100mL of oxalic acid (H2C2O4) requires 35mL of 0.04 M KMnO4 to titrate it to the endpoint. calculate the molarity of the oxalic acid
5H2C2O4 + 2KMnO4 + 3H2SO4 ==> 10CO2 + 8H2O + 2MnSO4 + K2SO4
millimoles KMnO4 = mL x M = 35 mL x 0.04 M = 1.4
millimoles H2C2O4 = 1.4 mmols KMnO4 x (5 mols H2C2O4/2 mols KMnO4) = 1.4 x 5/2 = 3.5
Then M H2C2O4 = millimoles/mL = 3.5/100 = ? M
Well, if the oxalic acid played hide and seek with the KMnO4 and required 35mL to be found, then the molarity of the oxalic acid can be calculated using the equation:
M1V1 = M2V2
Let's plug in the values we have:
M1 * 100mL = 0.04M * 35mL
Simplifying the equation:
M1 = (0.04M * 35mL) / 100mL
Calculating further:
M1 = 0.014M
So, the molarity of oxalic acid is 0.014M. If oxalic acid were a detective, it would have certainly been found by now!
To find the molarity of the oxalic acid solution, we can use the equation:
Molarity (mol/L) = (Volume of KMnO4 (L) × Molarity of KMnO4 (mol/L)) / Volume of Oxalic Acid (L)
Let's substitute the given values into the equation:
Volume of KMnO4 = 35 mL = 35/1000 L = 0.035 L
Molarity of KMnO4 = 0.04 M
Volume of Oxalic Acid = 100 mL = 100/1000 L = 0.1 L
Now we can calculate the molarity of the oxalic acid:
Molarity = (0.035 L × 0.04 M) / 0.1 L
Molarity = 0.0014 mol / 0.1 L
Molarity = 0.014 M
Therefore, the molarity of the oxalic acid solution is 0.014 M.
To calculate the molarity of the oxalic acid (H2C2O4), we can use the balanced chemical equation and the concept of stoichiometry.
First, let's write the balanced chemical equation for the reaction between oxalic acid (H2C2O4) and potassium permanganate (KMnO4):
2 KMnO4 + 5 H2C2O4 + 3 H2SO4 --> K2SO4 + 2 MnSO4 + 10 CO2 + 8 H2O
Based on the reaction stoichiometry, we can observe that it takes 5 moles of oxalic acid (H2C2O4) to react with 2 moles of potassium permanganate (KMnO4). This means that the ratio between the moles of oxalic acid and the moles of KMnO4 is 5:2.
Given that 35 mL of 0.04 M KMnO4 was used, we need to calculate the number of moles of KMnO4 used:
Moles of KMnO4 = volume (L) * molarity (M)
= 0.035 L * 0.04 M
= 0.0014 moles
Now, using the 5:2 stoichiometric ratio, we can calculate the number of moles of oxalic acid used:
Moles of H2C2O4 = (2/5) * moles of KMnO4
= (2/5) * 0.0014 moles
= 0.00056 moles
Finally, to calculate the molarity of the oxalic acid, divide the moles of oxalic acid by the volume of the solution in liters:
Molarity of H2C2O4 = Moles of H2C2O4 / volume (L)
= 0.00056 moles / 0.100 L
= 0.0056 M
Therefore, the molarity of the oxalic acid solution is 0.0056 M.