How to solve for part 2 Laboratory Report Name: Date: Partner\'s Name: Instructo

Question

How to solve for part 2 Laboratory Report Name: Date: Partner's Name: Instructor's Initials: Molarity of Stock KMnO, solution: Molarity of diluted KMnO, solution: 0uil Part 1 Trial 1 Trial 2 Trial 3 Initial Buret Reading of KMnO4 Final Buret Reading of KMnO4 135 35 115 35 137.15 Volume of KMnO4 used for Titration mmoles of Mino: 00135 mmoles of C,O [C2042 ] [Ba2+] Ksp 3 mol Am Shqw your calculations for Trial I below 000 cm3 e from the solubility you found in 10000M um ulations for the Kap O Sh The CRC Handbook of Chemistry and Physic30003ool 50 gu3 PAET OF TRI AL 1 CACAOS

Explanation / Answer

Titration of KMnO4 with BaC2O4

2 moles of KMnO4 reacts with 5 moles of BaC2O4

So,

Trial 1,

volume of KMnO4 used = 7.35 ml

molarity of KMnO4 = 0.001 M

So,

moles of KMnO4 used = molarity x volume = 0.001 M x 7.35 ml

                                                                     = 0.00735 mmol

moles of BaC2O4 reacted = 0.00735 mmol x 5/2 = 0.0184 mmol

Volume of BaC2O4 taken initially = 50 ml

molarity of [Ba2+] = [C2O4^2-] = 0.0184 mmol/50 ml

                                                  = 0.000368 mol/L

Ksp = [Ba2+][C2O4^2-]

        = (0.000368)(0.000368)

        = 1.354 x 10^-7

Similarly calculations for other two trials can be done.

Related Questions

Navigate

• Browse (All)
• Browse H
• Subjects

---

---

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