Question

b) If 28.93 mL of a potassium permanganate solution was used to titrate 25.00 mL of a sodium oxalate solution made by dissolving 1.0423 g of sodium oxalate in 250.0 mL of deionized water, calculate the molarity of the potassium permanganate.

Answer 1

Ansi- Griven that, the titration in between sodium onalate & potassium promoznati (KMn04) MnO4 + C2 042- - Mn 2+ + CO2 6 Now, we have to balance the above Emo Mn of + 8H7 - Mn 2+ + 4H20 - To balonec the Eam 6, add see on L.H.S. » MnO4 +677 +50 -> Mn2+ + 4H20 -☺. - Reduction Now (200² - 2002 + 200 ( added 200 on 0 RH.s. to balance the onidation Ean]. no. qesim a To balmes the Em B & @, ma Hiply Em withi' 2 Lo - maltiply sm 6 with 's' :2MM 0 + 16++106—2012++ 8H20 @→ 562042- -→ 10602 + 1908 2 mno+ 562042 +ight 2012+ + 10 CO2 +8H20 . Above Ean 6 is a balmeed Redox Reaction. Nous we have to calculate the noig moes con molarity q Naz 64 Given that, 1.0423 g g Na, C204 was dissolved in 250 mL q water. .: The molauity q sodium calate is 1.04239 1000 4 p : Melecular weight - 134 g/mol. 288me of Naz 2204 134glmal] = 0.0311M, q Na₂C204

Now, we need to calculate the molarity of KMnogin no. of moles q N , 2204 went into titration = 0.0311 moles Ie x 0.0254 C::25ML = 0.0252]. 1. = 7.75 x107 mcles of Na, C204 To determine the no. g moes a KMnog ender goes ftatim, We have to use no. q moles a Na₂CO 4 & balomeed Reactim. Since, H₂SO4 was added to Na, C4. oxalate is im acidie form. . 2 moles of Mnogo ². 7.75 110 moles q N92204 s moles q Ha C₂04 - 3.1 X104 mdles q Mn04. 4. 3.1810-4 moles o km nog was divided by 28.93 m to get the modacity oq kmnos ..3.18104 moles oq kmnoq 28.93810-3 = 0.107 M.q kmno4 i The Molarity q KMnO4 = 0.107 M.