Exercise 1: Using Buffers Data Table 1. Adding 0.1M HCl from D1 into A1. Drop Nu
ID: 998916 • Letter: E
Question
Exercise 1: Using Buffers
Data Table 1. Adding 0.1M HCl from D1 into A1.
Drop Number
pH of Solution
1
8
2
8
3
8
4
8
5
8
6
8
7
8
8
8
9
8
10
8
11
8
12
8
13
8
14
8
15
8
Data Table 2. Adding 0.1M NaOH from D6 into A6.
Drop Number
pH of Solution
1
8
2
8
3
8
4
8
5
8
6
8
7
8
8
8
9
8
10
8
11
8
12
8
13
8
14
8
15
8
Data Table 3. Adding 6M HCl from Pipet into B1.
Drop Number
pH of Solution
1
6
2
6
3
6
4
6
5
6
Data Table 4. Adding 6M NaOH from Pipet into B6.
Drop Number
pH of Solution
1
8
2
8
3
8
4
8
5
8
Data Table 5. Adding 0.1M HCl from D1 into C1.
Drop Number
pH of Solution
1
4
2
4
3
4
4
4
5
4
Data Table 6. Adding 0.1M NaOH from D6 into C6.
Drop Number
pH of Solution
1
10
2
10
3
10
4
10
5
10
Need help with these, stumped.
A. What do the data in Data Tables 5 and 6 show about the ability of water to behave as a buffer? Considering the definition of a buffer, elaborate on your answer.
B. In the procedures for Part C, “Adding Dilute Concentrations of Acid and Base to Distilled Water,” would it be useful to repeat procedures 2 and 3 but instead use concentrated NaOH and HCl? Explain why or why not?
C. One beaker contains 10.0 mL of acetic acid/sodium acetate buffer at maximum buffer capacity (equal concentrations of acetic acid and sodium acetate), and another contains 10.0 mL of pure water. Calculate the hydronium ion concentration and the pH after the addition of 0.25 mL of 0.10 M HCl to each one. What accounts for the difference in the hydronium ion concentrations? Explain this based on equilibrium concepts; in other words, saying that “one solution is a buffer” is not sufficient.
H3O+(aq) + C2H3O2-(aq) HC2H3O2(aq) +H2O
Ka of acetic acid = 1.8 x 10-5
Acetic acid/sodium acetate buffer: Acetic acid = 0.1M, Sodium acetate = 0.1M
Drop Number
pH of Solution
1
8
2
8
3
8
4
8
5
8
6
8
7
8
8
8
9
8
10
8
11
8
12
8
13
8
14
8
15
8
Explanation / Answer
For part A and B we would need the procedure that was followed and who is D1, A1, B6... (add these and ask on another Q&A please because we're only allowed to answer 1 question per Q&A):
Let's do part C:
The pH of the acetic acid buffer would be, following the equation (we will consider a 1M solution of both acetic acid and its salt):
pH = pka + log (salt / acid)
pH = -log (1.8 x 10-5) + log(1/1)
pH = 4.74
>When 0.25mL of 0.1 M HCl is added, the following is the reaction that occurs (OAc is the acetate ion):
NaOAc + HCl -> NaCl + HOAc
This means more HOAc (acetic acid) is generated and the ratio salt/acid will be different.
The HOAc produced/NaOAc reacted would be: (0.1mol/L)(0.00025L) = 0.000025 moles of HOAc produced and 0.00025moles of NaOAc reacted.
If there is 10mL of solution, this means we have: (1mol/L)(0.01L) = 0.01moles of HOAc and 0.01moles of NaOAc.
Then, we would have 0.01 + 0.00025 moles of HOAc and 0.01 - 0.00025 moles of NaOAc.
The pH of the solution will then be:
pH = pka + log (salt / acid)
pH = -log (1.8 x 10-5) + log(0.00975/0.01025)
pH = 4.72
(H3O+) = 10-pH
(H3O+) = 1.91 x 10-5 M
>For the addition of 0.25mL of HCl 0.1M to 10.0mL of water, we can calculate the (H3O+) directly using the formula:
C1V1 + C2V2 = C3V3
Where C1 = (H3O+) in pure water (1 x 10-7). V1 = 10.0mL of pure water. C2 = 0.1M. V2 = 0.25mL and V3 = 10.25mL (V1 + V2). Let's solve for C3 (H3O+):
(C1V1 + C2V2 )/V3 = (H3O+)
(H3O+) = 2.44 x 10-3 M
pH = -log (H3O+)
pH = 2.61
The pH in the buffer solution doesn't decrease drastically (from 4.74 to 4.72) like pure water's (from 7.00 to 2.61) because in the buffer solution we have an equilibrium between the HOAc/NaOAc (acetic acid and sodium acetate) while in pure water we don't. This equilibrium is the one that gives the buffer the capacity to resist sudden changes in pH.
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