What mass of solid potassium oxalate (K2C2O4), when dissolved in an acidic solution, would be titrated by exactly 50.0 mL of 0.0358 M potassium permanganate (KMnO4) solution?<\/p>\n\n\n\n
The correct answer and explanation is:<\/strong><\/mark><\/p>\n\n\n\n To solve this problem, we need to understand the redox titration<\/strong> between potassium permanganate (KMnO\u2084)<\/strong> and potassium oxalate (K\u2082C\u2082O\u2084)<\/strong> in an acidic solution<\/strong>.<\/p>\n\n\n\n The redox reaction between MnO\u2084\u207b<\/strong> and C\u2082O\u2084\u00b2\u207b<\/strong> in acidic medium is: 2MnO4\u2212+5C2O42\u2212+16H+\u21922Mn2++10CO2+8H2O2 \\text{MnO}_4^- + 5 \\text{C}_2\\text{O}_4^{2-} + 16 \\text{H}^+ \\rightarrow 2 \\text{Mn}^{2+} + 10 \\text{CO}_2 + 8 \\text{H}_2\\text{O}<\/p>\n\n\n\n This tells us:<\/p>\n\n\n\n Step 1: Moles of KMnO\u2084 used<\/strong> Moles of KMnO\u2084=M\u00d7V=0.0358\u2009mol\/L\u00d70.0500\u2009L=0.00179\u2009mol\\text{Moles of KMnO\u2084} = M \\times V = 0.0358 \\, \\text{mol\/L} \\times 0.0500 \\, \\text{L} = 0.00179 \\, \\text{mol}<\/p>\n\n\n\n Each mole of KMnO\u2084 provides 1 mole of MnO\u2084\u207b.<\/p>\n\n\n\n Step 2: Use mole ratio to find moles of oxalate (C\u2082O\u2084\u00b2\u207b)<\/strong> So: Moles of C2O42\u2212=0.00179\u00d752=0.004475\u2009mol\\text{Moles of C}_2\\text{O}_4^{2-} = 0.00179 \\times \\frac{5}{2} = 0.004475 \\, \\text{mol}<\/p>\n\n\n\n Step 3: Find molar mass of K\u2082C\u2082O\u2084<\/strong> K\u2082C\u2082O\u2084=2(39.10)+2(12.01)+4(16.00)=78.20+24.02+64.00=166.22\u2009g\/mol\\text{K\u2082C\u2082O\u2084} = 2(39.10) + 2(12.01) + 4(16.00) = 78.20 + 24.02 + 64.00 = 166.22 \\, \\text{g\/mol}<\/p>\n\n\n\n Step 4: Calculate mass of K\u2082C\u2082O\u2084<\/strong> Mass=moles\u00d7molar mass=0.004475\u00d7166.22=0.744\u2009g\\text{Mass} = \\text{moles} \\times \\text{molar mass} = 0.004475 \\times 166.22 = \\boxed{0.744} \\, \\text{g}<\/p>\n\n\n\n In this problem, potassium permanganate (KMnO\u2084) is used as a titrant to determine the amount of potassium oxalate (K\u2082C\u2082O\u2084) in an acidic medium. KMnO\u2084 is a strong oxidizing agent, and oxalate ions (C\u2082O\u2084\u00b2\u207b) are reducing agents. In an acidic environment, the permanganate ion (MnO\u2084\u207b) is reduced to Mn\u00b2\u207a, and oxalate is oxidized to carbon dioxide (CO\u2082). The stoichiometry of this redox reaction shows that 2 moles of permanganate react with 5 moles of oxalate.<\/p>\n\n\n\n First, we calculate how many moles of KMnO\u2084 are in 50.0 mL of 0.0358 M solution. This comes out to 0.00179 moles. Using the balanced chemical equation, we apply the 2:5 mole ratio to find the moles of oxalate required, which is 0.004475 moles.<\/p>\n\n\n\n Potassium oxalate (K\u2082C\u2082O\u2084) contains one mole of oxalate ion per mole of compound, so we assume a 1:1 ratio between K\u2082C\u2082O\u2084 and C\u2082O\u2084\u00b2\u207b. We then multiply the moles of potassium oxalate by its molar mass (166.22 g\/mol) to get the mass required: 0.744 grams.<\/p>\n\n\n\n Thus, 0.744 grams<\/strong> of solid K\u2082C\u2082O\u2084 would exactly react with 50.0 mL of 0.0358 M KMnO\u2084 solution in acidic conditions.<\/p>\n","protected":false},"excerpt":{"rendered":" What mass of solid potassium oxalate (K2C2O4), when dissolved in an acidic solution, would be titrated by exactly 50.0 mL of 0.0358 M potassium permanganate (KMnO4) solution? The correct answer and explanation is: To solve this problem, we need to understand the redox titration between potassium permanganate (KMnO\u2084) and potassium oxalate (K\u2082C\u2082O\u2084) in an acidic […]<\/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-18953","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/18953","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=18953"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/18953\/revisions"}],"predecessor-version":[{"id":18954,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/18953\/revisions\/18954"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=18953"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=18953"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=18953"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\n\n\n\nBalanced Redox Reaction in Acidic Medium<\/strong><\/h3>\n\n\n\n
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\n\n\n\nStep-by-step Calculation<\/strong><\/h3>\n\n\n\n
From the balanced equation: 2mol MnO4\u2212:5mol C2O42\u22122 \\text{mol MnO}_4^- : 5 \\text{mol C}_2\\text{O}_4^{2-}<\/p>\n\n\n\n
\n\n\n\n\u2705 Final Answer: 0.744 grams of potassium oxalate<\/strong><\/h3>\n\n\n\n
\n\n\n\n\ud83e\uddea Explanation Summary (300 Words)<\/h3>\n\n\n\n