Revision questions.. please help (1 Viewer)

[Cattle]

hey everyone i have been given a heap of revision questions to do but i dont understand half of them! any help would be greatly appreciated

1) a 0.20 mol/L solution of hydrogne peroxide in water is stable at low temperatures and in the absence of light. Iodide ions are believed to catalyse the decomposition of hydrogen peroxide by the following stepwise reaction
1- H2O2 + I -> H2O + IO
2- IO + H2OH -> H2O + O2 +I
a) write a balnced equation for the overall decomposition reaction
b) what observation would a student make if a few drops of sodium iodide solution were added to the hydrogne peroxide solution
c) what would the student have to demonstrate in order to confirm that the iodide ion was acting catalytically?


2)pure rubidium metal (0.85g) was allowed to react with water. when all reaction had finished the resulting solution was made up to a volume of 500ml with distilled water.
a)write a net ionic equation for the reaction of rubidium with water
b)given that rubidium hydroxide is a strong electrolyte, calculate the pH of the final solution
c) 25.0ml of the rubidium hydroxide solution were titrated with 0.01mol/L of acetic acid
i)write a balanced equation for the neutralisation reaction
ii) what indicator should be used
iii) calculate the volume of acetic acid that must be added to reach the equivalence point

 

[theworldisround]

Note: most of the questions here are out of the syllabus

1. a) Add equations one and two yields
2H2O2 +I +IO --> 2H2O + IO + I + O2
Cancelling out the I and IO on both sides lead to the net equation
2H2O2 --> 2H2O + O2

b) The student should observe the release of oxygen gas. This can be detected by an oxygen meter.

c)Absolutely no idea - but i'm sure that is NOT in the syllabus

2. a) 2Rb + 2H2O --> 2RbOH + H2

b) 2 moles of Rb produce 2 moles of RbOH from equation. No. of moles of Rb = 0.85 / 85.47 = 9.95 * 10^-3 moles
Cocentration = moles / volume = 9.95*10^-3 / 0.5 = 0.0198 M

pOH = -log 0.0198 = 1.7
Therefore pH = 12.3

c) CH3COOH + RbOH --> RbCH3COO + H2O
acetic acid + rubidium hydroxide --> rubidium acetate + water

Indicator used should be methyl orange

One mole of acetic acid reacts with one mole of rubidium hydroxide
We worked out from 2(b) that concentration of rubidium hydroxide was 0.0198M

Therefore 0.0198 * 0.025 = 0.01 * x (equating the moles)
x = 0.049 L = 49.5 mL