{"id":240239,"date":"2025-07-03T08:02:44","date_gmt":"2025-07-03T08:02:44","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=240239"},"modified":"2025-07-03T08:02:46","modified_gmt":"2025-07-03T08:02:46","slug":"what-is-the-ph-and-poh-of-a-solution-made-by-adding-water-to-15-grams-of-hydroiodic-acid-until-the-volume-of-the-solution-is-2500-ml","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/03\/what-is-the-ph-and-poh-of-a-solution-made-by-adding-water-to-15-grams-of-hydroiodic-acid-until-the-volume-of-the-solution-is-2500-ml\/","title":{"rendered":"What is the pH and pOH of a solution made by adding water to 15 grams of hydroiodic acid until the volume of the solution is 2500 mL"},"content":{"rendered":"\n<p>What is the pH and pOH of a solution made by adding water to 15 grams of hydroiodic acid until the volume of the solution is 2500 mL? (hint: you must calculate the molarity first)<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-1-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>To solve for the pH and pOH of the solution, we will follow these steps:<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: Find the molarity of the hydroiodic acid (HI) solution<\/h3>\n\n\n\n<p>First, we need to calculate the number of moles of HI in 15 grams. The molecular weight of HI is:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Iodine (I) = 126.9 g\/mol<\/li>\n\n\n\n<li>Hydrogen (H) = 1.008 g\/mol<\/li>\n\n\n\n<li>Therefore, the molar mass of HI = 126.9 + 1.008 = <strong>127.908 g\/mol<\/strong><\/li>\n<\/ul>\n\n\n\n<p>Now, calculate the moles of HI:Moles&nbsp;of&nbsp;HI=mass&nbsp;of&nbsp;HI&nbsp;(grams)molar&nbsp;mass&nbsp;(g\/mol)=15\u2009g127.908\u2009g\/mol\u22480.1177\u2009mol\\text{Moles of HI} = \\frac{\\text{mass of HI (grams)}}{\\text{molar mass (g\/mol)}} = \\frac{15 \\, \\text{g}}{127.908 \\, \\text{g\/mol}} \\approx 0.1177 \\, \\text{mol}Moles&nbsp;of&nbsp;HI=molar&nbsp;mass&nbsp;(g\/mol)mass&nbsp;of&nbsp;HI&nbsp;(grams)\u200b=127.908g\/mol15g\u200b\u22480.1177mol<\/p>\n\n\n\n<p>Next, calculate the molarity of the solution. Molarity is defined as moles of solute divided by the volume of the solution in liters:Molarity=0.1177\u2009mol2.5\u2009L=0.04708\u2009M\\text{Molarity} = \\frac{0.1177 \\, \\text{mol}}{2.5 \\, \\text{L}} = 0.04708 \\, \\text{M}Molarity=2.5L0.1177mol\u200b=0.04708M<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 2: Calculate the pH of the solution<\/h3>\n\n\n\n<p>Since HI is a strong acid, it dissociates completely in water:HI\u2192H++I\u2212\\text{HI} \\rightarrow \\text{H}^+ + \\text{I}^-HI\u2192H++I\u2212<\/p>\n\n\n\n<p>The concentration of hydrogen ions [H\u207a] is the same as the molarity of the acid, 0.04708 M. To find the pH, we use the formula:pH=\u2212log\u2061[H+]\\text{pH} = -\\log[\\text{H}^+]pH=\u2212log[H+]pH=\u2212log\u2061(0.04708)\u22481.327\\text{pH} = -\\log(0.04708) \\approx 1.327pH=\u2212log(0.04708)\u22481.327<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 3: Calculate the pOH of the solution<\/h3>\n\n\n\n<p>The relationship between pH and pOH is:pH+pOH=14\\text{pH} + \\text{pOH} = 14pH+pOH=14<\/p>\n\n\n\n<p>So:pOH=14\u2212pH=14\u22121.327=12.673\\text{pOH} = 14 &#8211; \\text{pH} = 14 &#8211; 1.327 = 12.673pOH=14\u2212pH=14\u22121.327=12.673<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Final Answer:<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>pH \u2248 1.33<\/li>\n\n\n\n<li>pOH \u2248 12.67<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation:<\/h3>\n\n\n\n<p>The pH and pOH values are derived from the concentration of hydrogen ions (H\u207a) and hydroxide ions (OH\u207b), respectively. The solution in question is strongly acidic due to the dissociation of hydroiodic acid, a strong acid. By calculating the molarity of HI and recognizing that it dissociates completely, we can directly relate the molarity to the concentration of hydrogen ions. Then, applying the formula for pH and using the relationship between pH and pOH, we obtain the final values of 1.33 for pH and 12.67 for pOH.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner5-113.jpeg\" alt=\"\" class=\"wp-image-240245\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>What is the pH and pOH of a solution made by adding water to 15 grams of hydroiodic acid until the volume of the solution is 2500 mL? (hint: you must calculate the molarity first) The Correct Answer and Explanation is: To solve for the pH and pOH of the solution, we will follow these [&hellip;]<\/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-240239","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/240239","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=240239"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/240239\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=240239"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=240239"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=240239"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}