What is the solubility (in grams\/L) of lead (II) chromate (PbCrO4, molar mass = 323 g\/mol, Ksp = 2*10^-14) in 0.45 M K2CrO4<\/p>\n\n\n\n
A 0.350 M solution of the weak acid Ha has pH 4.35 at 25 degrees C. What is the \u25b3G for the dissociation of the weak acid?<\/p>\n\n\n\n
The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n The solubility equilibrium of lead(II) chromate is: Since dissolution adds ( s ) amount of ( Pb^{2+} ) and ( s ) amount of ( CrO_4^{2-} ), the equilibrium expression becomes: The dissociation of the weak acid ( HA ) follows: Using the dissociation constant formula: To find Gibbs free energy: Explanation:<\/strong> What is the solubility (in grams\/L) of lead (II) chromate (PbCrO4, molar mass = 323 g\/mol, Ksp = 2*10^-14) in 0.45 M K2CrO4 A 0.350 M solution of the weak acid Ha has pH 4.35 at 25 degrees C. What is the \u25b3G for the dissociation of the weak acid? The Correct Answer and Explanation […]<\/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-186992","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/186992","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=186992"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/186992\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=186992"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=186992"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=186992"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Problem 1: Solubility of PbCrO\u2084 in 0.45 M K\u2082CrO\u2084<\/strong><\/h3>\n\n\n\n
[
PbCrO_4 (s) \\rightleftharpoons Pb^{2+} (aq) + CrO_4^{2-} (aq)
]
The solubility product constant is:
[
K_{sp} = [Pb^{2+}][CrO_4^{2-}]
]
Let ( s ) be the solubility of PbCrO\u2084. Since the solution contains 0.45 M ( K_2CrO_4 ), the chromate ion concentration is initially 0.45 M before any dissolution.<\/p>\n\n\n\n
[
K_{sp} = (s)(0.45 + s)
]
Given ( K_{sp} = 2.0 \\times 10^{-14} ), and since ( s ) is very small compared to 0.45, we approximate:
[
K_{sp} \\approx (s)(0.45)
]
Solving for ( s ):
[
s = \\frac{2.0 \\times 10^{-14}}{0.45} = 4.44 \\times 10^{-14} \\text{ M}
]
To convert to g\/L:
[
\\text{Solubility} = (4.44 \\times 10^{-14} \\text{ mol\/L}) \\times (323 \\text{ g\/mol}) = 1.43 \\times 10^{-11} \\text{ g\/L}
]<\/p>\n\n\n\n
\n\n\n\nProblem 2: Gibbs Free Energy for Weak Acid Dissociation<\/strong><\/h3>\n\n\n\n
[
HA \\rightleftharpoons H^+ + A^-
]
Given:<\/p>\n\n\n\n\n
[
K_a = \\frac{[H^+][A^-]}{[HA]}
]
Since equal amounts of ( H^+ ) and ( A^- ) are formed,
[
K_a = \\frac{(4.47 \\times 10^{-5})^2}{0.350} = 5.71 \\times 10^{-9}
]<\/p>\n\n\n\n
[
\\Delta G = -RT \\ln K_a
]
Using ( R = 8.314 ) J\/(mol\u00b7K) and ( T = 298 ) K:
[
\\Delta G = – (8.314)(298) \\ln (5.71 \\times 10^{-9})
]
[
= – (8.314)(298) (-18.38)
]
[
= 45.6 \\text{ kJ\/mol}
]
Thus, ( \\Delta G \\approx 45.6 ) kJ\/mol.<\/p>\n\n\n\n
Gibbs free energy change ((\\Delta G)) indicates whether a reaction is spontaneous. A positive ( \\Delta G ) (as we found here) means the dissociation of this weak acid is non-spontaneous under standard conditions. The equilibrium constant (( K_a )) is quite small, which means the acid does not ionize significantly in solution. This aligns with our large positive ( \\Delta G ), indicating that the reverse reaction (association of ( H^+ ) and ( A^- )) is favored over dissociation.<\/p>\n","protected":false},"excerpt":{"rendered":"