Experiment Results: Mixing Solution A with Solution B

In summary, the data suggests that the rate of reaction increases as the concentration of potassium iodate increases.
  • #1
Roxy
52
0
We did an experiment where we mixed Solution A with Solution B and found out the time. These are my results:

Solution A (0.020 mol/L potassium iodate solution plus distilled water)
10 drops potassium iodate plus 0 drops distilled water
9 drops potassium iodate plus 1 drops distilled water
8 drops potassium iodate plus 2 drops distilled water
7 drops potassium iodate plus 3 drops distilled water

Solution B (0.001 mol/L sodium bisulfate/ hydrochloric acid/starch)
10 drops
10 drops
10 drops
10 drops

Time is took to turn blue (s)
9
13
10
9

I need help with these questions:

1. How do I find concentration of potassium iodate? (do I divide 0.020 mol/L by something?)

2. And to find order of reaction with respect to potassium iodate I graph 1/time vs. concentration and square or cube it until I get a straight line right? But why do I want a straight line?
 
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  • #2
I would imagine that you would have been provided with specific instructions for this lab.
It'll probably be best to refer to your manual.

1. volume of potassium iodatexmolarity/total volume=concentration

2.1/time=the rate in this case, find the average slope value.
 
  • #3
Roxy said:
We did an experiment where we mixed Solution A with Solution B and found out the time. These are my results:

Solution A (0.020 mol/L potassium iodate solution plus distilled water)
10 drops potassium iodate plus 0 drops distilled water
9 drops potassium iodate plus 1 drops distilled water
8 drops potassium iodate plus 2 drops distilled water
7 drops potassium iodate plus 3 drops distilled water

Solution B (0.001 mol/L sodium bisulfate/ hydrochloric acid/starch)
10 drops
10 drops
10 drops
10 drops

Time is took to turn blue (s)
9
13
10
9

I need help with these questions:

1. How do I find concentration of potassium iodate? (do I divide 0.020 mol/L by something?)

Consider the case of 8 drops of iodate with 2 drops of water. Assume all drops have the same volume V.

# moles of iodate in 8 drops = molarity * volume of 8 drops = 0.02 * 8V.

After this has been mixed with 2 drops of water, the total volume becomes 10V.

Now concentration (molarity) = # moles/total volume = 0.02 *8V / 10V = 0.016 mol/L

2. And to find order of reaction with respect to potassium iodate I graph 1/time vs. concentration and square or cube it until I get a straight line right? But why do I want a straight line?
Write down the rate equation for a general n'th order reaction, and you will see why.
 
  • #4
rate equation is:

r= k[x]^n ...I don't get it

for this lab its

r = k[IO3]^2 right?
 
  • #5
The rate r is by definition = [itex]-d[x]/dt[/itex]. Substitute this above, and solve the reslting differential equation.
 
  • #6
Well, observe the experiment data, what happens to the rate when you double any of the reactants? If you wish to be sure calculate the slope of rate vs. concentration, for particular increasing/change in concentration (rate as the numerator). If the rate doubles, then you've got a first order relation, etc...
 

Related to Experiment Results: Mixing Solution A with Solution B

1. What were the results of mixing Solution A with Solution B?

The results of mixing Solution A with Solution B showed that a chemical reaction occurred, resulting in a new solution with different properties than the individual solutions.

2. How did the mixture of Solution A and Solution B compare to the individual solutions?

The mixture of Solution A and Solution B had different properties than the individual solutions, including a change in color, temperature, and chemical composition.

3. What factors can affect the outcome of mixing Solution A with Solution B?

The outcome of mixing Solution A with Solution B can be affected by various factors such as the concentrations of the solutions, temperature, and the presence of catalysts or inhibitors.

4. Can the experiment results be replicated?

Yes, the experiment results can be replicated by following the same procedures and using the same concentrations of Solution A and Solution B.

5. How can the results of this experiment be applied in real-world situations?

The results of this experiment can be applied in various industries, such as pharmaceuticals, agriculture, and manufacturing, to create new products or improve existing ones. It can also help scientists understand the properties and behaviors of different chemical substances.

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