The question is incomplete, the complete reaction equation is;
The concentration of a hydrogen peroxide solution can be determined by titration
with acidified potassium manganate(VII) solution. In this reaction the hydrogen
peroxide is oxidised to oxygen gas.
A 5.00 cm3 sample of the hydrogen peroxide solution was added to a volumetric flask
and made up to 250 cm3 of aqueous solution. A 25.0 cm3 sample of this diluted
solution was acidified and reacted completely with 24.35 cm3 of 0.0187 mol dm–3
potassium manganate(VII) solution.
Write an equation for the reaction between acidified potassium manganate(VII)
solution and hydrogen peroxide.
Use this equation and the results given to calculate a value for the concentration,
in mol dm–3, of the original hydrogen peroxide solution.
(If you have been unable to write an equation for this reaction you may assume that
3 mol of KMnO4 react with 7 mol of H2O2. This is not the correct reacting ratio.)
Answer:
2.275 M
Explanation:
The equation of the reaction is;
2 MnO4^-(aq) + 16 H^+(aq) + 5H2O2(aq) -------> 2Mn^+(aq) + 10H^+ (aq) + 8H2O(l)
Let;
CA= concentration of MnO4^- = 0.0187 mol dm–3
CB = concentration of H2O2 = ?
VA = volume of MnO4^- = 24.35 cm3
VB = volume of H2O2 = 25.0 cm3
NA = number of moles of MnO4^- = 2
NB = number of moles of H2O2 = 5
From;
CAVA/CBVB = NA/NB
CAVANB = CBVBNA
CB = CAVANB/VBNA
CB = 0.0187 * 24.35 * 5/25.0 * 2
CB = 0.0455 M
Since
C1V1 = C2V2
C1 = initial concentration of H2O2 solution = ?
V1 = initial volume of H2O2 solution = 5.0 cm3
C2 = final concentration of H2O2 solution= 0.0455 M
V2 = final volume of H2O2 solution = 250 cm3
C1 = C2V2/V1
C1 = 0.0455 * 250/5
C1 = 2.275 M
![pH=-log[H^{+}]=-log[H_{3}O]](https://tex.z-dn.net/?f=pH%3D-log%5BH%5E%7B%2B%7D%5D%3D-log%5BH_%7B3%7DO%5D%20%20)
. Using the value in formula, we get:![pH=-log[1.8*10^{-4}]=3.745](https://tex.z-dn.net/?f=pH%3D-log%5B1.8%2A10%5E%7B-4%7D%5D%3D3.745%20)
