Question

Please fill in the blank tables below, given the following information.

Experiment 1: Measure the pH of Acids

1. Take seven small, clean test tubes from the Containers shelf and place them on the workbench.

2. Double-click on the test tubes and label them with numbers 1 – 7.

3. Take 0.1 M hydrochloric acid from the Materials shelf and add 5 mL to test tube 1.

4. Take water from the Materials shelf and add 5 mL to test tube 1.

5. Take water from the Materials shelf and add 9 mL to test tubes 2 – 7.

6. Create a series of successively diluted acidic solutions as follows:

   1. Pour 1 mL from 1 into 2.

   2. Pour 1 mL from 2 into 3.

   3. Pour 1 mL from 3 into 4.

   4. Pour 1 mL from 4 into 5.

   5. Pour 1 mL from 5 into 6.

   6. Pour 1 mL from 6 into 7.

7. Take bromothymol blue from the Materials shelf and add 0.1 mL to each test tube. Observe the color of the solutions. Record your results to reference later.

8. Clear your station by dragging all of the test tubes to the recycling bin beneath the workbench.

9. Repeat steps 1 – 6.

10. Take methyl yellow from the Materials shelf and add 0.1 mL to each test tube. Observe the color of the solutions. Record your results.

11. Clear your station by emptying the test tubes into the waste, then placing the test tubes in the sink.

12. Repeat steps 1 – 6.

13. Take bromocresol green from the Materials shelf and add 0.1 mL to each test tube. Observe the color of the solutions. Record your results.

Experiment 2: Measure the pH of Bases

1. Take seven small, clean test tubes from the Containers shelf and place them on the workbench.

2. Double-click on the test tubes and label them with numbers 1 – 7.

3. Take 0.100 M sodium hydroxide from the Materials shelf and add 5 mL to test tube 1.

4. Take water from the Materials shelf and add 5 mL to test tube 1.

5. Take water from the Materials shelf and add 9 mL to test tubes 2 – 7.

6. Create a series of successively diluted basic solutions using the same dilution method you did in experiment 1:

   1. Pour 1 mL from 1 into 2.

   2. Pour 1 mL from 2 into 3.

   3. Pour 1 mL from 3 into 4.

   4. Pour 1 mL from 4 into 5.

   5. Pour 1 mL from 5 into 6.

   6. Pour 1 mL from 6 into 7.

7. Take bromothymol blue from the Materials shelf and add 0.1 mL to each test tube. Observe the color of the solutions. Record your results.

8. Clear your station by emptying the test tubes into the waste, then placing the test tubes in the sink.

9. Repeat steps 1 – 6.

10. Take alizarin yellow from the materials shelf and add 0.1 mL to each test tube. Observe the color of the solutions. Record your results.

11. Clear your station by emptying the test tubes into the waste, then placing the test tubes in the sink.

12. Repeat steps 1 – 6.

13. Take phenolphthalein and add 0.1 mL to each test tube. Observe the color of the solutions. Record your results.

Experiment 1 Data:

Test Tube #	Bromothymol Blue Color	Methyl Yellow Color	Bromocresol Green Color
1	yellow	red	yellow
2	yellow	red	yellow
3	yellow	orange	yellow
4	yellow	yellow	green
5	yellow	yellow	blue
6	green	yellow	blue
7	green	yellow	blue

EXPERIMENT 1: For each test tube, calculate the concentration of H₃O⁺ and pH. You can use the following formula to determine the concentration of HCl. M1×V1=M2×V2 where M₁ and V₁ are the molarity and volume of the first solution, respectively, and M₂ and V₂ are the molarity and volume of the second solution, respectively. Given that HCl is a strong acid, the H₃O⁺ concentration is equal to the HCl concentration, except at very low concentrations (test tubes 6 and 7) where the H₃O⁺ from the dissociation of water (1.00×10−7) becomes significant.

Test Tube #	HCl Concentration	[H3O+]	pH
1
2
3
4
5
6
7

Experiment 2 Data:

Test Tube #	Bromothymol Blue Color	Alizarin Yellow Color	Phenolphthalein Color
1	dark blue/violet	red	pink
2	dark blue/violet	orange	pink
3	dark blue/violet	orange	pink
4	dark blue/violet	yellow	light pink
5	dark blue/violet	yellow	light pink
6	dark blue/violet	yellow	grey
7	green	yellow	grey

EXPERIMENT 2: For each test tube, calculate the concentration of NaOH, OH^–, H₃O⁺, and pH. You can use the following formula to determine the concentration of NaOH. M1×V1=M2×V2 where M₁ and V₁ are the molarity and volume of the first solution, respectively, and M₂ and V₂ are the molarity and volume of the second solution, respectively. Given that NaOH is a strong base, the HO^– concentration is equal to the NaOH concentration, except at very low concentrations (test tubes 6 and 7) where the HO^– from the dissociation of water (1.00×10−7) becomes significant.

Test Tube #	NaOH Concentration	[OH-]	[H3O+]	pH
1
2
3
4
5
6
7

Answer 1

Experiment 1:

The test tube 1 serves as the stock solution for dilution to test tube 2.

The test tube 2 serves as the stock solution for dilution to test tube 3 and so on..

Test tube 1 is 2 times dilution of 0.1M HCl . Hence the concentration is 0.05 M = 5 x 10^-2 M

The test tube 2 is 10 times dilution of test tube 1. Hence the concentration is 0.005 M = 5 x 10^-3 M

The test tube 3 is 10 times dilution of test tube 2. Hence the concentration is 5 x 10^-4 M

The test tube 4 is 10 times dilution of test tube 3, Hence the concentration is 5 x 10^-5 M

The test tube 5 is 10 times dilution of test tube 4. Hence the concentration is 5 x 10^-6 M

The test tube 6 is 10 times dilution of test tube 5. Hence the concentration is 5 x 10^-7 M

The test tube 7 is 10 times dilution of test tube 6. Hence the concentration is 5 x 10^-8 M

Thus

pH of test tube 1 = - log 5 x 10^-2 = 1.3

pH of test tube 2 = - log 5 x 10^-3 = 2.3

pH of test tube 3 = - log 5 x 10^-4 = 3.3

pH of test tube 4 = - log 5 x 10^-5 = 4.3

pH of test tube 5 = - log 5 x 10^-6 = 5.3

pH of test tube 6 :

Here we cannot calculate as - log 5 x 10^-7 which is wrong as we have to consider the dissociation of water.

5 x 10^-7 M HCl dissociates as follows.

5 x 10^-7 HCl + 5 x 10^-7 H2O 5 x 10^-7 H3O+ + 5 x 10^-7 Cl-

              H2O + H2O H3O + + OH-

                                       x           x

Let x moles of H3O+ dissociated from H2O. Therefore the total H3O+ = [ x + 5 x 10^-7]

                                                         and the [OH-] = x

We know that [H3O+] [ OH-] = 10^-14 , the Kw

[x+ 5 x 10^-7] [x] = 10^-14

x^2 + 5 x 10^-7 x - 10^-14 = 0

The above is a quadratic equation, solving for x we have x = 0.1926 x 10^-7 M

Therefore the [H3O+] = x + 5 x 10^-7 = 5.1926 x 10^-7 M

Hence the pH = - log 5.1926 x 10^-7 = 6.28

similarly

pH of test tube 7 :

Here we cannot calculate as - log 5 x 10^-8 which is wrong as we have to consider the dissociation of water.

5 x 10^-8 M HCl dissociates as follows.

5 x 10^-8 HCl + 5 x 10^-8 H2O 5 x 10^-8 H3O+ + 5 x 10^-8 Cl-

              H2O + H2O H3O + + OH-

                                       x           x

Let x moles of H3O+ dissociated from H2O. Therefore the total H3O+ = [ x + 5 x 10^-8]

                                                         and the [OH-] = x

We know that [H3O+] [ OH-] = 10^-14 , the Kw

[x+ 5 x 10^-8] [x] = 10^-14

x^2 + 5 x 10^-8 x - 10^-14 = 0

The above is a quadratic equation, solving for x we have x = 7.808 x 10^-8 M

Therefore the [H3O+] = x + 5 x 10^-7 = 12.808 x 10^-8 M

                               = 1.2808 x 10^-7

Hence the pH = - log 1.2808 x 10^-7 = 6. 89

The above results are tabulated here:

Experimnet 2:

similar to the experiment 1

The concentrations of NaOH will be the same as the same dilution followed.

Test tube 1 we have NaOH conc 5 x 10^-2

Test tube 2 we have NaOH conc 5 x 10^-3 and so on.

Similarly the pOH of test tube 1 = 1.3

                   pOH of test tube 2 = 2.3

                   pOH of test tube 3 = 3.3

                   pOH of test tube 4 = 4.3

                   pOH of test tube 5 = 5.3

                   pOH of test tube 6 = 6.28 [ calculated by considering H2O dissociation]

                   pOH of testt ube 7 = 6.89

pH is then calculated by pH = 14 - pOH.

The results are given below: