{"id":26791,"date":"2025-06-19T18:40:25","date_gmt":"2025-06-19T18:40:25","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=26791"},"modified":"2025-06-19T18:40:27","modified_gmt":"2025-06-19T18:40:27","slug":"calculating-the-molarity-of-naoh-solution-to-a-very-good-approximation-at-the-endpoint-ep-the-number-of-moles-of-khp-that-have-been-neutralized-is-equal-to-the-number-of-moles-of-naoh-that-have-be","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/calculating-the-molarity-of-naoh-solution-to-a-very-good-approximation-at-the-endpoint-ep-the-number-of-moles-of-khp-that-have-been-neutralized-is-equal-to-the-number-of-moles-of-naoh-that-have-be\/","title":{"rendered":"Calculating the Molarity of NaOH Solution To a very good approximation, at the endpoint (ep), the number of moles of KHP that have been neutralized is equal to the number of moles of NaOH that have been added"},"content":{"rendered":"\n

Calculating the Molarity of NaOH Solution To a very good approximation, at the endpoint (ep), the number of moles of KHP that have been neutralized is equal to the number of moles of NaOH that have been added (the equivalence point). To determine the molarity of the NaOH solution, make the ratio: moles NaOH added (at ep) \/ L NaOH solution added (at ep) Sample calculation: A 1.413-g sample of KHP takes 19.43 mL of a NaOH solution to titrate it to a phenolphthalein endpoint. What is the molarity of the NaOH solution? The molar mass of KHP is 204.2 g\/mol. Data Sheet KHP Sample Mass of flask + KHP, g 44.727g___<\/strong> Mass of empty flask, g (tared) 44.192g<\/em><\/strong> Calculate moles of KHP (10 pts, show work here) 5 pts Mass of KHP sample, g Titration Data Final buret reading, mL 8.66mL<\/em> (endpoint) Initial buret reading, mL 0.01mL<\/em> 5 pts Volume of NaOH delivered, mL<\/strong> Moles of KHP titrated, mol 10 pts Moles of NaOH required, mol<\/strong> Show work: 15 pts Molarity of the NaOH solution, M _<\/strong><\/p>\n\n\n\n

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

Step 1: Calculate Mass of KHP Sample<\/strong><\/h3>\n\n\n\n

Mass of flask + KHP = 44.727 g<\/strong>
Mass of empty flask = 44.192 g<\/strong>Mass of KHP sample=44.727\u2009g\u221244.192\u2009g=0.535\u2009g\\text{Mass of KHP sample} = 44.727 \\, \\text{g} – 44.192 \\, \\text{g} = \\boxed{0.535 \\, \\text{g}}Mass of KHP sample=44.727g\u221244.192g=0.535g\u200b<\/p>\n\n\n\n

\n\n\n\n

Step 2: Calculate Moles of KHP<\/strong><\/h3>\n\n\n\n

Given:
Molar mass of KHP = 204.2 g\/molMoles of KHP=0.535\u2009g204.2\u2009g\/mol=0.002621\u2009mol\\text{Moles of KHP} = \\frac{0.535 \\, \\text{g}}{204.2 \\, \\text{g\/mol}} = \\boxed{0.002621 \\, \\text{mol}}Moles of KHP=204.2g\/mol0.535g\u200b=0.002621mol\u200b<\/p>\n\n\n\n

\n\n\n\n

Step 3: Volume of NaOH Delivered<\/strong><\/h3>\n\n\n\n

Final buret reading = 8.66 mL
Initial buret reading = 0.01 mLVolume of NaOH delivered=8.66\u22120.01=8.65\u2009mL=0.00865\u2009L\\text{Volume of NaOH delivered} = 8.66 – 0.01 = \\boxed{8.65 \\, \\text{mL}} = 0.00865 \\, \\text{L}Volume of NaOH delivered=8.66\u22120.01=8.65mL\u200b=0.00865L<\/p>\n\n\n\n

\n\n\n\n

Step 4: Moles of NaOH Required<\/strong><\/h3>\n\n\n\n

Since KHP is monoprotic (reacts 1:1 with NaOH),Moles of NaOH=Moles of KHP=0.002621\u2009mol\\text{Moles of NaOH} = \\text{Moles of KHP} = \\boxed{0.002621 \\, \\text{mol}}Moles of NaOH=Moles of KHP=0.002621mol\u200b<\/p>\n\n\n\n

\n\n\n\n

Step 5: Molarity of NaOH Solution<\/strong><\/h3>\n\n\n\n

Molarity=Moles of NaOHVolume of NaOH in L=0.0026210.00865=0.303\u2009M\\text{Molarity} = \\frac{\\text{Moles of NaOH}}{\\text{Volume of NaOH in L}} = \\frac{0.002621}{0.00865} = \\boxed{0.303 \\, \\text{M}}Molarity=Volume of NaOH in LMoles of NaOH\u200b=0.008650.002621\u200b=0.303M\u200b<\/p>\n\n\n\n

\n\n\n\n

Final Answer<\/strong><\/h3>\n\n\n\n

Molarity of the NaOH solution = 0.303 M<\/strong><\/p>\n\n\n\n

\n\n\n\n

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

In acid-base titration, a known mass of a monoprotic acid such as potassium hydrogen phthalate (KHP) is neutralized by a sodium hydroxide (NaOH) solution of unknown concentration. The reaction is:KHP+NaOH\u2192NaKP+H2O\\text{KHP} + \\text{NaOH} \\rightarrow \\text{NaKP} + \\text{H}_2\\text{O}KHP+NaOH\u2192NaKP+H2\u200bO<\/p>\n\n\n\n

This is a one-to-one mole reaction. At the equivalence point, moles of NaOH added are equal to the moles of KHP present. By determining how many moles of KHP were present in the sample, we can also know how many moles of NaOH were required to reach the endpoint, since they are equal.<\/p>\n\n\n\n

First, we determine the mass of KHP by subtracting the mass of the empty flask from the total mass of the flask and KHP. Then, using the molar mass of KHP (204.2 g\/mol), we convert this mass into moles.<\/p>\n\n\n\n

Next, we calculate the volume of NaOH solution used in the titration by subtracting the initial buret reading from the final reading. This gives the volume in milliliters, which is converted to liters to match the units used in the molarity formula.<\/p>\n\n\n\n

Since the moles of NaOH equal the moles of KHP in a 1:1 ratio, we use this value and divide it by the volume of NaOH solution in liters. This yields the molarity of the NaOH solution. Molarity is defined as moles per liter of solution, and this approach directly follows from that definition.<\/p>\n\n\n\n

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

Calculating the Molarity of NaOH Solution To a very good approximation, at the endpoint (ep), the number of moles of KHP that have been neutralized is equal to the number of moles of NaOH that have been added (the equivalence point). To determine the molarity of the NaOH solution, make the ratio: moles NaOH added […]<\/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-26791","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/26791","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=26791"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/26791\/revisions"}],"predecessor-version":[{"id":26820,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/26791\/revisions\/26820"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=26791"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=26791"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=26791"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}