Buffers Lab 4 - Lab 4 PDF

Preview

Summary

Lab 4...

Description

Buffers Lab Report: Name_Jesus Quintero___________

Date__October 20, 2020

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 what a buffer is. Include the relevant conceptual background along with balanced chemical reactions to show how buffers react with a strong acid or a strong base. A buffer is a water solution consisting of a weak acid and a salt of its conjugate base (or vice versa) in approximately equal concentration. A buffer is a solution that will resist a change in pH when a strong acid or base is added. Initial balanced reaction: HA (aq) + H2O (l) A- (aq) + H3O+ (aq) Reaction when strong acid is added: A- (aq) + H+ (aq)  HA (aq) Buffers work by applying Le Chatelier’s Principle to weak acid equilibrium. The buffer contains significant amounts of the conjugate base anions [A -], these ions combine with added acid to make more [HA] and keep H3O+ approximately constant.

Results and Calculations: Table 1: Addition of 0.20 M HCl to Water Volume (mL) of 0.20 M HCl added to 50 mL of water

Calculat ed pH

Measure d pH

7.00

6.50

3.70

3.35

3.40

2.67

3.22

2.40

3.10

2.34

2.70

2.24

0.0 1.0

Sample Calculation

x mol =0.20 M HCl 0.001 L 0.001

L∗0.20 moles HCl =0.0002 moles HCl 1L

2.0 pH =− log ( 0.0002 ) =3.70

3.0 4.0 5.0

Table 2: Addition of 0.20 M NaOH to Water Volume (mL)

of 0.20 M NaOH added to 50 mL of water

Calculat ed pH

Measure d pH

7.00

5.17

10.30

10.90

0.0 1.0

Sample Calculation

x mol =0.20 M NaOH 0.001 L 0.001

L∗0.20 moles NaOH 0.0002 moles NaOH 1L

10.60

11.80

10.78

11.99

pOH=−log ( 0.0002)=3.70

10.9

12.15

pH =14.00−3.70 =10.30

11.3

12.30

2.0 3.0 4.0 5.0

Table 3: Addition of 0.20 M HCl to 50. mL of the Buffer Volume (mL) of 0.20 M HCl added to 50 mL of water

Calculat ed pH

Measure d pH

4.74

5.33

0.0 5.30 1.0

Sample Calculation for pH after addition Of 5.0 mL of the acid

START ADD END

C2H3O2- + H+  HC2H3O2 500mmol 500mmol 1mmol 499mmol 501mmol

5.30 2.0 5.28 3.0 5.28 4.0 4.74 5.0

5.25

pH =4.74 +log

=4.74 ( 499 501 )

Table 4: Addition of 0.20 M NaOH to 50. mL of the Buffer Volume (mL) of 0.20 M NaOH added to 50 mL of water

Calculat ed pH

Measure d pH

4.74

5.23

Sample Calculation for pH after addition Of 5.0 mL of the base

HC2H3O2 + OH-  H2O + -

C 2H3O 2 START 500mmol 500mmol ADD 1mmol END 499mmol 501mmol

0.0 5.23 1.0 5.24 2.0 5.26 3.0 5.28

pH =4.74 +log

4.0 4.74

=4.74 ( 501 499 )

5.28

5.0

Table 5: Carbonate/Bicarbonate Buffer Masses added to 50.0 mL volumetric flask (g)

Desired pH

Measure d pH

N/A Na2CO3:_0.1 moles_____________

NaHCO3: 0.1 moles _______

Calculation of the masses: pH = pKa + log(base/acid) pH = 10.3 + log(0.1/0.1) = 10.3

10.3

Analysis 1. Step 2 of the activity has you boiling the water. Why does the act of boiling help neutralize the water? When water boils the degree of dissociation of water increases, neutralizing the water. 2. Quantitatively discuss the change in pH produced after the addition of each one mL increment of acid (HCl) to pure water as compared to the buffer solution. What conclusions can you reach? Since there is no buffer in water to neutralize the HCl being added, the change in pH is greater. In the buffer a conjugate base neutralizes the HCl being added to the solution maintaining the current pH level. The buffer should maintain the pH level until the 0.5 moles of conjugate base runs out. 3. Quantitatively discuss the change in pH produced after the addition of each one mL increment of base (NaOH) to pure water as compared to the buffer solution. What conclusions can you reach? NaOH being a strong base fully dissociates in water greatly increasing the pH level in the water. In the buffer solution there is a significant amount of a conjugate acid that neutralizes the NaOH as it is being added maintaining the pH level. The buffer should maintain a consistent pH level until the 0.5 moles of conjugate acid runs out. 4. Quantitatively show how would you prepare a solution of acetic acid and sodium acetate with a total concentration of 1.0 M buffered to a pH of 5.2? Show your work. To make a buffer of 1.0 M concentration and a pH of 5.2. We can add 0.26 moles of acetic acid to 1L of water and then add 0.74 moles of sodium acetate to the solution. pH = 4.74 + log(0.75/0.25) = 5.22 [since the volume is the same for both we can just use the moles of each as substitute for concentration.]

Submit your worksheet on time and to your TA in the dropbox on D2L....