I need help with my lab the whole thing please. i Dont know how to do the calcul

ID: 1043576 • Letter: I

Question

I need help with my lab the whole thing please. i Dont know how to do the calculations

ABSTRACT:

The aim of the experiment was to learn how successfully dilute stock (concentrated) solutions of 1.0 M NaOH

and 0.020 M KMnO4 solutions.

The general dilution equation is:

McVc = MdVd

In the equation above:

Mc is the molarity of the concentrated solution - was given

Vc is the volume of the concentrated solution - has to be calculated

Md is the molarity of the diluted solution - was given

Vd is the volume of the diluted solution - was given

To calculate the volume of the concentrated solution, we modify the equation above to the formula:

Vc =

In addition, we calculated the volume of water needed to dilute the solution completely according to formula

Vw =

In Part I of experiment, we tested pH of diluted NaOH solutions, which allow us to calculate molarity

of diluted solutions using following formula:

[OH-] =

We also need to learn the rules of Sig Fig in logarithms. Explain the rule in your worlds below:

In Part II of experiment, using a spectrometer we measured the absorbance of diluted KMnO4 solutions and we apply Beer’s Law to calculate molarity of diluted KMnO4.

Beer’s Law is : A = a*b*M

Where:

A is absorption of the solution (measured by us ) in nm

a is a is molar absorptivity coefficient = 2.24 x103 cm-1M-1

b is the path of light length in cuvette = 1.0 cm

M is the molarity of the sample M = mol/L

To calculate the molarity of the dilute potassium permanganate solutions we used following

M =

RESULTS: Table 8.1 Serial dilutions of NaOH solutions

Solution A:

was made by diluting 1.0 M NaOH. Required diluted molarity was 1.0 x10-1 M, required volume 25.0 mL

Solution B:

was made by diluting solution A. Required diluted molarity was 1.0 x10-3 M, required volume 20.0 mL

Solution C:

was made by diluting solution B. Required diluted molarity was 1.0 x10-5 M, required volume 20.0 mL

Solution

NaOH Mc

1.0 M

25.0 mL

1.0 X 10-1 M

20.0 mL

1.0 X 10-3 M

20.0 mL

1.0 X 10-5 M

1. For Table 8.1 show how did you calculate volume of concentrated NaOH solution and volume of water Vw, which has to be used in order to obtain requested Vd for the diluted solutions A,B,C. Fill empty boxes in a table.

Solution A

Solution B

Solution C

2. Calculate molarity of diluted NaOH solutions using your pH data. Show your calculations.

Report results in right SF as measured concentration (each wrong answer is -2 Pt)

Solution A: [OH-] =

Solution B: [OH-] =

Solution C: [OH-] =

3. Calculate % error for each solution using required molarity as accepted value.

Table 8.2 Serial dilutions of KMnO4 solutions

Please explain how you did serial dilutions for solutions D, E, F, G using example for solution D

Solution D

was made by diluting: 2.0 X 10-2 M KMnO4.Required diluted molarity was 2.0 x10-3 M, required volume 25.0 mL

Solution E

was made by: diluting

Solution F

was made by diluting:

Solution G

was made by diluting:

Solution

Absorbance

2.0 X 10-2 M

2.5 mL

22.5 mL

25.0 mL

2.0 X 10-3 M

2.0 X 10-4 M

1.0 X 10-4 M

1. For Table 8.1 show how did you calculate volume of concentrated KMnO4 solution and volume of water Vw, which has to be used in order to obtain requested Vd for the diluted solutions D,E,F,G. Fill empty boxes in a table.

Solution D

Vc =

Vw =

Solution E

Vc =

Vw =

Solution F

Vc =

Vw =

Solution G

Vc =

Vw =

1. Calculate molarity diluted KMnO4 Report results in right SF as measured molarity (wrong SF -2 pt)

D. M= A/ab =

E. M= A/ab =

F. M= A/ab =

G. M= A/ab =

2. Calculate % error for each solution using required molarity as accept value.

CONCLUSIONS:

PART I: Summary Table for dilutions of NaOH solution

Solution

Molarity

measured

Molarity

required

% Error

1.0x10-1

1.0x10-3

1.0x10-5

PART II: Summary Table for dilutions of KMnO4 solutions

Solution

Absorbance

Molarity

measured

Molarity required

% Error

2.0 X 10-3 M

2.0 X 10-4 M

1.0 X 10-4 M

5.0 X 10-5 M

Solution

NaOH Mc

1.0 M

2.5 mL 22.5 mL

25.0 mL

1.0 X 10-1 M

12.17

1.0 X 10-1 M

0.20 mL 19.8 mL

20.0 mL

1.0 X 10-3 M

9.60

1.0 X 10-3 M

0.20 mL 19.8 mL

20.0 mL

1.0 X 10-5 M

5.80

Explanation / Answer

Serial dilution of NaOH

Solution A,

Mc = 1.0 M Md = 1 x 10^-1 M

Vc = ? ml Vd = 25.0 ml

So,

Vc = Md.Vd/Mc = 1 x 10^-1 M x 25 ml/1.0 M = 2.5 ml

Vw = 25 - 2.5 = 22.5 ml

Solution B,

Mc = 1 x 10^-1 M Md = 1 x 10^-3 M

Vc = ? ml Vd = 20.0 ml

So,

Vc = Md.Vd/Mc = 1 x 10^-3 M x 20 ml/1 x 10^-1 M = 0.2 ml

Vw = 20 - 0.2 = 19.8 ml

----

Solution C,

Mc = 1 x 10^-3 M Md = 1 x 10^-5 M

Vc = ? ml Vd = 20.0 ml

So,

Vc = Md.Vd/Mc = 1 x 10^-5 M x 20 ml/1 x 10^-3 M = 0.2 ml

Vw = 20 - 0.2 = 19.8 ml

Solution A,

pH = 12.17

molarity [OH-] = inv.log[-(14 - 12.17)] = 0.015 M

%error = (0.015 - 0.01) x 100/0.01 = 50%

Solution B,

pH = 9.6

molarity [OH-] = inv.log[-(14 - 9.6)] = 4 x 10^-5 M

%error = (0.001 - 4 x 10^-5) x 100/0.001 = 96%

---

Solution C,

pH = 5.8

molarity [OH-] = inv.log[-(14 - 5.8)] = 6.31 x 10^-9 M

%error = (0.00001 - 6.31 x 10^-9) x 100/0.00001 = 99.94%

---

KMnO4 serial dilution

Table 8.1

Solution D,

Mc = 2 x 10^-2 M Md = 2 x 10^-3 M

Vc = ? ml Vd = 25.0 ml

So,

Vc = Md.Vd/Mc = 2 x 10^-3 M x 25 ml/2 x 10^-2 ml = 2.5 ml

Vw = 25 - 2.5 = 22.5 ml

Solution E,

Mc = 2 x 10^-3 M Md = 2 x 10^-4 M

Vc = ? ml Vd = 10.0 ml

So,

Vc = Md.Vd/Mc = 2 x 10^-4 M x 10 ml/2 x 10^-3 ml = 1.0 ml

Vw = 10 - 1.0 = 9.0 ml

Solution F,

Mc = 1 x 10^-4 M Md = 2 x 10^-4 M

Vc = ? ml Vd = 10.0 ml

So,

Vc = Md.Vd/Mc = 1 x 10^-4 M x 10 ml/2 x 10^-4 ml = 5.0 ml

Vw = 10 - 5.0 = 5.0 ml

Solution G,

Mc = 1 x 10^-4 M Md = 5 x 10^-5 M

Vc = ? ml Vd = 10.0 ml

So,

Vc = Md.Vd/Mc = 5 x 10^-5 M x 10 ml/1 x 10^-4 ml = 5.0 ml

Vw = 10 - 5.0 = 5.0 ml

1. molarity using absorbance

molarity = absorbance/2.24 x 10^3 M-1.cm-1 x 1.0 cm

For,

Solution D, molarity KMnO4 = 3.0/2.24 x 10^3 x 1.0 = 1.34 x 10^-3 M

%error = (2 x 10^-3 - 1.34 x 10^-3) x 100/2 x 10^-3 = 33.0%

Solution E, molarity KMnO4 = 0.384/2.24 x 10^3 x 1.0 = 1.70 x 10^-4 M

%error = (2 x 10^-4 - 1.70 x 10^-4) x 100/2 x 10^-4 = 13.0%

Solution F, molarity KMnO4 = 0.128/2.24 x 10^3 x 1.0 = 6.71 x 10^-5 M

%error = (1.0 x 10^-4 - 6.71 x 10^-5) x 100/1 x 10^-4 = 32.90%

Solution G, molarity KMnO4 = 0.053/2.24 x 10^3 x 1.0 = 2.37 x 10^-5 M

%error = (5.0 x 10^-5 - 2.37 x 10^-5) x 100/5 x 10^-5 = 52.6%

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