Question

Data Table 1. Adding 0.1M HCl from D1 into A1.

Drop Number	pH of Solution
1	5.0
2	5.0
3	5.0
4	5.0
5	5.0
6	5.0
7	5.0
8	5.0
9	5.0
10	5.0
11	5.0
12	5.0
13	5.0
14	5.0
15	5.0

Data Table 2. Adding 0.1M NaOH from D6 into A6.

Drop Number	pH of Solution
1	6.0
2	6.0
3	6.0
4	6.0
5	6.0
6	6.0
7	6.0
8	6.0
9	6.0
10	6.0
11	6.0
12	6.0
13	6.0
14	6.0
15	6.0

Data Table 3. Adding 6M HCl from Pipet into B1.

Drop Number	pH of Solution
1	4.0
2	4.0
3	3.0
4	2.0
5	2.0

Data Table 4. Adding 6M NaOH from Pipet into B6.

Drop Number	pH of Solution
1	6.0
2	6.0
3	6.0
4	6.0
5	6.0

Data Table 5. Adding 0.1M HCl from D1 into C1.

Drop Number	pH of Solution
1	2
2	2
3	2
4	2
5	2

Data Table 6. Adding 0.1M NaOH from D6 into C6.

Drop Number	pH of Solution
1	8.0
2	8.0
3	8.0
4	8.0
5	8.0

Questions

(5pts) Discuss if you saw a buffer. Define buffer and explain, with complete sentences, what you did to shift that buffer.

(10pts) A buffer was established with 2.0mL of 0.833M acetic acid. 0.5mL of 0.1M HCl is added. What is the new pH of the buffer? Ka of acetic acid = 1.8 x 10^-5 Show your work.

Discuss in a complete sentence how your theoretical calculation here is the same or different than what was obtained in Data Table 5.

Answer 1

A buffer solution is an aqueous solution of a mixture of a weak acid and its conjugate base, or vice versa.

Molarity of acetic acid = 0.833 M

Volume of acetic acid = 2.0 mL

Ka of acetic acid = 1.8 * 10^-5   

                         CH3COOH    CH3COO^-       + H⁺
Initial                  0.833                                      0                        0
Change                -x                                         +x                      +x
Final              0.833 - x                                      x                        x

Ka = [CH3COO^- ][H⁺ ] / [CH3COOH]

1.8*10^-5 = x² / (0.833 - x)

x = 0.00387

So, Initially 0.833 M acetic acid solution contains 0.00387 M CH3COO^- and 0.829 M CH3COOH.

Now, when HCl is added, it will react with CH3COO^-.

Molarity of HCl = 0.1 M

Volume of HCl added = 0.5 mL

CH3COO^- + H⁺ CH3COOH

Mmoles of CH3COO^- = 2.0 mL x 0.00387 M = 0.00774 mmol

Mmoles of CH3COOH = 2.0 mL x 0.829 M = 1.658 mmol

Mmoles of HCl added = 0.5 mL x 0.1 M = 0.05 mmol

Since, mmol HCl is greater than mmol CH3COO^- So, all the amount of CH3COO^- is converted into CH3COOH.

Mmoles of HCl left unreacted = 0.05 - 0.00774

= 0.04226 mmols

Total volume of solution = 2.0 + 0.5 = 2.5 mL

[HCl] = 0.04226 / 2.5 mL = 0.0169 M

pH = - log[H⁺ ]

= - log(0.0169)

= 1.77