common ion effect. This effect lowers the amount of SrF\u2082 that can dissolve because adding Sr\u00b2\u207a from Sr(NO\u2083)\u2082 shifts the dissolution equilibrium of SrF\u2082 to the left, decreasing its solubility.<\/p>\n","protected":false},"excerpt":{"rendered":" Write a balanced chemical equation describing the dissolution (dissolving) of strontium fluoride. Determine the molar solubility of SrF2 in a 0.5 mol\/L solution of Sr(NO3)2. (Strontium nitrate is a highly soluble salt) The Correct Answer and Explanation is : Balanced chemical equation for the dissolution of strontium fluoride: Strontium fluoride (SrF\u2082) dissociates in water as […]<\/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-156977","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/156977","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=156977"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/156977\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=156977"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=156977"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=156977"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\\text{SrF}_2 (s) \\rightleftharpoons \\text{Sr}^{2+} (aq) + 2\\text{F}^{-} (aq)
]<\/p>\n\n\n\nDetermining Molar Solubility in 0.5 mol\/L Sr(NO\u2083)\u2082 Solution<\/h3>\n\n\n\n
\n
K_{sp} = [\\text{Sr}^{2+}][\\text{F}^-]^2
]<\/p>\n\n\n\n\n
\n
[
[\\text{Sr}^{2+}] = 0.5 + s
]<\/li>\n\n\n\n
[
[\\text{F}^-] = 2s
]<\/li>\n<\/ul>\n\n\n\n
K_{sp} = (0.5 + s)(2s)^2
]<\/p>\n\n\n\n
K_{sp} \\approx (0.5)(4s^2)
]<\/p>\n\n\n\n
s^2 \\approx \\frac{K_{sp}}{2} \\times \\frac{1}{0.5}
]
[
s^2 = \\frac{2.6 \\times 10^{-9}}{2}
]
[
s = \\sqrt{1.3 \\times 10^{-9}}
]
[
s \\approx 1.14 \\times 10^{-5} \\, \\text{mol\/L}
]<\/p>\n\n\n\nExplanation:<\/h3>\n\n\n\n