{"id":29436,"date":"2025-06-21T04:22:27","date_gmt":"2025-06-21T04:22:27","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=29436"},"modified":"2025-06-21T04:22:32","modified_gmt":"2025-06-21T04:22:32","slug":"oxalic-acid-dihydrate-h2c2o4-a%c2%b7-2h2o-can-be-used-to-standardize-naoh-solution-through-acid-base-titration","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/oxalic-acid-dihydrate-h2c2o4-a%c2%b7-2h2o-can-be-used-to-standardize-naoh-solution-through-acid-base-titration\/","title":{"rendered":"Oxalic acid dihydrate (H2C2O4 \u00c2\u00b7 2H2O) can be used to standardize NaOH solution through acid-base titration."},"content":{"rendered":"\n

Oxalic acid dihydrate (H2C2O4 \u00c2\u00b7 2H2O) can be used to standardize NaOH solution through acid-base titration. To determine the molarity of an unknown NaOH solution, a student weighed out 0.56 g of solid oxalic acid dihydrate and dissolved the solid in 32.8 mL of DI water. The resulting solution was used in a titration with the unknown NaOH solution (3 drops of phenolphthalein were used as an indicator). The balanced chemical equation is given below. If the student used 12.84 mL of the unknown NaOH solution in the titration, what is the molarity of the NaOH solution? Keep the correct number of significant figures. H2C2O4 (aq) + 2NaOH (aq) –> Na2C2O4 (aq) + 2H2O (l)<\/p>\n\n\n\n

The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n

To find the molarity of the NaOH solution, we will follow these steps:<\/p>\n\n\n\n

\n\n\n\n

Step 1: Calculate moles of oxalic acid dihydrate (H\u2082C\u2082O\u2084 \u00b7 2H\u2082O)<\/strong><\/h3>\n\n\n\n

Molar mass of oxalic acid dihydrate (H\u2082C\u2082O\u2084 \u00b7 2H\u2082O):<\/strong><\/p>\n\n\n\n

    \n
  • H: 1.01 \u00d7 4 = 4.04<\/li>\n\n\n\n
  • C: 12.01 \u00d7 2 = 24.02<\/li>\n\n\n\n
  • O: 16.00 \u00d7 4 = 64.00<\/li>\n\n\n\n
  • H\u2082O: (2 \u00d7 (2.02 + 16.00)) = 36.04<\/li>\n<\/ul>\n\n\n\n

    Total molar mass:<\/strong>
    = 4.04 + 24.02 + 64.00 + 36.04
    = 128.10 g\/mol<\/strong><\/p>\n\n\n\n

    Now calculate the moles of oxalic acid dihydrate:Moles of H\u2082C\u2082O\u2084 \\cdotp 2H\u2082O=0.56 g128.10 g\/mol\u22480.00437 mol\\text{Moles of H\u2082C\u2082O\u2084 \u00b7 2H\u2082O} = \\frac{0.56 \\text{ g}}{128.10 \\text{ g\/mol}} \\approx 0.00437 \\text{ mol}Moles of H\u2082C\u2082O\u2084 \\cdotp 2H\u2082O=128.10 g\/mol0.56 g\u200b\u22480.00437 mol<\/p>\n\n\n\n

    \n\n\n\n

    Step 2: Use the balanced equation to find moles of NaOH<\/strong><\/h3>\n\n\n\n

    Balanced equation:
    H\u2082C\u2082O\u2084 + 2NaOH \u2192 Na\u2082C\u2082O\u2084 + 2H\u2082O<\/p>\n\n\n\n

    From the equation, 1 mole of oxalic acid reacts with 2 moles of NaOH<\/strong><\/p>\n\n\n\n

    So moles of NaOH = 0.00437 mol \u00d7 2 = 0.00874 mol<\/strong><\/p>\n\n\n\n

    \n\n\n\n

    Step 3: Use volume of NaOH to calculate molarity<\/strong><\/h3>\n\n\n\n

    Volume of NaOH solution = 12.84 mL = 0.01284 LMolarity (M)=0.00874 mol0.01284 L\u22480.681 M\\text{Molarity (M)} = \\frac{0.00874 \\text{ mol}}{0.01284 \\text{ L}} \\approx \\boxed{0.681 \\text{ M}}Molarity (M)=0.01284 L0.00874 mol\u200b\u22480.681 M\u200b<\/p>\n\n\n\n

    \n\n\n\n

    Explanation<\/strong><\/h3>\n\n\n\n

    The student used oxalic acid dihydrate to titrate an unknown NaOH solution. Oxalic acid is a diprotic acid, meaning it donates two protons per molecule, reacting with two hydroxide ions from NaOH. The student first converted the mass of oxalic acid dihydrate into moles using its molar mass. With the 1:2 mole ratio from the balanced equation, the number of moles of NaOH that reacted was determined. Finally, using the known volume of NaOH added, the molarity was calculated by dividing moles by liters. The final answer, 0.681 M<\/strong>, respects the correct significant figures from the measurements provided.<\/p>\n\n\n\n

    \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

    Oxalic acid dihydrate (H2C2O4 \u00c2\u00b7 2H2O) can be used to standardize NaOH solution through acid-base titration. To determine the molarity of an unknown NaOH solution, a student weighed out 0.56 g of solid oxalic acid dihydrate and dissolved the solid in 32.8 mL of DI water. The resulting solution was used in a titration with […]<\/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-29436","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/29436","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=29436"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/29436\/revisions"}],"predecessor-version":[{"id":29438,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/29436\/revisions\/29438"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=29436"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=29436"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=29436"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}