{"id":230839,"date":"2025-06-10T06:50:00","date_gmt":"2025-06-10T06:50:00","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=230839"},"modified":"2025-06-10T06:50:03","modified_gmt":"2025-06-10T06:50:03","slug":"you-need-1-00-l-of-0-05-m-ammonium-bicarbonate-buffer-ph-6-8","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/10\/you-need-1-00-l-of-0-05-m-ammonium-bicarbonate-buffer-ph-6-8\/","title":{"rendered":"You need 1.00 L of 0.05 M ammonium bicarbonate buffer pH 6.8."},"content":{"rendered":"\n<p>You need 1.00 L of 0.05 M ammonium bicarbonate buffer pH 6.8. Describe the steps to prepare this buffer including the amount of solid ammonium bicarbonate you need. The pKa values for ammonium bicarbonate are 6.35 and 9.25.<\/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 prepare <strong>1.00\u202fL of 0.05\u202fM ammonium bicarbonate buffer at pH\u202f6.80<\/strong>, follow these steps:<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">\ud83e\uddea Step\u2011by\u2011Step Preparation<\/h3>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Calculate total moles of ammonium bicarbonate needed<\/strong>\n<ul class=\"wp-block-list\">\n<li>Desired concentration: 0.05\u202fmol\/L \u2192 need 0.05\u202fmol in 1\u202fL.<\/li>\n\n\n\n<li>Molar mass of NH\u2084HCO\u2083 \u2248 79.06\u202fg\/mol \u2192 0.05\u00d779.06=3.953\u2005\u200ag 0.05 \\times 79.06 = 3.953\\;g0.05\u00d779.06=3.953g<\/li>\n\n\n\n<li><strong>Weigh out \u2248\u202f3.95\u202fg ammonium bicarbonate<\/strong>.<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>Dissolve and adjust volume<\/strong>\n<ul class=\"wp-block-list\">\n<li>Dissolve the 3.95\u202fg NH\u2084HCO\u2083 in ~800\u202fmL deionized water.<\/li>\n\n\n\n<li>Once fully dissolved, transfer to a 1\u202fL volumetric flask and make up to the mark with water.<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>Adjust pH to 6.80<\/strong>\n<ul class=\"wp-block-list\">\n<li>The effective buffer system is bicarbonate (HCO3\u2212\\mathrm{HCO_3^-}HCO3\u2212\u200b) \/ carbonic acid (H2CO3\\mathrm{H_2CO_3}H2\u200bCO3\u200b), with pKa \u2248 6.35.<\/li>\n\n\n\n<li>Using the Henderson\u2013Hasselbalch equation: pH=pKa+log\u2061[HCO3\u2212][H2CO3] \\text{pH} = \\text{p}K_a + \\log\\frac{[\\mathrm{HCO_3^-}]}{[\\mathrm{H_2CO_3}]}pH=pKa\u200b+log[H2\u200bCO3\u200b][HCO3\u2212\u200b]\u200b 6.80=6.35+log\u2061[HCO3\u2212][H2CO3]\u21d2[HCO3\u2212][H2CO3]=100.45\u22482.82 6.80 = 6.35 + \\log\\frac{[\\mathrm{HCO_3^-}]}{[\\mathrm{H_2CO_3}]} \\quad\\Rightarrow\\quad \\frac{[\\mathrm{HCO_3^-}]}{[\\mathrm{H_2CO_3}]} = 10^{0.45} \\approx 2.826.80=6.35+log[H2\u200bCO3\u200b][HCO3\u2212\u200b]\u200b\u21d2[H2\u200bCO3\u200b][HCO3\u2212\u200b]\u200b=100.45\u22482.82<\/li>\n\n\n\n<li>In your solution, most bicarbonate comes from the solid; to reach this ratio, <strong>slowly add dilute HCl (e.g., 0.1\u202fM)<\/strong> until the pH meter reads <strong>6.80<\/strong>.<\/li>\n\n\n\n<li>Stir thoroughly and let equilibrate.<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>Final volume check<\/strong>\n<ul class=\"wp-block-list\">\n<li>After pH adjustment, recheck the volume and top up to 1\u202fL if necessary.<\/li>\n\n\n\n<li>Re\u2011check the pH and fine\u2011tune with tiny increments of HCl or NaOH if needed.<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>Label and store<\/strong>\n<ul class=\"wp-block-list\">\n<li>Label as \u201c0.05\u202fM NH\u2084HCO\u2083 buffer, pH\u202f6.80.\u201d<\/li>\n\n\n\n<li>Store at 4\u202f\u00b0C if needed, though bicarbonate buffers are best used fresh due to CO\u2082 loss.<\/li>\n<\/ul>\n<\/li>\n<\/ol>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">\ud83e\uddec Explanation<\/h3>\n\n\n\n<p>Ammonium bicarbonate (NH\u2084HCO\u2083) dissociates into ammonium (NH4+\\mathrm{NH_4^+}NH4+\u200b) and bicarbonate (HCO3\u2212\\mathrm{HCO_3^-}HCO3\u2212\u200b). Its two relevant pKa values are 6.35 (carbonic acid \u21cc bicarbonate) and 9.25 (ammonium \u21cc ammonia). At pH 6.8\u2014which is near 6.35\u2014the primary buffering action comes from the bicarbonate\/carbonic acid pair, not ammonium\/ammonia. So we focus on the first pKa.<\/p>\n\n\n\n<p>The buffer\u2019s total molarity (0.05\u202fM) sets the sum [HCO3\u2212]+[H2CO3]=0.05\u2005\u200aM[\\mathrm{HCO_3^-}] + [\\mathrm{H_2CO_3}] = 0.05\\;M[HCO3\u2212\u200b]+[H2\u200bCO3\u200b]=0.05M. The Henderson\u2013Hasselbalch equation relates pH to the acid\/base ratio:6.80=6.35+log\u2061([HCO3\u2212][H2CO3])6.80 = 6.35 + \\log\\left(\\frac{[\\mathrm{HCO_3^-}]}{[\\mathrm{H_2CO_3}]}\\right)6.80=6.35+log([H2\u200bCO3\u200b][HCO3\u2212\u200b]\u200b)<\/p>\n\n\n\n<p>Solving gives a ratio [HCO3\u2212]:[H2CO3]\u22482.82[\\mathrm{HCO_3^-}]:[\\mathrm{H_2CO_3}] \\approx 2.82[HCO3\u2212\u200b]:[H2\u200bCO3\u200b]\u22482.82. Let x = [H2CO3][\\mathrm{H_2CO_3}][H2\u200bCO3\u200b], then [HCO3\u2212]=2.82x[\\mathrm{HCO_3^-}] = 2.82x[HCO3\u2212\u200b]=2.82x, and x+2.82x=0.05x + 2.82x = 0.05x+2.82x=0.05. This yields x\u22480.0131\u2005\u200aMx \u2248 0.0131\\;Mx\u22480.0131M and bicarbonate \u2248 0.0369\u202fM.<\/p>\n\n\n\n<p>By dissolving 0.05\u202fmol NH\u2084HCO\u2083 (3.95\u202fg) in 1\u202fL, you introduce 0.05\u202fM bicarbonate, which is higher than the target 0.0369\u202fM. To adjust, you must shift some bicarbonate into the acid form (H2CO3\\mathrm{H_2CO_3}H2\u200bCO3\u200b) by adding a small amount of strong acid (HCl), thus lowering [HCO3\u2212][\\mathrm{HCO_3^-}][HCO3\u2212\u200b] and raising [H2CO3][\\mathrm{H_2CO_3}][H2\u200bCO3\u200b] until the correct ratio is reached.<\/p>\n\n\n\n<p>Using a calibrated pH meter, you titrate until the pH stabilizes at 6.80. Because carbonic species can lose CO\u2082 over time, it\u2019s best to seal the buffer or prepare fresh each time. This approach ensures an accurate 0.05\u202fM buffer at pH\u202f6.8 with minimal error.<\/p>\n\n\n\n<p>2\/2<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner4-801.jpeg\" alt=\"\" class=\"wp-image-230840\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>You need 1.00 L of 0.05 M ammonium bicarbonate buffer pH 6.8. Describe the steps to prepare this buffer including the amount of solid ammonium bicarbonate you need. The pKa values for ammonium bicarbonate are 6.35 and 9.25. The Correct Answer and Explanation is: To prepare 1.00\u202fL of 0.05\u202fM ammonium bicarbonate buffer at pH\u202f6.80, follow [&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-230839","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/230839","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=230839"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/230839\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=230839"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=230839"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=230839"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}