Calculate the solubility of silver phosphate, Ag3PO3 –>3Ag+ + PO33- , in pure water.
Ksp = 2.6 x 10-18 correct answer: 1.8 x 10^-5 M please show me step by step, and as detailed as possible, how to solve this problem. the correct answer is already written above ,thank you so much \ud83d\ude42<\/p>\n\n\n\n
The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n To calculate the solubility of silver phosphate (Ag\u2083PO\u2084) in pure water, we will use its solubility product constant ((K_{sp})) and apply the equilibrium concepts.<\/p>\n\n\n\n Thus, the solubility of silver phosphate in pure water is approximately (1.8 \\times 10^{-5} \\, \\text{M}).<\/p>\n\n\n\n The key to solving this problem is recognizing that the solubility product constant ((K_{sp})) relates the concentrations of the ions in equilibrium. By setting up an ICE table and substituting the equilibrium concentrations into the (K_{sp}) expression, we can solve for the solubility of Ag\u2083PO\u2084 in water.<\/p>\n\n\n\n The solubility constant is very small ((K_{sp} = 2.6 \\times 10^{-18})), indicating that silver phosphate is only sparingly soluble in water. This low solubility means that only a small amount of Ag\u2083PO\u2084 dissolves to form ions in solution.<\/p>\n\n\n\n The solubility result, (1.8 \\times 10^{-5} \\, \\text{M}), tells us how much silver phosphate can dissolve to reach equilibrium at a given temperature.<\/p>\n","protected":false},"excerpt":{"rendered":" Calculate the solubility of silver phosphate, Ag3PO3 –>3Ag+ + PO33- , in pure water.Ksp = 2.6 x 10-18 correct answer: 1.8 x 10^-5 M please show me step by step, and as detailed as possible, how to solve this problem. the correct answer is already written above ,thank you so much \ud83d\ude42 The Correct Answer […]<\/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-195828","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/195828","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=195828"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/195828\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=195828"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=195828"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=195828"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Step-by-Step Solution:<\/h3>\n\n\n\n
\n
Silver phosphate dissociates in water as follows: [
Ag_3PO_4(s) \\rightleftharpoons 3Ag^+(aq) + PO_4^{3-}(aq)
] From the dissociation equation, you can see that for every 1 mole of silver phosphate that dissolves, 3 moles of silver ions (Ag\u207a) and 1 mole of phosphate ions (PO\u2084\u00b3\u207b) are produced.<\/li>\n\n\n\n
Let ( s ) be the molar solubility of Ag\u2083PO\u2084 in pure water, which is the amount that dissolves to form Ag\u207a and PO\u2084\u00b3\u207b. Ion Initial Concentration (M) Change (M) Equilibrium Concentration (M) (Ag^+) 0 +3s 3s (PO_4^{3-}) 0 +s s<\/li>\n\n\n\n
K_{sp} = [Ag^+]^3 [PO_4^{3-}]
] Substituting the equilibrium concentrations: [
K_{sp} = (3s)^3 \\cdot (s)
] This simplifies to: [
K_{sp} = 27s^4
]<\/li>\n\n\n\n
2.6 \\times 10^{-18} = 27s^4
]<\/li>\n\n\n\n
s^4 = \\frac{2.6 \\times 10^{-18}}{27} = 9.63 \\times 10^{-20}
] Next, take the fourth root of both sides: [
s = \\sqrt[4]{9.63 \\times 10^{-20}}
] Using a calculator: [
s \\approx 1.8 \\times 10^{-5} \\, \\text{M}
]<\/li>\n<\/ol>\n\n\n\nExplanation:<\/h3>\n\n\n\n