Question

While doing titration, a student recorded a burette reading of 10.0 mL for the neutralization of 10.0 mL NaHC2O4 (aq) with 0.1 M NaOH (aq). In a separate experiment, 10.0 mL of this NaHC2O4 (aq) solution could be completely oxidized by 10.0 mL of KMnO4 in an acidic medium. What would be the molarity of KMnO4 used by this student?

(a) 0.02 M
(b) 0.04 M
(c) 0.1 M
(d) 0.2 M

Answer 1

The molarity of KMnO4 used by the student is 0.1 M.

Step-by-step explanation:

To find the molarity of KMnO4 used in the separate experiment, we can use the concept of stoichiometry. Given that the volume of NaHC2O4(aq) solution is 10.0 mL and that it could be completely oxidized by 10.0 mL of KMnO4(aq) in an acidic medium, we can assume that the ratio of the two solutions is 1:1.

Therefore, the molarity of KMnO4 can be calculated using the formula:

Molarity (KMnO4) = (Volume of KMnO4 used / Volume of NaHC2O4 used) * Molarity of NaHC2O4

Plugging in the values, we get:

Molarity (KMnO4) = (10.0 mL / 10.0 mL) * 0.1 M = 0.1 M

Therefore, the molarity of KMnO4 used by this student would be 0.1 M.