In the laboratory you dissolve 16.4 g of aluminum iodide in a volumetric flask and add water to a total volume of 500 mL What is the concentration of the iodide anion?<\/p>\n\n\n\n
The correct answer and explanation is:<\/strong><\/mark><\/p>\n\n\n\n To find the concentration of the iodide anion (I\u207b)<\/strong>, we follow these steps:<\/p>\n\n\n\n The chemical formula is AlI\u2083<\/strong> (aluminum has a 3+ charge and iodide has a 1\u2212 charge, so they combine in a 1:3 ratio).<\/p>\n\n\n\n Moles of AlI\u2083=16.4\u2009g407.68\u2009g\/mol\u22480.04023\u2009mol\\text{Moles of AlI\u2083} = \\frac{16.4 \\, \\text{g}}{407.68 \\, \\text{g\/mol}} \\approx 0.04023 \\, \\text{mol}<\/p>\n\n\n\n AlI\u2083 (s)\u2192Al3+(aq)+3\u2009I\u2212(aq)\\text{AlI\u2083 (s)} \\rightarrow \\text{Al}^{3+} (aq) + 3 \\, \\text{I}^- (aq)<\/p>\n\n\n\n So, each mole of AlI\u2083 produces 3 moles of iodide ions (I\u207b)<\/strong>. Moles of I\u207b=0.04023\u2009mol\u00d73=0.1207\u2009mol\\text{Moles of I\u207b} = 0.04023 \\, \\text{mol} \\times 3 = 0.1207 \\, \\text{mol}<\/p>\n\n\n\n 500\u2009mL=0.500\u2009L500 \\, \\text{mL} = 0.500 \\, \\text{L}<\/p>\n\n\n\n [I\u207b]=0.1207\u2009mol0.500\u2009L=0.2414\u2009mol\/L\\text{[I\u207b]} = \\frac{0.1207 \\, \\text{mol}}{0.500 \\, \\text{L}} = 0.2414 \\, \\text{mol\/L}<\/p>\n\n\n\n [I\u207b]=0.241\u2009M\\boxed{\\text{[I\u207b]} = 0.241 \\, \\text{M}}<\/p>\n\n\n\n To determine the concentration of the iodide ion (I\u207b) in solution, we first look at the solute, aluminum iodide (AlI\u2083)<\/strong>. This compound dissociates completely in water, producing 1 Al\u00b3\u207a ion and 3 I\u207b ions<\/strong> per formula unit. This 1:3 dissociation ratio is key to the calculation.<\/p>\n\n\n\n Next, we find how many moles<\/strong> of AlI\u2083 are present in the solution. Using the formula mass (molar mass) of AlI\u2083 \u2014 which is the sum of the atomic masses of aluminum (26.98 g\/mol) and three iodines (3 \u00d7 126.90 g\/mol) \u2014 we get 407.68 g\/mol<\/strong>. By dividing the given mass (16.4 g) by this molar mass, we calculate the number of moles of AlI\u2083 to be about 0.0402 mol<\/strong>.<\/p>\n\n\n\n Each mole of AlI\u2083 provides 3 moles of iodide ions<\/strong>, so multiplying the moles of AlI\u2083 by 3 gives us 0.1207 mol of I\u207b<\/strong>.<\/p>\n\n\n\n To find the concentration, we divide the number of moles of I\u207b by the volume of the solution in liters. Since 500 mL is equal to 0.500 L, we perform the division: 0.1207\u2009mol0.500\u2009L=0.2414\u2009mol\/L\\frac{0.1207 \\, \\text{mol}}{0.500 \\, \\text{L}} = 0.2414 \\, \\text{mol\/L}<\/p>\n\n\n\n Thus, the concentration of iodide ion in the solution is 0.241 M<\/strong>, which expresses how many moles of I\u207b are present per liter of solution.<\/p>\n","protected":false},"excerpt":{"rendered":" In the laboratory you dissolve 16.4 g of aluminum iodide in a volumetric flask and add water to a total volume of 500 mL What is the concentration of the iodide anion? The correct answer and explanation is: To find the concentration of the iodide anion (I\u207b), we follow these steps: Step 1: Write the […]<\/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-26145","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/26145","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=26145"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/26145\/revisions"}],"predecessor-version":[{"id":26149,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/26145\/revisions\/26149"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=26145"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=26145"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=26145"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\n\n\n\nStep 1: Write the formula of aluminum iodide<\/strong><\/h3>\n\n\n\n
\n\n\n\nStep 2: Molar mass of AlI\u2083<\/strong><\/h3>\n\n\n\n
\n
Molar mass of AlI\u2083 = 26.98 + 380.70 = 407.68 g\/mol<\/strong><\/li>\n<\/ul>\n\n\n\n
\n\n\n\nStep 3: Calculate moles of AlI\u2083<\/strong><\/h3>\n\n\n\n
\n\n\n\nStep 4: Dissociation of AlI\u2083 in water<\/strong><\/h3>\n\n\n\n
\n\n\n\nStep 5: Convert 500 mL to liters<\/strong><\/h3>\n\n\n\n
\n\n\n\nStep 6: Calculate concentration of I\u207b<\/strong><\/h3>\n\n\n\n
\n\n\n\n\u2705 Final Answer:<\/strong><\/h3>\n\n\n\n
\n\n\n\nExplanation (300 words):<\/strong><\/h3>\n\n\n\n