Based on equation:
2MnO4¬¬(-) + 6H(+) +5H2O2 → 2Mn(2+) +8H2O + 5O2
A 50.0 mL solution of hydrogen peroxide of unknown concentration was diluted with 1.00L of water. 25mL of this diluted solution was then titrated with acidified potassium permanganate; 20.25mL 0f 0.02 mol/L permanganate was needed to completely neutralise this sample. Fine the concentration of the original hydrogen peroxide solution in mol/L.
I don't go along with the "to completely neutralize this sample" since this is a redo reaction and not an acid/base reaction. But I know what you mean.
mols KMnO4 = M x L = ?
mols H2O2 = mols KMnO4 x (5/2) = ?
That = mols in the 25 mL H2O2 and that times 40 = mols H2O2 in the 1.00L and that's the same as the mols in the 50 mL of the original solution.
(H2O2) = mols/0.05 = ?
To find the concentration of the original hydrogen peroxide solution in mol/L, we can use the stoichiometry of the balanced equation and the titration data.
First, let's determine the moles of permanganate used in the titration:
Moles of permanganate = concentration × volume
= 0.02 mol/L × 20.25 mL
= 0.000405 mol
Next, we need to determine the moles of hydrogen peroxide reacted with the permanganate. From the balanced equation, we know that the ratio of permanganate to hydrogen peroxide is 1:5.
Moles of hydrogen peroxide = (0.000405 mol permanganate) × (5 mol hydrogen peroxide/1 mol permanganate)
= 0.002025 mol
Since we know that 25 mL of the diluted hydrogen peroxide solution was titrated, we can use this volume to find the volume of the original hydrogen peroxide solution:
Volume of original hydrogen peroxide solution = (25 mL) × (1000 mL/1 L) × (1.00 L/1000 mL)
= 0.025 L
Finally, we can calculate the concentration of the original hydrogen peroxide solution using the moles and volume:
Concentration = moles/volume
= 0.002025 mol/0.025 L
= 0.081 mol/L
Therefore, the concentration of the original hydrogen peroxide solution is approximately 0.081 mol/L.
To find the concentration of the original hydrogen peroxide solution, we can use the balanced equation and stoichiometry.
First, let's determine the number of moles of permanganate used in the titration. We are given that 20.25 mL of 0.02 mol/L permanganate was used. We can calculate the number of moles of permanganate used as follows:
Moles of permanganate = volume (in L) x concentration
= 0.02025 L x 0.02 mol/L
= 0.000405 mol
According to the balanced equation, the stoichiometric ratio between hydrogen peroxide and permanganate is 5:2. Therefore, the number of moles of hydrogen peroxide must be in the ratio 2:5.
Moles of hydrogen peroxide = (moles of permanganate x 5) / 2
= (0.000405 mol x 5) / 2
= 0.0010125 mol
Next, we need to calculate the number of moles of hydrogen peroxide in the 25 mL of diluted solution. Since the diluted solution was prepared by diluting the original solution 50 times, we can calculate the number of moles of hydrogen peroxide in the 25 mL as follows:
Moles of hydrogen peroxide in diluted solution = (moles of hydrogen peroxide x Dilution factor)
= 0.0010125 mol x (1/50)
= 0.00002025 mol
Now, we can find the concentration of the original hydrogen peroxide solution in mol/L by dividing the moles of hydrogen peroxide in the diluted solution by the volume of the diluted solution.
Concentration of original hydrogen peroxide solution = moles of hydrogen peroxide / volume of diluted solution
= 0.00002025 mol / 1.00 L
= 0.00002025 mol/L
Therefore, the concentration of the original hydrogen peroxide solution is 0.00002025 mol/L or 0.02025 mmol/L.