{"id":41398,"date":"2025-06-28T08:37:34","date_gmt":"2025-06-28T08:37:34","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=41398"},"modified":"2025-06-28T08:37:35","modified_gmt":"2025-06-28T08:37:35","slug":"determine-the-number-of-grams-of-pbi2-that-will-dissolve-in-0-500-l-of-water","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/determine-the-number-of-grams-of-pbi2-that-will-dissolve-in-0-500-l-of-water\/","title":{"rendered":"Determine the number of grams of PbI2 that will dissolve in 0.500 L of water"},"content":{"rendered":"\n<p>Determine the number of grams of PbI2 that will dissolve in 0.500 L of water: (2 pts) b. Determine the number of grams of PbI2 that will dissolve in 0.500 L of 0.0634 M NaI: (3 pts) 3. A solution contains the following anions at a concentration of 0.10 M: Cl-, Br-, and CrO4^2-. Determine in what order the anions will precipitate if Ag+ is used as the precipitating agent: (4 pts) 4. A solution contains 0.020 M Mg^2+ and 0.10 M Ca^2+. Can 99.90% of Mg^2+ be precipitated by the addition of NaOH without precipitation of Ca^2+? (4 pts<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"700\" height=\"189\" src=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/image-439.png\" alt=\"\" class=\"wp-image-41417\" srcset=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/image-439.png 700w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/image-439-300x81.png 300w\" sizes=\"auto, (max-width: 700px) 100vw, 700px\" \/><\/figure>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-vivid-cyan-blue-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>Let&#8217;s solve the chemistry problems shown in the image step-by-step:<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>2a. Determine the number of grams of PbI\u2082 that will dissolve in 0.500 L of water.<\/strong><br>We use the solubility product (Ksp) of PbI\u2082.<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Ksp of PbI\u2082 = 7.1 \u00d7 10\u207b\u2079<\/li>\n\n\n\n<li>PbI\u2082 \u21cc Pb\u00b2\u207a + 2 I\u207b<br>Let the molar solubility of PbI\u2082 be <em>s<\/em> mol\/L.<br>Then, [Pb\u00b2\u207a] = <em>s<\/em>, [I\u207b] = 2<em>s<\/em><br>Ksp = [Pb\u00b2\u207a][I\u207b]\u00b2 = s \u00d7 (2s)\u00b2 = 4s\u00b3<br>So, 4s\u00b3 = 7.1 \u00d7 10\u207b\u2079<br>s\u00b3 = 1.775 \u00d7 10\u207b\u2079<br>s \u2248 1.21 \u00d7 10\u207b\u00b3 mol\/L<\/li>\n<\/ul>\n\n\n\n<p>Now multiply by volume:<br>mol PbI\u2082 = 1.21 \u00d7 10\u207b\u00b3 mol\/L \u00d7 0.500 L = 6.05 \u00d7 10\u207b\u2074 mol<\/p>\n\n\n\n<p>Molar mass of PbI\u2082 = 207.2 + 2(126.9) \u2248 461.0 g\/mol<br>Mass = 6.05 \u00d7 10\u207b\u2074 mol \u00d7 461.0 g\/mol \u2248 <strong>0.279 g<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>2b. Determine the number of grams of PbI\u2082 that will dissolve in 0.500 L of 0.0634 M NaI.<\/strong><br>Common ion effect: [I\u207b] = 0.0634 M (initial), so:<\/p>\n\n\n\n<p>Ksp = [Pb\u00b2\u207a][I\u207b]\u00b2<br>Let [Pb\u00b2\u207a] = x<br>Then, 7.1 \u00d7 10\u207b\u2079 = x \u00d7 (0.0634)\u00b2<br>x = 7.1 \u00d7 10\u207b\u2079 \/ (0.0634)\u00b2 \u2248 1.77 \u00d7 10\u207b\u2076 mol\/L<\/p>\n\n\n\n<p>mol PbI\u2082 = 1.77 \u00d7 10\u207b\u2076 mol\/L \u00d7 0.500 L = 8.85 \u00d7 10\u207b\u2077 mol<br>Mass = 8.85 \u00d7 10\u207b\u2077 mol \u00d7 461.0 g\/mol \u2248 <strong>0.000408 g<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>3. Order of precipitation with Ag\u207a:<\/strong><br>Solubility rules and Ksp values:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>AgCl (Ksp \u2248 1.8 \u00d7 10\u207b\u00b9\u2070)<\/li>\n\n\n\n<li>AgBr (Ksp \u2248 5.0 \u00d7 10\u207b\u00b9\u00b3)<\/li>\n\n\n\n<li>AgI (Ksp \u2248 8.3 \u00d7 10\u207b\u00b9\u2077)<\/li>\n\n\n\n<li>Ag\u2082CrO\u2084 (Ksp \u2248 1.1 \u00d7 10\u207b\u00b9\u00b2)<\/li>\n<\/ul>\n\n\n\n<p>The lower the Ksp, the earlier it precipitates:<\/p>\n\n\n\n<p><strong>Order: AgI \u2192 AgBr \u2192 Ag\u2082CrO\u2084 \u2192 AgCl<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>4. Precipitation of Mg(OH)\u2082 and Ca(OH)\u2082:<\/strong><\/p>\n\n\n\n<p>Ksp Mg(OH)\u2082 = 5.6 \u00d7 10\u207b\u00b9\u00b2<br>Ksp Ca(OH)\u2082 = 6.5 \u00d7 10\u207b\u2076<\/p>\n\n\n\n<p>To precipitate 99.90% of Mg\u00b2\u207a:<\/p>\n\n\n\n<p>Remaining Mg\u00b2\u207a = 0.0001 \u00d7 0.020 M = 2.0 \u00d7 10\u207b\u2076 M<br>Ksp = [Mg\u00b2\u207a][OH\u207b]\u00b2 \u2192 5.6 \u00d7 10\u207b\u00b9\u00b2 = (2.0 \u00d7 10\u207b\u2076)[OH\u207b]\u00b2<br>[OH\u207b]\u00b2 = 2.8 \u00d7 10\u207b\u2076 \u2192 [OH\u207b] \u2248 1.67 \u00d7 10\u207b\u00b3 M<\/p>\n\n\n\n<p>Check Ca(OH)\u2082:<br>[OH\u207b] = 1.67 \u00d7 10\u207b\u00b3, [Ca\u00b2\u207a] = 0.10 M<br>Q = [Ca\u00b2\u207a][OH\u207b]\u00b2 = 0.10 \u00d7 (1.67 \u00d7 10\u207b\u00b3)\u00b2 = 2.8 \u00d7 10\u207b\u2077<br>Since Q &lt; Ksp (6.5 \u00d7 10\u207b\u2076), <strong>Ca\u00b2\u207a does not precipitate<\/strong><\/p>\n\n\n\n<p><strong>Yes, 99.90% of Mg\u00b2\u207a can be precipitated without Ca\u00b2\u207a precipitation.<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"852\" height=\"1024\" src=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-1335.jpeg\" alt=\"\" class=\"wp-image-41427\" srcset=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-1335.jpeg 852w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-1335-250x300.jpeg 250w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-1335-768x923.jpeg 768w\" sizes=\"auto, (max-width: 852px) 100vw, 852px\" \/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Determine the number of grams of PbI2 that will dissolve in 0.500 L of water: (2 pts) b. Determine the number of grams of PbI2 that will dissolve in 0.500 L of 0.0634 M NaI: (3 pts) 3. A solution contains the following anions at a concentration of 0.10 M: Cl-, Br-, and CrO4^2-. Determine [&hellip;]<\/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-41398","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/41398","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=41398"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/41398\/revisions"}],"predecessor-version":[{"id":41429,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/41398\/revisions\/41429"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=41398"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=41398"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=41398"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}