{"id":197092,"date":"2025-03-06T15:12:26","date_gmt":"2025-03-06T15:12:26","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=197092"},"modified":"2025-03-06T15:12:29","modified_gmt":"2025-03-06T15:12:29","slug":"the-ionization-of-p-nitrophenol-is-shown-below-pka-7-0","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/03\/06\/the-ionization-of-p-nitrophenol-is-shown-below-pka-7-0\/","title":{"rendered":"The ionization of p-nitrophenol is shown below (pKa= 7.0)"},"content":{"rendered":"\n<p>The ionization of p-nitrophenol is shown below (pKa= 7.0)<\/p>\n\n\n\n<p>a) At pH 7, what are the relative concentrations of ionized and un-ionized p-nitrophenol?<\/p>\n\n\n\n<p>b) If enough concentrated hydrochloric acid is added to a solution of p-nitrophenol to lower the pH from 7 to 5, what will happen to the relative concentrations of the ionized and un-ionized forms?<\/p>\n\n\n\n<p>c) A solution of p-nitrophenol at pH 8.3 was found to have an A400 of 0.550. What is the total concentration (in um) of p-nitrophenol (ionized plus un-ionized) in the solution? The molar extinction coefficient of p-nitrophenol is 19,500M-1cm-1 and the pKa is 7.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/03\/image-381.png\" alt=\"\" class=\"wp-image-197093\"\/><\/figure>\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<h3 class=\"wp-block-heading\"><strong>Solution: Ionization of p-Nitrophenol<\/strong><\/h3>\n\n\n\n<h4 class=\"wp-block-heading\"><strong>(a) Relative Concentrations at pH 7<\/strong><\/h4>\n\n\n\n<p>The Henderson-Hasselbalch equation is used to determine the ratio of ionized to un-ionized p-nitrophenol:<\/p>\n\n\n\n<p>[<br>\\text{pH} = \\text{pKa} + \\log \\left( \\frac{[\\text{A}^-]}{[\\text{HA}]} \\right)<br>]<\/p>\n\n\n\n<p>At pH = 7.0 and pKa = 7.0:<\/p>\n\n\n\n<p>[<br>7.0 = 7.0 + \\log \\left( \\frac{[\\text{A}^-]}{[\\text{HA}]} \\right)<br>]<\/p>\n\n\n\n<p>[<br>\\log \\left( \\frac{[\\text{A}^-]}{[\\text{HA}]} \\right) = 0<br>]<\/p>\n\n\n\n<p>[<br>\\frac{[\\text{A}^-]}{[\\text{HA}]} = 10^0 = 1<br>]<\/p>\n\n\n\n<p>This means the ionized and un-ionized forms are present in equal concentrations:<br>[<br>[\\text{A}^-] = [\\text{HA}]<br>]<\/p>\n\n\n\n<h4 class=\"wp-block-heading\"><strong>(b) Effect of Lowering pH to 5<\/strong><\/h4>\n\n\n\n<p>If the pH is lowered from 7 to 5, the equation becomes:<\/p>\n\n\n\n<p>[<br>5.0 = 7.0 + \\log \\left( \\frac{[\\text{A}^-]}{[\\text{HA}]} \\right)<br>]<\/p>\n\n\n\n<p>[<br>\\log \\left( \\frac{[\\text{A}^-]}{[\\text{HA}]} \\right) = -2<br>]<\/p>\n\n\n\n<p>[<br>\\frac{[\\text{A}^-]}{[\\text{HA}]} = 10^{-2} = 0.01<br>]<\/p>\n\n\n\n<p>This means the un-ionized form ((\\text{HA})) is now 100 times more abundant than the ionized form ((\\text{A}^-)). Adding HCl shifts the equilibrium towards the un-ionized form.<\/p>\n\n\n\n<h4 class=\"wp-block-heading\"><strong>(c) Determining Total Concentration at pH 8.3<\/strong><\/h4>\n\n\n\n<p>At pH 8.3, p-nitrophenol is mostly ionized. Using Beer&#8217;s Law:<\/p>\n\n\n\n<p>[<br>A = \\varepsilon c l<br>]<\/p>\n\n\n\n<p>where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>( A = 0.550 ) (absorbance at 400 nm),<\/li>\n\n\n\n<li>( \\varepsilon = 19,500 ) M(^{-1})cm(^{-1}) (molar extinction coefficient),<\/li>\n\n\n\n<li>( l = 1.0 ) cm (path length),<\/li>\n\n\n\n<li>( c ) = concentration of ionized form.<\/li>\n<\/ul>\n\n\n\n<p>Solving for ( c ):<\/p>\n\n\n\n<p>[<br>c = \\frac{A}{\\varepsilon l} = \\frac{0.550}{(19,500)(1.0)}<br>]<\/p>\n\n\n\n<p>[<br>c = 2.82 \\times 10^{-5} M = 28.2 \\, \\mu M<br>]<\/p>\n\n\n\n<p>Since at pH 8.3, approximately 95% of p-nitrophenol is ionized, the total concentration is:<\/p>\n\n\n\n<p>[<br>\\frac{28.2 \\, \\mu M}{0.95} \\approx 29.7 \\, \\mu M<br>]<\/p>\n\n\n\n<p>Thus, the total p-nitrophenol concentration is <strong>29.7 \u03bcM<\/strong>.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Explanation<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>At pH 7, the ionized and un-ionized forms are present in equal amounts.<\/li>\n\n\n\n<li>Lowering the pH to 5 shifts the equilibrium toward the un-ionized form, making it 100 times more prevalent than the ionized form.<\/li>\n\n\n\n<li>At pH 8.3, most p-nitrophenol is ionized, and its concentration is calculated using Beer&#8217;s Law. Since ionized p-nitrophenol absorbs at 400 nm, we use absorbance data to estimate its total concentration.<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>The ionization equilibrium of p-nitrophenol.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/03\/image-382.png\" alt=\"\" class=\"wp-image-197094\"\/><\/figure>\n\n\n\n<p>Here is the scientific illustration showing the ionization equilibrium of p-nitrophenol. It depicts the un-ionized (HA) and ionized (A\u207b) forms, along with the effect of pH on the equilibrium shift.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>The ionization of p-nitrophenol is shown below (pKa= 7.0) a) At pH 7, what are the relative concentrations of ionized and un-ionized p-nitrophenol? b) If enough concentrated hydrochloric acid is added to a solution of p-nitrophenol to lower the pH from 7 to 5, what will happen to the relative concentrations of the ionized and [&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-197092","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/197092","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=197092"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/197092\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=197092"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=197092"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=197092"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}