{"id":196283,"date":"2025-03-03T20:21:17","date_gmt":"2025-03-03T20:21:17","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=196283"},"modified":"2025-03-03T20:21:20","modified_gmt":"2025-03-03T20:21:20","slug":"write-the-equation-for-the-dissociation-of-aluminum-hydroxide","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/03\/03\/write-the-equation-for-the-dissociation-of-aluminum-hydroxide\/","title":{"rendered":"Write the equation for the dissociation of aluminum hydroxide"},"content":{"rendered":"\n

a) Write the equation for the dissociation of aluminum hydroxide. 12 b) If the pH is 9.85, what is the concentration of the aluminum hydroxide solution? 13. a) Write the equation for the dissociation of calcium hydroxide. b) If the pH is 11.64 and you have 2.55 L of solution, how many grams of calcium hydroxide are in the solution? The molecular weight of calcium hydroxide is 74.1 g\/mole.<\/p>\n\n\n\n

The correct answer and explanation is:<\/mark><\/strong><\/p>\n\n\n\n

Solutions to the Questions:<\/strong><\/h3>\n\n\n\n

1. Dissociation of Aluminum Hydroxide (Al(OH)\u2083)<\/strong><\/h4>\n\n\n\n

The equation for the dissociation of aluminum hydroxide in water is: Al(OH)3(s)\u21ccAl3+(aq)+3OH\u2212(aq)Al(OH)_3 (s) \\rightleftharpoons Al^{3+} (aq) + 3OH^- (aq)<\/p>\n\n\n\n

Since aluminum hydroxide is only sparingly soluble in water, its solubility is governed by its solubility product constant, KspK_{sp}.<\/p>\n\n\n\n

1b. Finding the Concentration of Al(OH)\u2083 Given pH = 9.85<\/strong><\/h4>\n\n\n\n
    \n
  1. Find the pOH:<\/strong> pOH=14\u2212pH=14\u22129.85=4.15pOH = 14 – pH = 14 – 9.85 = 4.15<\/li>\n\n\n\n
  2. Find the Hydroxide Ion Concentration:<\/strong> [OH\u2212]=10\u2212pOH=10\u22124.15=7.08\u00d710\u22125\u00a0M[OH^-] = 10^{-pOH} = 10^{-4.15} = 7.08 \\times 10^{-5} \\text{ M}<\/li>\n\n\n\n
  3. Determine the Solubility of Al(OH)\u2083:<\/strong> Since each aluminum hydroxide molecule dissociates into 3 hydroxide ions: S=[OH\u2212]3=7.08\u00d710\u221253=2.36\u00d710\u22125\u00a0MS = \\frac{[OH^-]}{3} = \\frac{7.08 \\times 10^{-5}}{3} = 2.36 \\times 10^{-5} \\text{ M}<\/li>\n<\/ol>\n\n\n\n

    Thus, the concentration of Al(OH)\u2083<\/strong> in solution is 2.36\u00d710\u221252.36 \\times 10^{-5} M<\/strong>.<\/p>\n\n\n\n

    \n\n\n\n

    2. Dissociation of Calcium Hydroxide (Ca(OH)\u2082)<\/strong><\/h4>\n\n\n\n

    The equation for the dissociation of calcium hydroxide in water is: Ca(OH)2(s)\u21ccCa2+(aq)+2OH\u2212(aq)Ca(OH)_2 (s) \\rightleftharpoons Ca^{2+} (aq) + 2OH^- (aq)<\/p>\n\n\n\n

    2b. Finding the Mass of Ca(OH)\u2082 Given pH = 11.64 and Volume = 2.55 L<\/strong><\/h4>\n\n\n\n
      \n
    1. Find the pOH:<\/strong> pOH=14\u221211.64=2.36pOH = 14 – 11.64 = 2.36<\/li>\n\n\n\n
    2. Find the Hydroxide Ion Concentration:<\/strong> [OH\u2212]=10\u2212pOH=10\u22122.36=4.37\u00d710\u22123\u00a0M[OH^-] = 10^{-pOH} = 10^{-2.36} = 4.37 \\times 10^{-3} \\text{ M}<\/li>\n\n\n\n
    3. Find the Calcium Hydroxide Concentration:<\/strong> Since Ca(OH)\u2082 dissociates into 2 hydroxide ions per molecule<\/strong>, the solubility of Ca(OH)\u2082<\/strong> is: S=[OH\u2212]2=4.37\u00d710\u221232=2.18\u00d710\u22123\u00a0MS = \\frac{[OH^-]}{2} = \\frac{4.37 \\times 10^{-3}}{2} = 2.18 \\times 10^{-3} \\text{ M}<\/li>\n\n\n\n
    4. Find the Moles in 2.55 L Solution:<\/strong> moles\u00a0of\u00a0Ca(OH)2=S\u00d7V=(2.18\u00d710\u22123\u00a0M)\u00d7(2.55\u00a0L)\\text{moles of } Ca(OH)_2 = S \\times V = (2.18 \\times 10^{-3} \\text{ M}) \\times (2.55 \\text{ L}) =5.56\u00d710\u22123\u00a0moles= 5.56 \\times 10^{-3} \\text{ moles}<\/li>\n\n\n\n
    5. Find the Mass of Ca(OH)\u2082:<\/strong> Using the molar mass of calcium hydroxide (74.1\u00a0g\/mol74.1 \\text{ g\/mol}): Mass=moles\u00d7molar\u00a0mass\\text{Mass} = \\text{moles} \\times \\text{molar mass} =(5.56\u00d710\u22123\u00a0moles)\u00d7(74.1\u00a0g\/mol)= (5.56 \\times 10^{-3} \\text{ moles}) \\times (74.1 \\text{ g\/mol}) =0.412\u00a0g= 0.412 \\text{ g}<\/li>\n<\/ol>\n\n\n\n

      Thus, 0.412 g of Ca(OH)\u2082<\/strong> is present in the 2.55 L solution.<\/p>\n\n\n\n

      \n\n\n\n

      Explanation <\/strong><\/h3>\n\n\n\n

      Understanding the dissociation of aluminum hydroxide (Al(OH)\u2083<\/strong>) and calcium hydroxide (Ca(OH)\u2082<\/strong>) in water is crucial for calculating their solubility and mass in solution.<\/p>\n\n\n\n

      When Al(OH)\u2083<\/strong> dissolves, it releases Al3+Al^{3+}<\/strong> and OH\u2212OH^-<\/strong> ions, with a 1:3 ratio. Given a pH of 9.85<\/strong>, the hydroxide ion concentration can be determined using the relation pOH=14\u2212pHpOH = 14 – pH. This helps in calculating the molar solubility of aluminum hydroxide by dividing the hydroxide ion concentration by 3. Since aluminum hydroxide is sparingly soluble, its concentration is very low, around 2.36\u00d710\u221252.36 \\times 10^{-5} M<\/strong>.<\/p>\n\n\n\n

      For Ca(OH)\u2082<\/strong>, the dissociation releases Ca2+Ca^{2+}<\/strong> and OH\u2212OH^-<\/strong> ions in a 1:2 ratio. Given a pH of 11.64<\/strong>, we find the pOH and then the hydroxide ion concentration. Since calcium hydroxide dissociates into two hydroxide ions per molecule<\/strong>, we divide by 2 to obtain the molar solubility of 2.18\u00d710\u221232.18 \\times 10^{-3} M<\/strong>. Multiplying this by the given solution volume (2.55 L<\/strong>) gives the moles of calcium hydroxide, which is converted into mass using its molar mass (74.174.1 g\/mol). This yields 0.412 g of Ca(OH)\u2082<\/strong> in the solution.<\/p>\n\n\n\n

      Understanding such dissociation calculations is essential in analytical chemistry, environmental science, and industrial applications. These concepts are widely used in water treatment, chemical manufacturing, and pH control in various processes.<\/p>\n\n\n\n

      \n\n\n\n

      Now, I’ll generate an image illustrating these dissociation reactions.<\/p>\n\n\n\n

      I’ve generated an educational chemistry diagram illustrating the dissociation of aluminum hydroxide and calcium hydroxide in water, including their chemical equations and labeled molecular structures. Let me know if you need any modifications!<\/p>\n","protected":false},"excerpt":{"rendered":"

      a) Write the equation for the dissociation of aluminum hydroxide. 12 b) If the pH is 9.85, what is the concentration of the aluminum hydroxide solution? 13. a) Write the equation for the dissociation of calcium hydroxide. b) If the pH is 11.64 and you have 2.55 L of solution, how many grams of calcium […]<\/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-196283","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/196283","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=196283"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/196283\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=196283"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=196283"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=196283"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}