{"id":183317,"date":"2025-01-16T07:44:49","date_gmt":"2025-01-16T07:44:49","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=183317"},"modified":"2025-01-16T07:44:51","modified_gmt":"2025-01-16T07:44:51","slug":"the-ph-of-a-solution-of-hn3-ka1-910-5-and-nan3-is-4-86","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/01\/16\/the-ph-of-a-solution-of-hn3-ka1-910-5-and-nan3-is-4-86\/","title":{"rendered":"The pH of a solution of HN3 (ka=1.9*10^-5) and NaN3 is 4.86"},"content":{"rendered":"\n<p>The pH of a solution of HN3 (ka=1.9*10^-5) and NaN3 is 4.86.<br>What is the molarity of NAN3 if the molarity of HN3 is 0.020M?<\/p>\n\n\n\n<p>have posted this question 2 times only correct answer will get a lifesaver rating!!<\/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 solve this problem, we use the <strong>Henderson-Hasselbalch equation<\/strong> for buffer solutions:<\/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>Where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>(\\text{pH} = 4.86) (given)<\/li>\n\n\n\n<li>(\\text{pKa} = -\\log K_a = -\\log(1.9 \\times 10^{-5}))<\/li>\n\n\n\n<li>([\\text{HA}] = [\\text{HN}_3] = 0.020\\, \\text{M}) (given)<\/li>\n\n\n\n<li>([\\text{A}^-] = [\\text{NaN}_3]) (unknown)<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: Calculate (\\text{pKa})<\/h3>\n\n\n\n<p>[<br>\\text{pKa} = -\\log(1.9 \\times 10^{-5}) = 4.72<br>]<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 2: Substitute values into the Henderson-Hasselbalch equation<\/h3>\n\n\n\n<p>[<br>4.86 = 4.72 + \\log \\left( \\frac{[\\text{NaN}_3]}{0.020} \\right)<br>]<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 3: Solve for (\\frac{[\\text{NaN}_3]}{0.020})<\/h3>\n\n\n\n<p>[<br>4.86 &#8211; 4.72 = \\log \\left( \\frac{[\\text{NaN}_3]}{0.020} \\right)<br>]<\/p>\n\n\n\n<p>[<br>0.14 = \\log \\left( \\frac{[\\text{NaN}_3]}{0.020} \\right)<br>]<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 4: Remove the logarithm<\/h3>\n\n\n\n<p>[<br>10^{0.14} = \\frac{[\\text{NaN}_3]}{0.020}<br>]<\/p>\n\n\n\n<p>[<br>1.38 = \\frac{[\\text{NaN}_3]}{0.020}<br>]<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 5: Solve for ([\\text{NaN}_3])<\/h3>\n\n\n\n<p>[<br>[\\text{NaN}_3] = 1.38 \\times 0.020 = 0.0276\\, \\text{M}<br>]<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Final Answer:<\/h3>\n\n\n\n<p>The molarity of (\\text{NaN}_3) is:<\/p>\n\n\n\n<p>[<br>\\boxed{0.0276 \\, \\text{M}}<br>]<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation (300 words)<\/h3>\n\n\n\n<p>The problem involves a buffer solution containing a weak acid (HN(_3)) and its conjugate base (NaN(_3)). Such solutions resist changes in pH, and the Henderson-Hasselbalch equation is used to relate the pH, the acid dissociation constant ((K_a)), and the concentrations of the acid ((HA)) and its conjugate base ((A^-)).<\/p>\n\n\n\n<p>First, the (K_a) of HN(_3) is used to calculate the pKa ((pKa = -\\log K_a)), which measures the strength of the weak acid. Knowing the pH (4.86) and the concentration of HN(_3) (0.020 M), we rearranged the Henderson-Hasselbalch equation to solve for the ratio of the base (([\\text{NaN}_3])) to the acid (([\\text{HN}_3])).<\/p>\n\n\n\n<p>After isolating the ratio, the logarithmic function was converted to exponential form, allowing us to directly solve for ([\\text{NaN}_3]). The result, 0.0276 M, indicates the concentration of NaN(_3) required to maintain the given pH of the solution.<\/p>\n\n\n\n<p>This calculation demonstrates the importance of the conjugate base in stabilizing the solution&#8217;s pH. If the concentration of NaN(_3) changes, the buffer&#8217;s effectiveness at maintaining pH decreases.<\/p>\n\n\n\n<p><\/p>\n","protected":false},"excerpt":{"rendered":"<p>The pH of a solution of HN3 (ka=1.9*10^-5) and NaN3 is 4.86.What is the molarity of NAN3 if the molarity of HN3 is 0.020M? have posted this question 2 times only correct answer will get a lifesaver rating!! The Correct Answer and Explanation is : To solve this problem, we use the Henderson-Hasselbalch equation for [&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-183317","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/183317","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=183317"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/183317\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=183317"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=183317"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=183317"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}