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<\/figure>\n\n\n\nEquvalents OH 2.0 2.5 3.0 Answer the following questions regarding preparation of this buffer: BIO 101 Spring 2019 (a) Identify acid and its conjugate base in this buffer system. (b) Which pKa value is relevant for the preparation of this buffer? (c) Determine the weight in grams of sodium dihydrogen phosphate (NaH2PO4; 138g\/mol) and disodium hydrogen phosphate (Na2HPO4; 142g\/mol) needed to prepare this buffer 5. Phosphate buffer (a) What is the pH of a mixture of 0.042 M NaH2P04 and 0.058 M Na2HPO4? (b) If1 mL of 10 M NaOH is added to a liter of the buffer prepared in (a), how much will the pH change? (e) If1 mL of 10 M NaOH is added to a liter of pure water at pH 7, what is the final pH? Compare this with the answer in (b)<\/p>\n\n\n\n
The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n In this buffer system, we are working with the phosphate buffer<\/strong> system. The acid and its conjugate base are:<\/p>\n\n\n\n In this system, NaH\u2082PO\u2084 (H\u2082PO\u2084\u207b) acts as the acid, and Na\u2082HPO\u2084 (HPO\u2084\u00b2\u207b) is its conjugate base. The acid dissociates to release H\u207a ions, while the conjugate base accepts these ions to neutralize excess acid.<\/p>\n\n\n\n For a buffer with a pH of 7.00<\/strong>, the relevant pKa is the one closest to 7. In the case of phosphate, pKa\u2082<\/strong> = 7.2<\/strong>. This corresponds to the dissociation of H\u2082PO\u2084\u207b (which is the acid) into HPO\u2084\u00b2\u207b (the conjugate base). This pKa is the key factor for the buffer system because it governs the equilibrium between H\u2082PO\u2084\u207b and HPO\u2084\u00b2\u207b that will help maintain the pH.<\/p>\n\n\n\n To prepare a 1 L buffer solution<\/strong> at pH 7.00 with a total phosphate concentration of 0.100 M<\/strong>, you need to determine the molar amounts of NaH\u2082PO\u2084 and Na\u2082HPO\u2084 required to achieve this buffer. The required ratio of NaH\u2082PO\u2084 to Na\u2082HPO\u2084 is determined using the Henderson-Hasselbalch equation<\/strong>:<\/p>\n\n\n\n [ Substitute pH = 7.00, pKa = 7.2, and solve for the ratio of base to acid:<\/p>\n\n\n\n [ From the ratio, we know that the molarity of Na\u2082HPO\u2084<\/strong> should be 0.63 times the molarity of NaH\u2082PO\u2084<\/strong>. Let:<\/p>\n\n\n\n [ Given that the total concentration is 0.100 M, we can write:<\/p>\n\n\n\n [ Thus, the molarity of Na\u2082HPO\u2084<\/strong> (conjugate base) is 0.0613 M, and the molarity of NaH\u2082PO\u2084<\/strong> (acid) is:<\/p>\n\n\n\n [ Now, calculate the amounts of NaH\u2082PO\u2084 and Na\u2082HPO\u2084 required for 1 liter (since molarity is moles per liter, the total moles needed are just the molarity times the volume):<\/p>\n\n\n\n Finally, using the molar masses of NaH\u2082PO\u2084 (138 g\/mol) and Na\u2082HPO\u2084 (142 g\/mol), the required masses are:<\/p>\n\n\n\n So, to prepare 1 L of the buffer at pH 7.00, you need 5.32 g of NaH\u2082PO\u2084<\/strong> and 8.70 g of Na\u2082HPO\u2084<\/strong>.<\/p>\n\n\n\n [ [ So, the pH is approximately 7.34<\/strong>.<\/p>\n\n\n\n In comparison, adding NaOH to the buffer solution in part (b) results in a smaller pH change, showing the buffering capacity of the system.<\/p>\n","protected":false},"excerpt":{"rendered":" You decided to prepare a phosphate buffer from solid sodium dihydrogen phosphate (NaH2PO4) and disodium hydrogen phosphate (Na2HPO4) and you need 1L of the buffer at pH 7.00 with a total phosphate concentration (NaH2PO4 + Na2HPO) of 0.100 M. Hint: Phosphoric acid (H3PO4), a triprotic acid, has 3 pKa values: 2.14, Equvalents OH 2.0 2.5 […]<\/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-189442","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/189442","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=189442"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/189442\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=189442"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=189442"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=189442"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}(a) Acid and Conjugate Base in the Buffer System<\/h3>\n\n\n\n
\n
(b) Relevant pKa Value for Buffer Preparation<\/h3>\n\n\n\n
(c) Determining the Weights of Sodium Dihydrogen Phosphate (NaH\u2082PO\u2084) and Disodium Hydrogen Phosphate (Na\u2082HPO\u2084)<\/h3>\n\n\n\n
\\text{pH} = \\text{pKa} + \\log \\left(\\frac{[\\text{Base}]}{[\\text{Acid}]}\\right)
]<\/p>\n\n\n\n
7.00 = 7.2 + \\log \\left(\\frac{[\\text{Na}_2\\text{HPO}_4]}{[\\text{NaH}_2\\text{PO}_4]}\\right)
]
[
\\log \\left(\\frac{[\\text{Na}_2\\text{HPO}_4]}{[\\text{NaH}_2\\text{PO}_4]}\\right) = -0.2
]
[
\\frac{[\\text{Na}_2\\text{HPO}_4]}{[\\text{NaH}_2\\text{PO}_4]} = 10^{-0.2} \\approx 0.63
]<\/p>\n\n\n\n
[\\text{Na}_2\\text{HPO}_4] = x \\quad \\text{and} \\quad [\\text{NaH}_2\\text{PO}_4] = 0.63x
]<\/p>\n\n\n\n
x + 0.63x = 0.100
]
[
1.63x = 0.100
]
[
x = \\frac{0.100}{1.63} \\approx 0.0613 \\, \\text{M}
]<\/p>\n\n\n\n
0.63 \\times 0.0613 \\approx 0.0386 \\, \\text{M}
]<\/p>\n\n\n\n\n
\n
Additional Questions:<\/h3>\n\n\n\n
\n
\\text{pH} = 7.2 + \\log \\left(\\frac{0.058}{0.042}\\right)
]<\/p>\n\n\n\n
\\text{pH} = 7.2 + \\log(1.381) \\approx 7.2 + 0.141 \\approx 7.34
]<\/p>\n\n\n\n\n