A 25 ml solution containing oxalate, C204²‐ was titrated with KMnO4 solution in acidic medium. C2O4²- ions were oxidized to CO2 and KMnO4 ions were reduced to Mn2+ ions. If 30.50 ml of 1.25M KMnO4 solution is required for the solution. Calculate the concentration of the solution containing C2O4²- ions.
2[MnO4]^- + 5[C2O4]^2- + 16H^+ ===> 2Mn^2+ + 10CO2 + 8H2O
moles KMnO4 = M x L = 1.25 M x 0.03050 L = 0.03813
mols C2O4^2- = 0.03813 moles MnO4^2- x (5 mols C2O4^2-/2 mols MnO4^-) = 0.03813 x 5/2 = 0.09531, then
(C2O4^2-) = mols/L = 0.09531/0.025 = ? M
Check my arithmetic.
To calculate the concentration of the solution containing C2O4²- ions, we can use the concept of stoichiometry and the balanced chemical equation of the reaction between oxalate (C2O4²-) and KMnO4.
The balanced chemical equation is:
5C2O4²- + 2MnO4- + 16H+ -> 10CO2 + 2Mn²+ + 8H2O
From the equation, we can see that 5 moles of C2O4²- react with 2 moles of MnO4-. Therefore, the ratio of C2O4²- to MnO4- is 5:2.
Given that 30.50 ml of 1.25 M KMnO4 solution is required, we can calculate the number of moles of KMnO4 used in the titration:
moles of KMnO4 = Molarity × Volume
moles of KMnO4 = 1.25 M × 0.03050 L = 0.038125 mol
According to the stoichiometry, the number of moles of C2O4²- in the solution is half the number of moles of KMnO4 used:
moles of C2O4²- = (0.038125 mol) × (5/2) = 0.0953125 mol
Now, we can calculate the concentration of the solution containing C2O4²- ions:
Concentration of C2O4²- = moles of C2O4²- / Volume of Solution
Concentration of C2O4²- = 0.0953125 mol / 0.025 L = 3.8125 M
Therefore, the concentration of the solution containing C2O4²- ions is 3.8125 M.
To calculate the concentration of the solution containing C2O4²- ions, we can use the concept of stoichiometry and the balanced chemical equation.
The balanced chemical equation for the reaction between oxalate (C2O4²-) and KMnO4 in acidic medium is:
5C2O4²- + 2MnO4- + 16H+ → 10CO2 + 2Mn²+ + 8H2O
From the balanced equation, we can see that 5 moles of C2O4²- ions react with 2 moles of KMnO4.
Given the volume and concentration of KMnO4 solution used, we can determine the number of moles of KMnO4 used.
Volume of KMnO4 solution used = 30.50 ml = 0.03050 L
Concentration of KMnO4 solution = 1.25 M
Number of moles of KMnO4 = Volume × Concentration = 0.03050 L × 1.25 M = 0.038125 moles
Since the stoichiometry relationship indicates that 5 moles of C2O4²- ions react with 2 moles of KMnO4, we can calculate the number of moles of C2O4²- ions present in the solution as follows:
Number of moles of C2O4²- ions = (Number of moles of KMnO4 × 5) / 2
= (0.038125 × 5) / 2
= 0.0953125 moles
Finally, to determine the concentration of the solution containing C2O4²- ions, we divide the number of moles by the volume of the solution:
Concentration = Number of moles / Volume
= 0.0953125 moles / 0.025 L
= 3.8125 M
Therefore, the concentration of the solution containing C2O4²- ions is 3.8125 M.