To calculate the amount of sodium dihydrogen phosphate (NaH\u2082PO\u2084) and sodium hydrogen phosphate (Na\u2082HPO\u2084) required to prepare 50 mL of 20 mM phosphate buffer at pH 7.2, we will use the Henderson-Hasselbalch equation<\/strong> and basic stoichiometry.<\/p>\n\n\n\n pH=pKa+log\u2061([base][acid])\\text{pH} = \\text{p}K_a + \\log\\left(\\frac{[\\text{base}]}{[\\text{acid}]}\\right)pH=pKa\u200b+log([acid][base]\u200b)<\/p>\n\n\n\n Here:<\/p>\n\n\n\n Substituting:7.2=7.2+log\u2061([base][acid])\u21d2log\u2061([base][acid])=0\u21d2[base][acid]=17.2 = 7.2 + \\log\\left(\\frac{[\\text{base}]}{[\\text{acid}]}\\right) \\Rightarrow \\log\\left(\\frac{[\\text{base}]}{[\\text{acid}]}\\right) = 0 \\Rightarrow \\frac{[\\text{base}]}{[\\text{acid}]} = 17.2=7.2+log([acid][base]\u200b)\u21d2log([acid][base]\u200b)=0\u21d2[acid][base]\u200b=1<\/p>\n\n\n\n So, equal concentrations<\/strong> of base and acid are needed.<\/p>\n\n\n\n We want a total concentration<\/strong> of 20 mM for the buffer in 50 mL<\/strong>.<\/p>\n\n\n\n Since base and acid concentrations are equal:[base]=[acid]=20 mM2=10 mM[\\text{base}] = [\\text{acid}] = \\frac{20\\ \\text{mM}}{2} = 10\\ \\text{mM}[base]=[acid]=220 mM\u200b=10 mM<\/p>\n\n\n\n Volume = 50 mL = 0.050 L<\/p>\n\n\n\n Moles of each component:moles of base (Na\u2082HPO\u2084)=10\u00d710\u22123 mol\/L\u00d70.050 L=5.0\u00d710\u22124 mol\\text{moles of base (Na\u2082HPO\u2084)} = 10 \\times 10^{-3}\\ \\text{mol\/L} \\times 0.050\\ \\text{L} = 5.0 \\times 10^{-4}\\ \\text{mol}moles of base (Na\u2082HPO\u2084)=10\u00d710\u22123 mol\/L\u00d70.050 L=5.0\u00d710\u22124 molmoles of acid (NaH\u2082PO\u2084)=5.0\u00d710\u22124 mol\\text{moles of acid (NaH\u2082PO\u2084)} = 5.0 \\times 10^{-4}\\ \\text{mol}moles of acid (NaH\u2082PO\u2084)=5.0\u00d710\u22124 mol<\/p>\n\n\n\n Molar mass:<\/p>\n\n\n\n mass of Na\u2082HPO\u2084=5.0\u00d710\u22124\u00d7142=0.071 g\\text{mass of Na\u2082HPO\u2084} = 5.0 \\times 10^{-4} \\times 142 = 0.071\\ \\text{g}mass of Na\u2082HPO\u2084=5.0\u00d710\u22124\u00d7142=0.071 gmass of NaH\u2082PO\u2084=5.0\u00d710\u22124\u00d7120=0.060 g\\text{mass of NaH\u2082PO\u2084} = 5.0 \\times 10^{-4} \\times 120 = 0.060\\ \\text{g}mass of NaH\u2082PO\u2084=5.0\u00d710\u22124\u00d7120=0.060 g<\/p>\n\n\n\n Buffers are solutions that resist changes in pH when small amounts of acid or base are added. A phosphate buffer commonly uses sodium dihydrogen phosphate (NaH\u2082PO\u2084) and disodium hydrogen phosphate (Na\u2082HPO\u2084) as the weak acid and its conjugate base. The pKa for this buffer system is 7.2, which is ideal for maintaining a near-neutral pH.<\/p>\n\n\n\n To prepare a buffer of a specific pH, the Henderson-Hasselbalch equation is used. This equation relates the pH of the buffer to the ratio of base to acid. If the desired pH is equal to the pKa, then the ratio of base to acid becomes 1. This means equal amounts, in terms of concentration, of the weak acid and its conjugate base must be present.<\/p>\n\n\n\n In this question, the target is a 20 mM phosphate buffer in a total volume of 50 mL. Since the base and acid concentrations are equal, each must contribute 10 mM. To find the mass of each compound required, we calculate the number of moles using the molarity formula and then convert to grams using the molar mass of each salt.<\/p>\n\n\n\n For 10 mM in 50 mL, this equates to 0.0005 mol of each compound. Multiplying these mole values by their respective molar masses gives the required mass. The results show that 0.060 g of sodium dihydrogen phosphate and 0.071 g of disodium hydrogen phosphate are needed to make the buffer solution. This precise balance will ensure that the solution maintains a stable pH of 7.2.<\/p>\n\n\n\n Calculate the amount of sodium dihydrogen phosphate, NaH2PO4 (molecular weight = 120 g\/mol) and hydrated sodium hydrogen phosphate, Na2HPO4 (molecular weight = 142 g\/mol) needed to make 50 mL of 20 mM pH 7.2 buffer. (pKa of the buffer = 7.2). The Correct Answer and Explanation is: To calculate the amount of sodium dihydrogen phosphate […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-39092","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/39092","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=39092"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/39092\/revisions"}],"predecessor-version":[{"id":39094,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/39092\/revisions\/39094"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=39092"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=39092"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=39092"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\n\n\n\nStep 1: Understand the Henderson-Hasselbalch Equation<\/h3>\n\n\n\n
\n
\n\n\n\nStep 2: Total Buffer Concentration<\/h3>\n\n\n\n
\n\n\n\nStep 3: Calculate Moles Needed<\/h3>\n\n\n\n
\n\n\n\nStep 4: Convert Moles to Grams<\/h3>\n\n\n\n
\n
\n\n\n\nFinal Answer<\/strong>:<\/h3>\n\n\n\n
\n
\n\n\n\nExplanation <\/h3>\n\n\n\n
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<\/figure>\n","protected":false},"excerpt":{"rendered":"