{"id":35081,"date":"2025-06-24T06:36:52","date_gmt":"2025-06-24T06:36:52","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=35081"},"modified":"2025-06-24T06:36:54","modified_gmt":"2025-06-24T06:36:54","slug":"how-many-grams-of-sodium-acetate-ch3co2na-mw-82-0-g-mol-should-be-present-in-1-0-l-of-0-62-mole-acetic-acid-ch3co2h-solution-to-achieve-a-buffer-of-ph-4-61","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/how-many-grams-of-sodium-acetate-ch3co2na-mw-82-0-g-mol-should-be-present-in-1-0-l-of-0-62-mole-acetic-acid-ch3co2h-solution-to-achieve-a-buffer-of-ph-4-61\/","title":{"rendered":"How many grams of sodium acetate (CH3CO2Na, Mw = 82.0 g\/mol) should be present in 1.0 L of 0.62 mole acetic acid (CH3CO2H) solution to achieve a buffer of pH = 4.61"},"content":{"rendered":"\n<p>How many grams of sodium acetate (CH3CO2Na, Mw = 82.0 g\/mol) should be present in 1.0 L of 0.62 mole acetic acid (CH3CO2H) solution to achieve a buffer of pH = 4.61? Ka = 1.78 \u00c3\u2014 10\u00e2\u20ac\u201c5 for benzoic acid. Please keep your answer to one decimal place.<\/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><strong>Correct Answer: 23.3 grams of sodium acetate<\/strong><\/p>\n\n\n\n<p>To calculate how many grams of sodium acetate (CH\u2083CO\u2082Na) are needed to make a buffer solution with acetic acid (CH\u2083CO\u2082H) at pH = 4.61, we use the <strong>Henderson-Hasselbalch equation<\/strong>: pH=pKa+log\u2061([A\u2212][HA])\\text{pH} = \\text{p}K_a + \\log\\left(\\frac{[\\text{A}^-]}{[\\text{HA}]}\\right)<\/p>\n\n\n\n<p>Where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>pH=4.61\\text{pH} = 4.61 (target pH)<\/li>\n\n\n\n<li>HA\\text{HA} is the weak acid, which is acetic acid<\/li>\n\n\n\n<li>A\u2212\\text{A}^- is the conjugate base, which is sodium acetate<\/li>\n\n\n\n<li>Ka=1.78\u00d710\u22125\\text{Ka} = 1.78 \\times 10^{-5} for acetic acid<\/li>\n\n\n\n<li>[HA]=0.62[\\text{HA}] = 0.62 M (given concentration of acetic acid)<\/li>\n<\/ul>\n\n\n\n<p><strong>Step 1: Calculate pKa<\/strong> pKa=\u2212log\u2061(1.78\u00d710\u22125)=4.75\\text{p}K_a = -\\log(1.78 \\times 10^{-5}) = 4.75<\/p>\n\n\n\n<p><strong>Step 2: Use the Henderson-Hasselbalch equation<\/strong> 4.61=4.75+log\u2061([A\u2212]0.62)4.61 = 4.75 + \\log\\left(\\frac{[\\text{A}^-]}{0.62}\\right)<\/p>\n\n\n\n<p><strong>Step 3: Solve for [A\u2212][\\text{A}^-]<\/strong> 4.61\u22124.75=log\u2061([A\u2212]0.62)\u21d2\u22120.14=log\u2061([A\u2212]0.62)4.61 &#8211; 4.75 = \\log\\left(\\frac{[\\text{A}^-]}{0.62}\\right) \\Rightarrow -0.14 = \\log\\left(\\frac{[\\text{A}^-]}{0.62}\\right)<\/p>\n\n\n\n<p>Raise 10 to both sides: 10\u22120.14=[A\u2212]0.62\u21d20.72=[A\u2212]0.6210^{-0.14} = \\frac{[\\text{A}^-]}{0.62} \\Rightarrow 0.72 = \\frac{[\\text{A}^-]}{0.62} [A\u2212]=0.72\u00d70.62=0.4464\u2009mol\/L[\\text{A}^-] = 0.72 \\times 0.62 = 0.4464 \\, \\text{mol\/L}<\/p>\n\n\n\n<p><strong>Step 4: Convert moles to grams<\/strong><br>In 1.0 L of solution: mol&nbsp;of&nbsp;CH\u2083CO\u2082Na=0.4464\u2009mol\\text{mol of CH\u2083CO\u2082Na} = 0.4464 \\, \\text{mol} grams=0.4464\u2009mol\u00d782.0\u2009g\/mol=36.6\u2009g\\text{grams} = 0.4464 \\, \\text{mol} \\times 82.0 \\, \\text{g\/mol} = 36.6 \\, \\text{g}<\/p>\n\n\n\n<p><strong>Correction<\/strong>: Earlier you were told the Ka of <strong>benzoic acid<\/strong>. However, since we are using <strong>acetic acid<\/strong>, and you wrote CH\u2083CO\u2082H, the correct Ka is for <strong>acetic acid<\/strong>, not benzoic acid. The Ka of <strong>acetic acid<\/strong> is 1.8 \u00d7 10\u207b\u2075, and its pKa is about 4.74 to 4.75. So the calculation above remains correct.<\/p>\n\n\n\n<p><strong>Final Answer: 36.6 grams of sodium acetate should be added<\/strong> to 1.0 L of 0.62 M acetic acid to prepare a buffer solution at pH 4.61.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>How many grams of sodium acetate (CH3CO2Na, Mw = 82.0 g\/mol) should be present in 1.0 L of 0.62 mole acetic acid (CH3CO2H) solution to achieve a buffer of pH = 4.61? Ka = 1.78 \u00c3\u2014 10\u00e2\u20ac\u201c5 for benzoic acid. Please keep your answer to one decimal place. The correct answer and explanation is: Correct [&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-35081","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/35081","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=35081"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/35081\/revisions"}],"predecessor-version":[{"id":35086,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/35081\/revisions\/35086"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=35081"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=35081"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=35081"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}