[Rachel.k]

Hi guys i have some prac work to hand in and i am stuck with 2 questions which i am not sure how to work out.

Question 1:

In a standardisation of a thiosulfate solution, excess potassium iodide was added to a 25.00mL aliquot of a solution prepared by dissolving 3.424g potassium iodate in sufficient water to make 1.000 litre. After acidification it required 23.70mL of the thiosulfate solution to reach the end point. What was the thiosulfate concentration?

This is what i did. Found molarity of potassium iodate first being 3.424g/214 = 0.016M  ( not sure i the mw is correct here or should i use 166 for potassium iodide)

the i found the moles being 0.016 x 25/1000 = 4.0 x 10e-4 moles

here is where i am stuck. What would the equation be for this. And what ratio do i use to calculate moles of thiosulfate then the molarity.

Question 2:

Reaction was: 2MnO4- + 5C6H5CH2OH + 6H+ = 2Mn2+ + 5 C6H5CHO + 8H2O

Step 3: Pipette 10ml of reaction mixture in an beaker and add 1 ml of KMnO3. Thus volume for each final reaction mix is 11/10


Conc of H+ before mixing = 0.05M
Conc of benzyl alcohol before mixiing = 6.00x 10e-2M

Value of kobs from ln(absorbance) vs time = -0.1896

What i have to do is calculate k from this equation

kobs= k(5C6H5CH2OH)e1(H+)e1

what is the difference between kobs and k. Is there any?

kobs is the observed k value. The reaction is first order.

I have made the graphs and i get a linear ln(abs) vs time. My slope is -0.1896.

rate of reaction is not given and is not actually needed. The equation is need to use is the one highlighted above. My answer comes to -76.61 . But i think i was told that kobs should be the same as k. Ohhhhhhhhhhhhhh

[Borek]

http://www.titrations.info/iodometric-titration