{"id":27616,"date":"2025-06-20T04:09:49","date_gmt":"2025-06-20T04:09:49","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=27616"},"modified":"2025-06-20T04:09:52","modified_gmt":"2025-06-20T04:09:52","slug":"what-mast-0-cjnoh-must-added-to-olo-lom-hf-solution-to-begin-erecipikation-al-cafm","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/what-mast-0-cjnoh-must-added-to-olo-lom-hf-solution-to-begin-erecipikation-al-cafm\/","title":{"rendered":"What mast 0\/ CJ(NO;h; must added to [ OLo )LOM HF solution to begin erecipikation al Cafm"},"content":{"rendered":"\n

What mast 0\/ CJ(NO;h; must added to [ OLo )LOM HF solution to begin erecipikation al Cafm? (You MJY AssUme there hn0 votume change on the additton ol calaum t trate | Cafiakt 4.010 u Hk<\/em>7210 Describe hot You would scparte JMmorin [unethy Amenehydratint andeyridine Using Ion echanga cmonatogtjol. (Use Lhe follawing tablo for naeded valbes ) Aqueous-Equilibrium Constants TABLB Diesociallon Constants for F ABaraz 25C Hcnmn ommne Amecole R5: Anuyn Denatnutuen u 54<\/em>10 ‘ Etuelinmt 2 67 * 10″ tdtalit 13710- Anentennc aee 1140m Melneee flan L7ro- nnn acer<\/p>\n\n\n\n

The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n

To solve this question, we need to approach it in two parts<\/strong>:<\/p>\n\n\n\n

\n\n\n\n

Part 1: How much Ca(NO\u2083)\u2082 must be added to 0.010 M HF to begin precipitation of CaF\u2082?<\/strong><\/h3>\n\n\n\n

We are given:<\/p>\n\n\n\n

    \n
  • CaF2\u21ccCa2++2F\u2212CaF\u2082 \u21cc Ca\u00b2\u207a + 2F\u207bCaF2\u200b\u21ccCa2++2F\u2212<\/li>\n\n\n\n
  • Ksp of CaF\u2082 = 4.0 \u00d7 10\u207b\u00b9\u00b9<\/strong><\/li>\n\n\n\n
  • Initial [HF] = 0.010 M<\/strong><\/li>\n\n\n\n
  • Ka of HF = 6.8 \u00d7 10\u207b\u2074<\/strong><\/li>\n<\/ul>\n\n\n\n

    Step 1: Calculate [F\u207b] from HF<\/strong><\/h4>\n\n\n\n

    HF is a weak acid, so it only partially dissociates:HF\u21ccH++F\u2212HF \u21cc H\u207a + F\u207bHF\u21ccH++F\u2212<\/p>\n\n\n\n

    Using the expression for Ka:Ka=[H+][F\u2212][HF]Ka = \\frac{[H\u207a][F\u207b]}{[HF]}Ka=[HF][H+][F\u2212]\u200b<\/p>\n\n\n\n

    Assume initial [HF] = 0.010 M, let x be [H\u207a] and [F\u207b] at equilibrium:Ka=x20.010\u2212xKa = \\frac{x^2}{0.010 – x}Ka=0.010\u2212xx2\u200b<\/p>\n\n\n\n

    Since Ka is small, assume 0.010\u2212x\u22480.0100.010 – x \u2248 0.0100.010\u2212x\u22480.010:6.8\u00d710\u22124=x20.010\u21d2x2=6.8\u00d710\u22126\u21d2x=[F\u2212]\u22482.61\u00d710\u22123M6.8 \u00d7 10\u207b\u2074 = \\frac{x^2}{0.010} \\Rightarrow x^2 = 6.8 \u00d7 10\u207b\u2076 \\Rightarrow x = [F\u207b] \u2248 2.61 \u00d7 10\u207b\u00b3 M6.8\u00d710\u22124=0.010×2\u200b\u21d2x2=6.8\u00d710\u22126\u21d2x=[F\u2212]\u22482.61\u00d710\u22123M<\/p>\n\n\n\n

    \n\n\n\n

    Step 2: Use Ksp to find required [Ca\u00b2\u207a]<\/strong><\/h4>\n\n\n\n

    CaF2\u21ccCa2++2F\u2212CaF\u2082 \u21cc Ca\u00b2\u207a + 2F\u207bCaF2\u200b\u21ccCa2++2F\u2212Ksp=[Ca2+][F\u2212]2Ksp = [Ca\u00b2\u207a][F\u207b]\u00b2Ksp=[Ca2+][F\u2212]24.0\u00d710\u221211=[Ca2+](2.61\u00d710\u22123)2\u21d2[Ca2+]=4.0\u00d710\u2212116.81\u00d710\u22126\u21d2[Ca2+]\u22485.87\u00d710\u22126M4.0 \u00d7 10\u207b\u00b9\u00b9 = [Ca\u00b2\u207a](2.61 \u00d7 10\u207b\u00b3)^2 \\Rightarrow [Ca\u00b2\u207a] = \\frac{4.0 \u00d7 10\u207b\u00b9\u00b9}{6.81 \u00d7 10\u207b\u2076} \\Rightarrow [Ca\u00b2\u207a] \u2248 5.87 \u00d7 10\u207b\u2076 M4.0\u00d710\u221211=[Ca2+](2.61\u00d710\u22123)2\u21d2[Ca2+]=6.81\u00d710\u221264.0\u00d710\u221211\u200b\u21d2[Ca2+]\u22485.87\u00d710\u22126M<\/p>\n\n\n\n

    Answer:<\/strong> You must add Ca(NO\u2083)\u2082 to produce a [Ca\u00b2\u207a] \u2248 5.87 \u00d7 10\u207b\u2076 M<\/strong> in the solution to begin precipitation of CaF\u2082.<\/p>\n\n\n\n

    \n\n\n\n

    Part 2: Separation of Aniline, Aminoacetophenone, and Pyridine using Ion Exchange Chromatography<\/strong><\/h3>\n\n\n\n

    Given: Their pKa values<\/strong><\/p>\n\n\n\n

      \n
    • Aniline<\/strong> (weak base): pKa \u2248 4.6<\/li>\n\n\n\n
    • Aminoacetophenone<\/strong> (stronger base): pKa \u2248 9.7<\/li>\n\n\n\n
    • Pyridine<\/strong> (basic): pKa \u2248 5.2<\/li>\n<\/ul>\n\n\n\n

      Ion exchange chromatography separates compounds based on charge<\/strong>. We use cation exchange resin<\/strong> for separating basic compounds.<\/p>\n\n\n\n

      Step-by-step separation plan:<\/strong><\/h4>\n\n\n\n
        \n
      1. Acidic mobile phase (pH \u2248 2):<\/strong>\n
          \n
        • All bases are protonated and become positively charged.<\/li>\n\n\n\n
        • All will bind to the cation exchange resin.<\/li>\n<\/ul>\n<\/li>\n\n\n\n
        • Gradual increase in pH:<\/strong>\n
            \n
          • First eluted:<\/strong> The weakest base<\/strong> (Aniline, pKa 4.6) becomes neutral first and elutes.<\/li>\n\n\n\n
          • Second:<\/strong> Pyridine<\/strong> (pKa 5.2) elutes as it becomes neutral.<\/li>\n\n\n\n
          • Last:<\/strong> Aminoacetophenone<\/strong> (pKa 9.7) remains protonated longer and elutes at higher pH.<\/li>\n<\/ul>\n<\/li>\n<\/ol>\n\n\n\n

            Conclusion:<\/strong> By adjusting the pH of the mobile phase from low to high, the three compounds can be separated based on their different protonation states and affinities for the cation exchange resin.<\/p>\n\n\n\n

            \n\n\n\n

            This approach uses equilibrium principles, acid-base dissociation constants, and solubility rules to solve both the precipitation and separation parts of the question.<\/p>\n\n\n\n

            \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

            What mast 0\/ CJ(NO;h; must added to [ OLo )LOM HF solution to begin erecipikation al Cafm? (You MJY AssUme there hn0 votume change on the additton ol calaum t trate | Cafiakt 4.010 u Hk7210 Describe hot You would scparte JMmorin [unethy Amenehydratint andeyridine Using Ion echanga cmonatogtjol. (Use Lhe follawing tablo for naeded […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-27616","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/27616","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=27616"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/27616\/revisions"}],"predecessor-version":[{"id":27618,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/27616\/revisions\/27618"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=27616"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=27616"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=27616"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}