{"id":39088,"date":"2025-06-26T19:43:06","date_gmt":"2025-06-26T19:43:06","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=39088"},"modified":"2025-06-26T19:43:10","modified_gmt":"2025-06-26T19:43:10","slug":"how-many-grams-of-sodium-dihydrogen-phosphate-nah2po4-m120-g-mol-must-be-added-to-the-solution-containing-0-05-mole-of-disodium-hydrogen-phosphate-na2hpo4-in-order-to-produce","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/how-many-grams-of-sodium-dihydrogen-phosphate-nah2po4-m120-g-mol-must-be-added-to-the-solution-containing-0-05-mole-of-disodium-hydrogen-phosphate-na2hpo4-in-order-to-produce\/","title":{"rendered":"How many grams of sodium dihydrogen phosphate (NaH2PO4, M=120 g\/mol) must be added to the solution containing 0.05 mole of disodium hydrogen phosphate (Na2HPO4) in order to produce,"},"content":{"rendered":"\n<p>How many grams of sodium dihydrogen phosphate (NaH2PO4, M=120 g\/mol) must be added to the solution containing 0.05 mole of disodium hydrogen phosphate (Na2HPO4) in order to produce, after dilution with water to the final volume of 1 L, a phosphate buffer of pH = 7.612? (Non-anonymous question@) (2 Points) Dissociation constants for the phosphoric acid: K1 = 6.3 x 10^-3; K2 = 6.2 x 10^-8; K3 = 5.0 x 10^-13.<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-vivid-cyan-blue-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>To solve this problem, we need to apply the <strong>Henderson-Hasselbalch equation<\/strong>: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<p>We are given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>pH = 7.612<\/strong><\/li>\n\n\n\n<li>The base is HPO42\u2212\\text{HPO}_4^{2-}HPO42\u2212\u200b (from Na\u2082HPO\u2084), which is the conjugate base of H2PO4\u2212\\text{H}_2\\text{PO}_4^{-}H2\u200bPO4\u2212\u200b<\/li>\n\n\n\n<li>The acid is H2PO4\u2212\\text{H}_2\\text{PO}_4^{-}H2\u200bPO4\u2212\u200b (from NaH\u2082PO\u2084)<\/li>\n\n\n\n<li>Moles of base = 0.05 mol<\/li>\n\n\n\n<li>Final volume = 1 L, so mol = mol\/L<\/li>\n<\/ul>\n\n\n\n<p>We are dealing with the second dissociation of <strong>phosphoric acid<\/strong>:H2PO4\u2212\u21ccHPO42\u2212+H+\\text{H}_2\\text{PO}_4^- \\rightleftharpoons \\text{HPO}_4^{2-} + \\text{H}^+H2\u200bPO4\u2212\u200b\u21ccHPO42\u2212\u200b+H+<\/p>\n\n\n\n<p>For this equilibrium:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>pKa=\u2212log\u2061(6.2\u00d710\u22128)\u22487.21\\text{p}K_a = -\\log(6.2 \\times 10^{-8}) \\approx 7.21pKa\u200b=\u2212log(6.2\u00d710\u22128)\u22487.21<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: Use Henderson-Hasselbalch Equation<\/h3>\n\n\n\n<p>7.612=7.21+log\u2061(0.05[acid])7.612 = 7.21 + \\log \\left( \\frac{0.05}{[\\text{acid}]} \\right)7.612=7.21+log([acid]0.05\u200b)<\/p>\n\n\n\n<p>Subtract 7.21 from both sides:0.402=log\u2061(0.05[acid])0.402 = \\log \\left( \\frac{0.05}{[\\text{acid}]} \\right)0.402=log([acid]0.05\u200b)<\/p>\n\n\n\n<p>Now take the inverse log:100.402\u22482.5310^{0.402} \\approx 2.53100.402\u22482.530.05[acid]=2.53\\frac{0.05}{[\\text{acid}]} = 2.53[acid]0.05\u200b=2.53<\/p>\n\n\n\n<p>Now solve for the concentration of the acid:[acid]=0.052.53\u22480.01976&nbsp;mol[\\text{acid}] = \\frac{0.05}{2.53} \\approx 0.01976\\ \\text{mol}[acid]=2.530.05\u200b\u22480.01976&nbsp;mol<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 2: Convert to grams<\/h3>\n\n\n\n<p>We now know we need approximately 0.01976 mol of sodium dihydrogen phosphate (NaH\u2082PO\u2084). Use its molar mass to get grams:mass=0.01976&nbsp;mol\u00d7120&nbsp;g\/mol=2.3712&nbsp;g\\text{mass} = 0.01976\\ \\text{mol} \\times 120\\ \\text{g\/mol} = 2.3712\\ \\text{g}mass=0.01976&nbsp;mol\u00d7120&nbsp;g\/mol=2.3712&nbsp;g<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Final Answer:<\/h3>\n\n\n\n<p>2.37&nbsp;g&nbsp;of&nbsp;NaH2PO4\\boxed{2.37\\ \\text{g of NaH}_2\\text{PO}_4}2.37&nbsp;g&nbsp;of&nbsp;NaH2\u200bPO4\u200b\u200b<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation <\/h3>\n\n\n\n<p>To prepare a phosphate buffer at pH 7.612, you need a specific ratio of the conjugate base HPO42\u2212\\text{HPO}_4^{2-}HPO42\u2212\u200b and its conjugate acid H2PO4\u2212\\text{H}_2\\text{PO}_4^-H2\u200bPO4\u2212\u200b. This ratio is calculated using the Henderson-Hasselbalch equation, which relates pH, pKa, and the ratio of base to acid. Since the relevant equilibrium is the second dissociation of phosphoric acid (Ka\u2082), we use its value Ka=6.2\u00d710\u22128K_a = 6.2 \\times 10^{-8}Ka\u200b=6.2\u00d710\u22128, and compute pKa=7.21\\text{p}K_a = 7.21pKa\u200b=7.21.<\/p>\n\n\n\n<p>The goal is to make the ratio of base to acid equal to the ratio that yields pH 7.612. Plugging into the equation, we find that the base to acid ratio must be about 2.53. With 0.05 mol of base already present, we calculate how much acid is needed to achieve this ratio, which turns out to be approximately 0.01976 mol.<\/p>\n\n\n\n<p>We then convert the required moles of acid to grams using its molar mass (120 g\/mol). Multiplying gives us 2.37 grams. This means that, to achieve the target pH in a 1-liter buffer solution, we must add 2.37 grams of sodium dihydrogen phosphate to the solution already containing 0.05 mol of disodium hydrogen phosphate.<\/p>\n\n\n\n<p>This buffer resists changes in pH by neutralizing added acids or bases, making it useful in many chemical and biological processes.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"1024\" height=\"1024\" src=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner5-226.jpeg\" alt=\"\" class=\"wp-image-39089\" srcset=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner5-226.jpeg 1024w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner5-226-300x300.jpeg 300w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner5-226-150x150.jpeg 150w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner5-226-768x768.jpeg 768w\" sizes=\"auto, (max-width: 1024px) 100vw, 1024px\" \/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>How many grams of sodium dihydrogen phosphate (NaH2PO4, M=120 g\/mol) must be added to the solution containing 0.05 mole of disodium hydrogen phosphate (Na2HPO4) in order to produce, after dilution with water to the final volume of 1 L, a phosphate buffer of pH = 7.612? (Non-anonymous question@) (2 Points) Dissociation constants for the phosphoric [&hellip;]<\/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-39088","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/39088","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=39088"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/39088\/revisions"}],"predecessor-version":[{"id":39090,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/39088\/revisions\/39090"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=39088"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=39088"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=39088"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}