{"id":187857,"date":"2025-02-06T07:34:52","date_gmt":"2025-02-06T07:34:52","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=187857"},"modified":"2025-02-06T07:34:54","modified_gmt":"2025-02-06T07:34:54","slug":"determine-the-ph-of-each-of-the-following-solutions","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/02\/06\/determine-the-ph-of-each-of-the-following-solutions\/","title":{"rendered":"Determine the pH of each of the following solutions"},"content":{"rendered":"\n<p>Determine the pH of each of the following solutions. Express your answer to two decimal places.<\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li>0.15M KCHO2<\/li>\n<\/ol>\n\n\n\n<p>2 .0.24M CH3NH3I<\/p>\n\n\n\n<p>3 .0.16M KI<\/p>\n\n\n\n<p><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-6-color\"><strong>The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n<p>To determine the pH of each solution, we need to analyze the nature of the solute and its behavior in water.<\/p>\n\n\n\n<p><strong>1. 0.15 M KCHO\u2082 (Potassium Formate):<\/strong><\/p>\n\n\n\n<p>otassium formate (KCHO\u2082) is a salt derived from formic acid (HCHO\u2082), a weak acid, and potassium hydroxide (KOH), a strong base.n aqueous solution, KCHO\u2082 dissociates completely into potassium ions (K\u207a) and formate ions (CHO\u2082\u207b).he formate ion can hydrolyze in water, accepting a proton to form formic acid and hydroxide ions (OH\u207b), making the solution basic.<br>The hydrolysis reaction is:<\/p>\n\n\n\n<p>HO\u2082\u207b + H\u2082O \u21cc HCHO\u2082 + OH\u207b<br>To find the pH:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Determine the ( K_b ) of the formate ion:<\/strong> ( K_b = \\frac{K_w}{K_a} )<br>iven ( K_a ) of formic acid is ( 1.8 \\times 10^{-4} ):<br>( K_b = \\frac{1.0 \\times 10^{-14}}{1.8 \\times 10^{-4}} \\approx 5.56 \\times 10^{-11} )<\/li>\n\n\n\n<li><strong>Set up the expression for ( K_b ):<\/strong> ( K_b = \\frac{[HCHO\u2082][OH\u207b]}{[CHO\u2082\u207b]} )<br>ssume ( x ) is the concentration of OH\u207b produced:<br>( 5.56 \\times 10^{-11} = \\frac{x^2}{0.15 &#8211; x} )<br>ince ( K_b ) is small, ( x ) is negligible compared to 0.15 M:<br>( 5.56 \\times 10^{-11} = \\frac{x^2}{0.15} )<br>( x^2 = 5.56 \\times 10^{-11} \\times 0.15 )<br>( x^2 = 8.34 \\times 10^{-12} )<br>( x = \\sqrt{8.34 \\times 10^{-12}} \\approx 9.13 \\times 10^{-6} )<br>hus, [OH\u207b] \u2248 ( 9.13 \\times 10^{-6} ) M<\/li>\n\n\n\n<li><strong>Calculate pOH:<\/strong> ( \\text{pOH} = -\\log[OH\u207b] )<br>( \\text{pOH} = -\\log(9.13 \\times 10^{-6}) \\approx 5.04 )<\/li>\n\n\n\n<li><strong>Determine pH:<\/strong> ( \\text{pH} = 14 &#8211; \\text{pOH} )<br>( \\text{pH} = 14 &#8211; 5.04 = 8.96 )<br>herefore, the pH of 0.15 M KCHO\u2082 is approximately <strong>8.96<\/strong>.<br><strong>2. 0.24 M CH\u2083NH\u2083I (Methylammonium Iodide):<\/strong><\/li>\n<\/ul>\n\n\n\n<p>ethylammonium iodide (CH\u2083NH\u2083I) is a salt formed from methylamine (CH\u2083NH\u2082), a weak base, and hydroiodic acid (HI), a strong acid.n water, it dissociates into methylammonium ions (CH\u2083NH\u2083\u207a) and iodide ions (I\u207b).he methylammonium ion can donate a proton to water, acting as a weak acid:<br>H\u2083NH\u2083\u207a + H\u2082O \u21cc CH\u2083NH\u2082 + H\u2083O\u207a<br>To find the pH:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Determine the ( K_a ) of the methylammonium ion:<\/strong> ( K_a = \\frac{K_w}{K_b} )<br>iven ( K_b ) of methylamine is ( 4.4 \\times 10^{-4} ):<br>( K_a = \\frac{1.0 \\times 10^{-14}}{4.4 \\times 10^{-4}} \\approx 2.27 \\times 10^{-11} )<\/li>\n\n\n\n<li><strong>Set up the expression for ( K_a ):<\/strong> ( K_a = \\frac{[CH\u2083NH\u2082][H\u2083O\u207a]}{[CH\u2083NH\u2083\u207a]} )<br>ssume ( x ) is the concentration of H\u2083O\u207a produced:<br>( 2.27 \\times 10^{-11} = \\frac{x^2}{0.24 &#8211; x} )<br>ince ( K_a ) is small, ( x ) is negligible compared to 0.24 M:<br>( 2.27 \\times 10^{-11} = \\frac{x^2}{0.24} )<br>( x^2 = 2.27 \\times 10^{-11} \\times 0.24 )<br>( x^2 = 5.45 \\times 10^{-12} )<br>( x = \\sqrt{5.45 \\times 10^{-12}} \\approx 7.38 \\times 10^{-6} )<br>hus, [H\u2083O\u207a] \u2248 ( 7.38 \\times 10^{-6} ) M<\/li>\n\n\n\n<li><strong>Calculate pH:<\/strong> ( \\text{pH} = -\\log[H\u2083O\u207a] )<\/li>\n<\/ul>\n","protected":false},"excerpt":{"rendered":"<p>Determine the pH of each of the following solutions. Express your answer to two decimal places. 2 .0.24M CH3NH3I 3 .0.16M KI The Correct Answer and Explanation is : To determine the pH of each solution, we need to analyze the nature of the solute and its behavior in water. 1. 0.15 M KCHO\u2082 (Potassium [&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-187857","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/187857","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=187857"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/187857\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=187857"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=187857"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=187857"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}