Acid Ionization Lab Report PDF

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Acid Ionization Lab Report: Date: 19th April 2017

Name: Caitlin Bettenay

There is not a formal lab report for this lab. Complete the below pages and submit them to your TA before leaving lab. Briefly describe Ka. How does Ka relate to acid strength? Describe how you can use the pH of an aqueous acid solution and its initial concentration to determine the Ka. Ka, the acid ionization constant, is the equilibrium constant for chemical reactions involving weak acids in aqueous solution. The numerical value of Ka is used to predict the extent of acid dissociation. A large Ka value indicates a stronger acid (more of the acid dissociates) and a small Ka value indicates a weaker acid (less of the acid dissociates) (Libretexts, 2017). It is defined for the reaction below: HA(aq) + H2O(l)

H3O+(aq) + A–(aq)

With an equilibrium constant expression of:

Figure 1: Ka Equilibrium Expression A larger equilibrium constant means that the reaction has a greater tendency to drive towards the products side (more H3O+ and A–, less HA). Thus, the acid has a stronger tendency to donate its H+, which is the definition of a stronger acid. The pH and the initial concentration of a solution can be used in order to determine the Ka value. The Ka value can be found by first, writing the dissociation equation for the acid: HA ⇌ H+ + A¯. Then, by writing the equilibrium expression which can be seen from Figure 1 above. Then, solve for Ka by using the pH to calculate the [H+] value. From the dissociation equation, there is a 1:1 ratio between [H+] and [A-]: pH = -log [H+], therefore [H+] = 10¯pH and [A-] = 10¯pH Finally, [HA] is given and is the concentration, substitute the values to solve for Ka.

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Data: Insert a data table below that captures all of the relevant information. Make sure to include the concentrations of the HC2H3O2 solutions you made and the volume of the stock solution and water used to make the solutions. Include the pH of vinegar. Table 1: Table displaying raw data and calculations including the Assigned Concentrations of HC2H3O2 Solutions, Measured pH value, Ka Expression, Volume of 2 M Acetic Acid, [H+]eq, C2H3O2-]eq, [HC2H3O2]eq and finally the Ka Calculations. Assigned Concentration: Measured pH:

0.75 M

0.85

2.76

2.68

Ka Expression:

Volume of 2 M Acetic Acid: [H+]eq

+¿¿ H ¿ −¿ ¿ C2 H 3 O 2 ¿ ¿ K a=¿ 18.75 mL

+¿¿ H ¿ −¿ ¿ C2 H 3 O 2 ¿ ¿ K a=¿ 21.25 mL

1.7 X 10-3 M

1.4 X 10-3 M

[C2H3O2-]eq

1.7 X 10-3 M

1.4 X 10-3 M

[HC2H3O2]eq

0.7483 M

0.8486 M

Ka Calculation

K a=

( 1.7 ×10−3 ) ( 1.7 ×10−3) 0.7483 −6 K a=3.9 × 10

K a=

( 1.4 × 10−3 )( 1.4 ×10−3) 0.8486 −6 K a=2.3 × 10

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Results: 1. Report your calculated Ka value for the first concentration used. Include a copy of the ICE table.

2. Report your calculated Ka value for the second concentration used. Include a copy of the ICE table.

3. Report the class average Ka. 3

4. Compare the class average experimentally determined Ka with the accepted value at 25 °C (1.8 x 10-5) and calculate the percent error.

5. Comment on whether the initial HC2H3O2 concentration have any effect on Ka? Briefly explain. The initial concentration of HC2H3O2 has no effect on the Ka value, as long as outside conditions, such as temperature, are not altered. This is due to the fact that Ka is a constant that is independent of concentration. However, if the concentration were to increase, then that would result in making the solution more acidic.

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6. Calculate the percent ionization of acetic acid for each concentration you used. Percent Ionization Calculations:

7. What effect does initial HC2H3O2 concentration seem to have on the percent ionization? Percent ionization varies as a function of the HC2H3O concentration. Therefore, the fraction of molecules which ionize, depends on how many acid molecules there are per liter of solution. As some, but not all of the acid molecules are ionized, this means that HC2H3O2molecules are present in solution at the same time as the negative ions and the positive hydrogen ions.

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8. Calculate the concentration of acetic acid in the vinegar solution The pH recorded for the Vinegar solution was found to be 2.34pH as can be seen from the data Table 1. Furthermore, the average value of Ka between the first and second concentrations of HC2H3O2 will be used:

For Vinegar:

9. In the past, many students have listed that the accidental addition of too much acetic acid contributed greatly to the difference between the experimental value and the accepted value. Suppose that Student A was supposed to make a 0.18 M solution by diluting 9.0 mL of 2.0 M acetic acid to 100.0 mL. The expected pH for this solution is 2.74. The Ka of acetic acid is 1.8 x 10-5. a. What would be the expected pH if a Student A accidentally diluted 9.1 mL (instead of 9.0 mL) of the acid to 100 mL? If the initial concentration of the solution is higher than it’s supposed to be, then the concentration of hydrogen at the equilibrium is also going to be higher than what it is expected to be. Then, with a higher concentration of hydrogen, the pH value will be lower than expected, which would mean that the solution is more acidic than it should to be.

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M 1 V 1= M 2 V 2 M 1 × 100=2.0 × 9.1 M 1=0.182 M CALCUALTE : +¿¿ H ¿ Calculate ¿ +¿¿ H ² ¿ ¿ 1.8 ×10−5 =¿ H ¿ +¿ ¿ ¿ H ¿ +¿ ¿ ¿ H ¿ +¿ ¿ ¿ pH =−log(1.81 × 10−3 ) pH =2.74

b. If student A measured the above calculated pH, what would be the resultant Ka of acetic acid given that they expected the acid to have an initial concentration of 0.18 M? The calculated pH value is lower than it is expected to be. This means that the solution is more acidic than it is supposed to be. The stronger the acid, the larger the Kavalue is. Thus, the resultant Ka value is going to be larger than it should be.

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H ¿ +¿ ¿ ¿² ¿ ¿ K a=¿ −6 3 .276−10 K a= 0.18 −5 K a=1.82 ×10

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