{"id":1112,"date":"2019-12-03T14:18:06","date_gmt":"2019-12-03T14:18:06","guid":{"rendered":"http:\/\/chemdictionary.org\/?p=1112"},"modified":"2019-12-03T14:18:13","modified_gmt":"2019-12-03T14:18:13","slug":"halide-test","status":"publish","type":"post","link":"https:\/\/chemdictionary.org\/halide-test\/","title":{"rendered":"Halide Test"},"content":{"rendered":"\n<h2>What is halide?<\/h2>\n\n\n\n<p>A halide is a dual-phase in\nwhich one part is a halogen atom and the other part is an element or radical\nthat is less electronegative or more electropositive than that of halogen to\nmake a <strong>fluoride, chloride, bromide,\niodide, astatide<\/strong> or hypothetically <strong>tennesside<\/strong>\ncompound. Under suitable conditions, the alkali metals combine directly with\nhalogens forming halides by using this general formula<\/p>\n\n\n\n<p><strong>MX (X= F, Cl, Br or I)<\/strong><\/p>\n\n\n\n<p>All metals present in Group 1\nform halides that are white solids at room temperature.<\/p>\n\n\n\n<p><strong>Halide\ntest<\/strong><\/p>\n\n\n\n<p>We will discuss halide ions\n(F, Cl, Br, I) tests by using silver nitrate and ammonia.<\/p>\n\n\n\n<p><strong>By\nUsing Silver Nitrate Solution<\/strong><\/p>\n\n\n\n<p>We need a solution of halide\nions. The step is adding a dilute nitric acid to acidify the solution. The\nnitric acid starts reacting with and removes the other ions present that might\nform precipitates with silver nitrate.&nbsp; <\/p>\n\n\n\n<p>The second step is to add\nsilver nitrate solution and the following products will be identified by the\nhalide<\/p>\n\n\n\n<figure class=\"wp-block-table\"><table class=\"\"><tbody><tr><td>\n  <strong>Ions present<\/strong>\n  <\/td><td>\n  <strong>Observation<\/strong>\n  <\/td><\/tr><tr><td>\n  <strong>F<sup>&#8211;<\/sup><\/strong>\n  <\/td><td>\n  No\n  precipitate\n  <\/td><\/tr><tr><td>\n  <strong>Cl<sup>&#8211;<\/sup><\/strong>\n  <\/td><td>\n  White\n  precipitate\n  <\/td><\/tr><tr><td>\n  <strong>Br<sup>&#8211;<\/sup><\/strong>\n  <\/td><td>\n  A\n  very pale cream precipitate\n  <\/td><\/tr><tr><td>\n  <strong>I<sup>&#8211;<\/sup><\/strong>\n  <\/td><td>\n  A\n  very pale yellow precipitate\n  <\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p>The formed precipitates of chloride, bromide and iodide precipitates are shown in the photograph below<\/p>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter\"><img loading=\"lazy\" width=\"237\" height=\"225\" src=\"https:\/\/chemdictionary.org\/wp-content\/uploads\/2019\/11\/image.png\" alt=\"\" class=\"wp-image-1114\"\/><\/figure><\/div>\n\n\n\n<p>The precipitates of chloride\nare identified easily but the rest of the two are quite similar to each other.\nThey can be differentiated only in a side-by-side comparison. If the\nprecipitates are exposed to light, all of them will change their colors i.e.\ngrey or purple tints. The fluoride ions will not show any precipitates and the\nabsence of a precipitate ion is unhelpful unless it is well known that a\nhalogen is present otherwise, it will show that the chloride, bromide, or iodide\nis absent. <\/p>\n\n\n\n<h2>The chemistry of the test<\/h2>\n\n\n\n<p>The precipitates that are\nformed are insoluble silver halides: silver chloride, silver bromide or silver\niodide. The following equation explains the formation<\/p>\n\n\n\n<p>Ag<sup>+<\/sup><sub>(aq)\n<\/sub>+ Cl<sup>&#8211;<\/sup><sub>(aq)<\/sub> \u2192\nAgCl<sub>(s)<\/sub><\/p>\n\n\n\n<p>Ag<sup>+<\/sup><sub>(aq) <\/sub>+ Br<sup>&#8211;<\/sup> <sub>(aq)<\/sub>\n\u2192 AgBr<sub>(s)<\/sub><\/p>\n\n\n\n<p>Ag<sup>+<\/sup><sub>(aq) <\/sub>+ I<sup>&#8211;<\/sup><sub>(aq)<\/sub>\n\u2192 AgI(<sub>s)<\/sub><\/p>\n\n\n\n<p>Silver fluoride is soluble so\nno precipitate is formed.<\/p>\n\n\n\n<p>Ag<sup>+<\/sup>\n<sub>(aq) <\/sub>+ F<sup>&#8211;<\/sup><sub> (aq)&nbsp;\n<\/sub>\u2192&nbsp; Ag<sup>+<\/sup><sub>(aq)<\/sub> + F<sup>&#8211; <\/sup><sub>(aq)<\/sub><\/p>\n\n\n\n<h2>Using ammonia solution to confirm the precipitate<\/h2>\n\n\n\n<p>A\nsolution of ammonia is added to the precipitates<\/p>\n\n\n\n<figure class=\"wp-block-table\"><table class=\"\"><tbody><tr><td>\n  <strong>Original Precipitate<\/strong>\n  <\/td><td>\n  <strong>Observation<\/strong>\n  <\/td><\/tr><tr><td>\n  AgCl\n  <\/td><td>\n  Precipitate\n  dissolve to give a colorless solution\n  <\/td><\/tr><tr><td>\n  AgBr\n  <\/td><td>\n  Precipitate\n  formed remain nearly unchanged by using dilute ammonia solution, while it dissolves\n  in concentrated ammonia solution to give a colorless solution\n  <\/td><\/tr><tr><td>\n  AgI\n  <\/td><td>\n  Precipitate\n  formed is insoluble in ammonia solution of any concentration\n  <\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<h3><strong>Explanation\n<\/strong><\/h3>\n\n\n\n<p>There are no insoluble ionic\ncompounds. In water, if the concentrations of the ions in solution exceed a\ncertain value, a precipitate will be formed which will be different for every\ndifferent compound. This value can be considered as a solubility product. The\nexpression for the solubility product of silver halides is given as<\/p>\n\n\n\n<p><strong>K<sub>sp<\/sub> = [Ag<sup>+<\/sup>] [X<sup>&#8211;<\/sup>]<\/strong><\/p>\n\n\n\n<p>The square brackets express\nthe molar concentrations having units of mol\/L.<\/p>\n\n\n\n<h3>Conditions of forming precipitates<\/h3>\n\n\n\n<ul><li>There will be no precipitate formation if the actual\nconcentration of the ions in a solution will produce less value than that of\nsolubility product.<\/li><li>There will be a precipitate formation if the\nproduct of the concentrations would exceed the value of the solubility product.<\/li><\/ul>\n\n\n\n<p>Necessarily the product of the\nionic concentrations is never greater than the solubility product vale. Every\ntime enough solid is precipitated to lower the ionic product to the value of\nthe solubility product. The table given below enlists the solubility products\nfrom silver chloride to silver iodide (a solubility product of silver fluoride can\u2019t\nnot be described as it is too soluble).<\/p>\n\n\n\n<figure class=\"wp-block-table\"><table class=\"\"><tbody><tr><td>\n  &nbsp;\n  <\/td><td>\n  <strong>K<sub>sp<\/sub> (mol<sup>2<\/sup>dm<sup>-6<\/sup>)<\/strong>\n  <\/td><\/tr><tr><td>\n  AgCl\n  <\/td><td>\n  1.8&nbsp; x 10-10\n  <\/td><\/tr><tr><td>\n  AgBr\n  <\/td><td>\n  7.7\n  x 10-13\n  <\/td><\/tr><tr><td>\n  AgI\n  <\/td><td>\n  8.3\n  x10-17\n  <\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p>You will notice that all\ncompounds are very insoluble but become even less soluble as you go down from\nthe chloride to iodide.<\/p>\n\n\n\n<h2>The Role of Ammonia<\/h2>\n\n\n\n<p>The ammonia comes in contact with silver ions to give a complex ion known as the diamminesilver(I) ion, [Ag(NH<sub>3<\/sub>)<sub>2<\/sub>]<sup> +<\/sup>. This reaction is reversible but the complex is very stable and the equilibrium position lies proficiently to the right. The equation for this reaction is given below:<\/p>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter\"><img loading=\"lazy\" width=\"539\" height=\"207\" src=\"https:\/\/chemdictionary.org\/wp-content\/uploads\/2019\/11\/image-1.png\" alt=\"\" class=\"wp-image-1115\" srcset=\"https:\/\/chemdictionary.org\/wp-content\/uploads\/2019\/11\/image-1.png 539w, https:\/\/chemdictionary.org\/wp-content\/uploads\/2019\/11\/image-1-300x115.png 300w\" sizes=\"(max-width: 539px) 100vw, 539px\" \/><\/figure><\/div>\n\n\n\n<p>A solution that is in contact\nwith one of the silver halide precipitates will contain a minute concentration\nof dissolved silver ions. The consequence of adding the ammonia is to lower\nthis concentration still further. If the adjusted silver ion concentration is\nmultiplied by the halide ion concentration is less than the solubility product,\nsome of the precipitates will be dissolved to restore the equilibrium. <\/p>\n\n\n\n<p>&nbsp;If the ammonia is concentrated this will take\nplace with silver chloride and with silver bromide. The more concentrated\nammonia pushes the equilibrium even further to the right that lowers the\nconcentration of silver ions even more.<\/p>\n\n\n\n<p>The silver iodide is very\ninsoluble that the ammonia will not lower the silver ion concentration enough\nfor the precipitate to dissolve.<\/p>\n\n\n\n<p><strong>An\nalternative Test using a concentrated Sulphuric Acid<\/strong><\/p>\n\n\n\n<p>By adding a concentrated\nSulphuric acid to a solid sample of any one of the halide gives the results as\nfollows<\/p>\n\n\n\n<figure class=\"wp-block-table\"><table class=\"\"><tbody><tr><td>\n  <strong>Ion present<\/strong>\n  <\/td><td>\n  <strong>Observation<\/strong>\n  <\/td><\/tr><tr><td>\n  F<sup>&#8211;<\/sup>\n  <\/td><td>\n  Steamy\n  acidic fumes (of HF)\n  <\/td><\/tr><tr><td>\n  Cl<sup>&#8211;<\/sup>\n  <\/td><td>\n  Steamy\n  acidic fumes (of HCl)\n  <\/td><\/tr><tr><td>\n  Br<sup>&#8211;<\/sup>\n  <\/td><td>\n  Steamy\n  acidic fumes (of Hbr) contaminated with brown bromine vapor\n  <\/td><\/tr><tr><td>\n  I<sup>&#8211;<\/sup>\n  <\/td><td>\n  Some HI\n  fumes are seen with large amounts of purple iodine vapor and a red compound\n  in the reaction vessel\n  <\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p>The only confusion that can occur\nbetween fluoride and chloride is that they behave identically under these\nconditions. They can be distinguished by dissolving the original solid in water\nand then testing with a silver nitrate solution. The chloride results in white\nprecipitate while the fluoride produces none.<\/p>\n\n\n\n<h4>References <\/h4>\n\n\n\n<ul><li><a href=\"https:\/\/www.chemguide.co.uk\/inorganic\/group7\/testing.html\">https:\/\/www.chemguide.co.uk\/inorganic\/group7\/testing.html<\/a><\/li><li><a href=\"https:\/\/chem.libretexts.org\/Bookshelves\/Inorganic_Chemistry\/Supplemental_Modules_(Inorganic_Chemistry)\/Descriptive_Chemistry\/Elements_Organized_by_Block\/2_p-Block_Elements\/Group_17%3A_The_Halogens\/1Group_17%3A_General_Reactions\/Testing_for_Halide_Ions\">https:\/\/chem.libretexts.org\/Bookshelves\/Inorganic_Chemistry\/Supplemental_Modules_(Inorganic_Chemistry)\/Descriptive_Chemistry\/Elements_Organized_by_Block\/2_p-Block_Elements\/Group_17%3A_The_Halogens\/1Group_17%3A_General_Reactions\/Testing_for_Halide_Ions<\/a><\/li><li><a href=\"https:\/\/chem.libretexts.org\/Bookshelves\/Organic_Chemistry\/Supplemental_Modules_(Organic_Chemistry)\/Alkyl_Halides\/Reactivity_of_Alkyl_Halides\/Reaction_of_Alkyl_Halides_with_Silver_Nitrate\">https:\/\/chem.libretexts.org\/Bookshelves\/Organic_Chemistry\/Supplemental_Modules_(Organic_Chemistry)\/Alkyl_Halides\/Reactivity_of_Alkyl_Halides\/Reaction_of_Alkyl_Halides_with_Silver_Nitrate<\/a><\/li><li><a href=\"https:\/\/en.wikipedia.org\/wiki\/Halide\">https:\/\/en.wikipedia.org\/wiki\/Halide<\/a><\/li><\/ul>\n","protected":false},"excerpt":{"rendered":"<p>What is halide? A halide is a dual-phase in which one part is a halogen atom and the other part is an element or radical that is less electronegative or more electropositive than that of halogen to make a fluoride, chloride, bromide, iodide, astatide or hypothetically tennesside compound. Under suitable conditions, the alkali metals combine &#8230; <a title=\"Halide Test\" class=\"read-more\" href=\"https:\/\/chemdictionary.org\/halide-test\/\" aria-label=\"More on Halide Test\">Read more<\/a><\/p>\n","protected":false},"author":6,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"rating_form_position":"","rating_results_position":"","mr_structured_data_type":""},"categories":[4],"tags":[],"multi-rating":{"mr_rating_results":[{"adjusted_star_result":3,"star_result":3,"total_max_option_value":5,"adjusted_score_result":3,"score_result":3,"percentage_result":60,"adjusted_percentage_result":60,"count":2,"post_id":1112}]},"_links":{"self":[{"href":"https:\/\/chemdictionary.org\/wp-json\/wp\/v2\/posts\/1112"}],"collection":[{"href":"https:\/\/chemdictionary.org\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/chemdictionary.org\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/chemdictionary.org\/wp-json\/wp\/v2\/users\/6"}],"replies":[{"embeddable":true,"href":"https:\/\/chemdictionary.org\/wp-json\/wp\/v2\/comments?post=1112"}],"version-history":[{"count":3,"href":"https:\/\/chemdictionary.org\/wp-json\/wp\/v2\/posts\/1112\/revisions"}],"predecessor-version":[{"id":1304,"href":"https:\/\/chemdictionary.org\/wp-json\/wp\/v2\/posts\/1112\/revisions\/1304"}],"wp:attachment":[{"href":"https:\/\/chemdictionary.org\/wp-json\/wp\/v2\/media?parent=1112"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/chemdictionary.org\/wp-json\/wp\/v2\/categories?post=1112"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/chemdictionary.org\/wp-json\/wp\/v2\/tags?post=1112"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}