..ooo Verizon 6:25 PM 8396 Back Titration of 7UP post lab.docx Titration of 7UP

Question

..ooo Verizon 6:25 PM 8396 Back Titration of 7UP post lab.docx Titration of 7UP Alicia Palermo Lab Partners: Elizabeth Alamillo, Michael Avery, Bain Caton, Garmikly LaFrance TA: Kyle Flanagan Section:602 June 20th, 2016 Data Reaction equations HCI and NaOH: HCI+NaOH à H2O+NaCl H3PO4 and NaOH: H3PO4+NaOH à NaH2PO4+H20 H3PO4+2NaOH à Na2HPO4+2H20 C6H807 and NaOH: C6H807+NaOH à C6H7Na07+H20 C6H7Na07+NaOH à C6H62Na07+H20 C6H62Na07+NaOH à C6H63NaO07+H20 Trial 1 Trial 2 Mass of HCI 2.5x104 2.5x104 (moles) Initial burette11715 reading of NaOH mL) Final burette reading of NaOH (mL) 43.1 .14x10-37.04x10-3 NaOH 8.41x10-3 8.38x10-3 Avg [NaOH H3PO4 7 .5x 10-362 x 10-83.6 x 10-13

Explanation / Answer

From the graphs we can get the pH values for first and second end points of the titration,the points were sudden change in pH is seen is the inflection point.the pH corresponding to the point is taken

From graph of H3PO4 titration,

First end point at 8.5 ml ,corresponding pH=5

pH=-log[H3O+]

[H3O+]=10^-pH=10^-5

[OH-]=kw/[H3O+}=10^-14/10^-5=10^-9

Second end point at 17ml ,pH=9

[H3O+]=10^-pH=10^-9

[OH_}=10^-14/10^-9=10^-5

Third end point,pH=12

[OH=]=10^-14/10^-12=10^-2 M

[OH-]=10^-2 M

[H3PO4]=3[OH-]=3*10^-2M

Ka=[Na3PO4]/[OH-]^3[H3PO4]

From 3 NaOH +H3PO4<-->Na3PO4+H2O

Ka=ka1*ka2*ka3=1.7*10^-22

[H3PO4]=3*10^-2M

From citric acid titration graph,

First end point, pH=4

Second end point pH=5.6

Third end point=9

[H3O+]=10^-9=

[OH-]=10^-14/10[^-9=10^-5

[citic acid]=3*10^-5 M

From 3NaOH+H3C2O7 <---.Na3C2O7+H2O

In 240 ml=0.24 L  moles of citric acid=0.24*3*10^-5 moles=0.24*3*10^-5 mole*192.12g/mol=138.33*10^-5 g=138.33 *10^-5 *10^3 mg=1.38 mg

Related Questions

Navigate

• Browse (All)
• Browse #
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.