You dissolve 3.25 g of sodium oxalate (Na2C2O4) in sufficient water to produce a final solution volume of 100.0 mL. Assuming all of the solid dissolves, calculate the solution pH.<\/p>\n\n\n\n
The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n To calculate the pH of the solution formed by dissolving sodium oxalate (Na\u2082C\u2082O\u2084) in water, we need to follow several steps.<\/p>\n\n\n\n Sodium oxalate is a salt, and when it dissolves in water, it dissociates into sodium ions (Na\u207a) and oxalate ions (C\u2082O\u2084\u00b2\u207b): Sodium ions (Na\u207a) do not affect the pH because they are spectator ions, so we focus on the oxalate ions (C\u2082O\u2084\u00b2\u207b), which are the conjugate base of oxalic acid (H\u2082C\u2082O\u2084).<\/p>\n\n\n\n Oxalate ions can undergo hydrolysis in water to form hydroxide ions (OH\u207b), thus affecting the pH of the solution: The hydrolysis of C\u2082O\u2084\u00b2\u207b is a weak base reaction. To calculate the pH, we need the base dissociation constant (Kb) for oxalate, which can be derived from the acid dissociation constant (Ka) of oxalic acid. Oxalic acid is a diprotic acid with two dissociation constants. For simplicity, we use the value for the second dissociation (which is the relevant one for the C\u2082O\u2084\u00b2\u207b ion), and it is:<\/p>\n\n\n\n [ Using the relationship between Ka and Kb for conjugate acid-base pairs:<\/p>\n\n\n\n [ [ Now, we calculate the concentration of OH\u207b produced from the hydrolysis of C\u2082O\u2084\u00b2\u207b. First, determine the molarity of C\u2082O\u2084\u00b2\u207b in the solution. The molar mass of Na\u2082C\u2082O\u2084 is: Next, we use the Kb expression for hydrolysis: Assuming the concentration of OH\u207b formed is ( x ), we have: Since ( x ) is very small compared to 0.243, we approximate: Solving for ( x ) (the concentration of OH\u207b): The concentration of OH\u207b is ( 6.7 \\times 10^{-6} \\, \\text{M} ), so we can calculate the pOH: Finally, the pH is related to the pOH by: The pH of the solution is approximately 8.82<\/strong>, indicating that the solution is mildly basic.<\/p>\n","protected":false},"excerpt":{"rendered":" You dissolve 3.25 g of sodium oxalate (Na2C2O4) in sufficient water to produce a final solution volume of 100.0 mL. Assuming all of the solid dissolves, calculate the solution pH. The Correct Answer and Explanation is : To calculate the pH of the solution formed by dissolving sodium oxalate (Na\u2082C\u2082O\u2084) in water, we need to […]<\/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-188624","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/188624","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=188624"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/188624\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=188624"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=188624"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=188624"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}1. Dissociation of Sodium Oxalate:<\/h3>\n\n\n\n
[
\\text{Na\u2082C\u2082O\u2084} \\rightarrow 2\\text{Na}^+ + \\text{C\u2082O\u2084}^{2-}
]<\/p>\n\n\n\n2. Oxalate Ion and Its Hydrolysis:<\/h3>\n\n\n\n
[
\\text{C\u2082O\u2084}^{2-} + \\text{H\u2082O} \\rightleftharpoons \\text{HC\u2082O\u2084}^- + \\text{OH}^-
]<\/p>\n\n\n\n
\\text{Ka}_2 = 5.4 \\times 10^{-5}
]<\/p>\n\n\n\n
\\text{Kb} = \\frac{K_w}{\\text{Ka}_2}
]
where ( K_w = 1.0 \\times 10^{-14} ) (the ion product of water at 25\u00b0C).<\/p>\n\n\n\n
\\text{Kb} = \\frac{1.0 \\times 10^{-14}}{5.4 \\times 10^{-5}} = 1.85 \\times 10^{-10}
]<\/p>\n\n\n\n3. Calculate the Concentration of OH\u207b:<\/h3>\n\n\n\n
[
\\text{Molar mass of Na\u2082C\u2082O\u2084} = 2(22.99) + 2(12.01) + 4(16.00) = 134.00 \\, \\text{g\/mol}
]
The moles of Na\u2082C\u2082O\u2084 in 3.25 g is:
[
\\text{moles of Na\u2082C\u2082O\u2084} = \\frac{3.25 \\, \\text{g}}{134.00 \\, \\text{g\/mol}} = 0.0243 \\, \\text{mol}
]
The concentration of C\u2082O\u2084\u00b2\u207b in the solution is:
[
\\text{Concentration of C\u2082O\u2084\u00b2\u207b} = \\frac{0.0243 \\, \\text{mol}}{0.1 \\, \\text{L}} = 0.243 \\, \\text{M}
]<\/p>\n\n\n\n
[
\\text{Kb} = \\frac{[\\text{HC\u2082O\u2084}^-][\\text{OH}^-]}{[\\text{C\u2082O\u2084}^{2-}]}
]<\/p>\n\n\n\n
[
1.85 \\times 10^{-10} = \\frac{x^2}{0.243 – x}
]<\/p>\n\n\n\n
[
1.85 \\times 10^{-10} = \\frac{x^2}{0.243}
]<\/p>\n\n\n\n
[
x^2 = (1.85 \\times 10^{-10})(0.243)
]
[
x^2 = 4.49 \\times 10^{-11}
]
[
x = \\sqrt{4.49 \\times 10^{-11}} = 6.7 \\times 10^{-6} \\, \\text{M}
]<\/p>\n\n\n\n4. Calculate the pOH and pH:<\/h3>\n\n\n\n
[
\\text{pOH} = -\\log(6.7 \\times 10^{-6}) = 5.18
]<\/p>\n\n\n\n
[
\\text{pH} = 14 – \\text{pOH} = 14 – 5.18 = 8.82
]<\/p>\n\n\n\nConclusion:<\/h3>\n\n\n\n