Method 1: Titration with Potassium Permanganate 5 H2O2 + 2 MnO4- + 6 H+ --> 8 H2
ID: 880629 • Letter: M
Question
Method 1: Titration with Potassium Permanganate
5 H2O2 + 2 MnO4- + 6 H+ --> 8 H2O + 5 O2 + 2 Mn2+
Method 2: Catalytic Decomposition to Water and Oxygen
2 H2O2 --> 2 H2O + O2
Method 3: Oxidation of Iodide to Iodine; Spectroscopic measurement of TriIodide ion.
IO3- + 5 I- + 6 H+ --> 3 H2O + 3 I2 (standard)
H2O2 + 2 I- + 2 H+ --> 2 H2O + I2 (unknown)
I- + I2 --> I3-
Method 1:
You weigh out Potassium Permanganate,
Mass of KMnO4 (MW=158.034).......... 0.1736 g
You dissolve it completely in water.......... 46.00 mL
(a) Molarity of MnO4- solution........................... ________________ M
You pipet out your UNKNOWN............. 0.90 mL
You acidify and titrate it with your KMnO4 solution until the permanganate color persists.
Volume of permanganate solution.......... 11.39 mL
(b) Molarity of the HOOH in the UNKNOWN...................................._____________ M
Method 2:
You pipet your UNKNOWN into a reaction tube.... 4.40 mL
Add a "boat" containing Manganese Dioxide, MnO2
Set up the gas collection apparatus shown on the right.
You sink the boat and allow the oxygen generated to collect in the collection flask. The water temperature is................. 27 oC
The Barometric Pressure is............... 761.9 mm Hg
Weight of the collection Flask with remaining water.... 154.81 g.
Weight of collection flask filled with water........... 200.70 g.
DensityH2O ................................. 0.996 g/cc
(c) Partial Pressure of Oxygen collected................................. ____________ mm Hg
(d) Volume of oxygen collected ...................................... ____________ mL
(e) Millimoles of oxygen gas generated.................................. ____________ mmol
(f) Molarity of Hydrogen Peroxide in UNKNOWN.......................... ____________ M
Method 3:
STANDARD: Put into a clean 250 mL volumetric flask exactly ..... 1.00 ml
of 0.95 mM iodate STANDARD.
Add an excess of 1M KI and 1M Sulfuric Acid and let stand for 5 minutes. Dilute to volume.
UNKNOWN: Pipet into a clean 250 mL volumetric flask 1.00 mL of your UNKNOWN and dilute to volume with distilled water.
Put into a clean 250 mL volumetric flask exactly ........ 1.00 mL
of this diluted H2O2 UNKNOWN.Add an excess of 1M KI and 1M Sulfuric Acid and let stand for 5 minutes. Dilute to volume (250 mL). Measure STANDARD and UNKNOWN spectroscopically at 360 nm using the same cuvet throughout.
% Transmission of BLANK.................................. 98.1 %
% Transmission of STANDARD................................ 48.5 %
% Transmission of UNKNOWN................................. 34.7 %
(g) Absorbance of STANDARD.................................... ___________________
(h) Absorbance of UNKNOWN.................................... ___________________
(i) Molarity of original UNKNOWN................................. ___________________ M
Explanation / Answer
Method - 1:
(a) Mass of KMnO4 = 0.1736 g
Molecular mass of KMnO4 = 158.034 gmol-1
Moles of KMnO4 = (Mass of KMnO4) / (Molecular mass of KMnO4) = 0.1736 g / 158.034 gmol-1 = 0.001098 mol
Volume of water = 46.00 mL = 46.00 mL * (1L / 1000mL) = 0.04600 L
Hence concentration of KMnO4 solution
= (Moles of KMnO4 ) / Volume = 0.001098 mol / 0.04600 L = 0.02387 M (answer)
(b) Given the volume of the unkown solution = 0.90 mL = 0.90 mL * (1L / 1000mL) = 0.00090 L
Volume of KMnO4 required to titrate 0.90mL unknown solution = 11.39 mL = 11.39 mL * (1L / 1000mL) = 0.01139L
Concentration of KMnO4 solution as calculated in (a) = 0.02387M
Hence moles of KMnO4 in 0.01139L KMnO4 solution = M*V = 0.02387M* 0.01139L = 2.719*10-4 mol
In the given balanced chemical reaction
5 H2O2 + 2 MnO4- + 6 H+ ---> 8 H2O + 5 O2 + 2 Mn2+
5 mol 2 mol
2 moles of KMnO4 reacts with 5 moles of H2O2.
Hence 2.719*10-4 mol of KMnO4 that will react with the moles of H2O2
= (5 moles of H2O2 / 2 moles of KMnO4) *(2.719*10-4 mol of KMnO4 ) = (5/2)*(2.719*10-4 mol H2O2
= 6.797*10-4 mol H2O2
Hence molarity of H2O2 in the unknown solution = moles of H2O2 / volume of unknown solution
= 6.797*10-4 mol H2O2 / 0.00090L = 0.7553 M (answer)