buret and with A solution of KMnO4 of unknown concentration is poured into a sta

Question

buret and with A solution of KMnO4 of unknown concentration is poured into a standardized oven-dried potassium oxalate ( molar mass 166.0) according to the following equation: solid, 16 H"(aq) + 2 Mno;(aq) + 5 KrGo,(s)--2 Mn"(aq) + 10 ??,(g) + 8 H2O + 10K+(aq) The solid KC,0, is weighed, dissolved in 25 mL of water, and 5 mL of H SO, is added to acidify the solution in the flask. The solution is then titrated in 3 trials with the KMnO4 from the buret until a persistent pale color remains. The data table for the 3 different trials is below. Volume of KMnO, solution Mass of solid K,C20 0.250 g 0.250 g 0.250 g Trial 31.15 mL 31.05 mL 31.05 mL (a) Is the C,0, ion undergoing oxidation or reduction? Explain. (b) Calculate the moles of K,C,0, reacted in Trial 2 (c) Calculate the molar concentration (M) of the KMnO, solution. Please show your work (d) Which experimental error below could account for the different volume of KMnO4 solution in Trial 1? Explain. i. The buret is rinsed with distilled H2O but not with the KMnO4, solution ii. Some solid K,C,04 remains in the "weighing boat."

Explanation / Answer

By using the equation M1V1/M2V2 = 2/5

M1 (molarity of K­­MnO4) = (0.050 M x 30 mL x 2) / (31.05 mL x 5) = 0.019 M

The molarity of KMnO4 solution is 0.019 M

i. The buret is rinsed with distilled water, so some leftover water on the buret wall, diluted the concentration of KMnO4, which leads in the increased volume in first titration, the second and third becomes accurate due to rinsing effect of first titration.

ii. It is not due to the weighing and transfer error, in that case the volume of KMnO4 will be less than trial 2 and trial 3 as the amount of oxalate is less.

Related Questions

Navigate

• Browse (All)
• Browse B
• Subjects

---

---

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