Using the diluted solution in beaker B, pour 5 mL of this solution into beaker C
ID: 498584 • Letter: U
Question
Using the diluted solution in beaker B, pour 5 mL of this solution into beaker C. Dilute the solution with 45 mL of DI H_2O. Repeat the same procedure (steps) for the weak acid (0.1 M CH_3COOH solution), the strong base (0.1M NaOH), and the weak base (0.1 M NH_3). Calculation of the expected pHs of the weak acid and base are not required but recommended. Extra points for correct results. Show calculations only for one pH value of each. For the determination of [H_3O^+] and (OH^-], in table 3 show calculations only for the HCI solutions (in beakers A, B, C, and D).Explanation / Answer
Calculation of pH
For 0.1 M HCl
A : 0.1 M HCl
pH = -log[H+] = -log(0.1) = 1
B : 5 ml of 0.1 M HCl + 45 ml DI water
[H+] = 0.1 x 5/50 = 0.01 M
pH = -log[H+] = -log(0.01) = 2
C : 5 ml B + 45 ml DI water
[H+] = 0.01 x 5/50 = 0.001 M
pH = -log[H+] = -log(0.001) = 3
D: : 5 ml C + 45 ml DI water
[H+] = 0.001 x 5/50 = 0.0001 M
pH = -log[H+] = -log(0.0001) = 4
For 0.1 M CH3COOH
A : 0.1 M CH3COOH
1.8 x 10^-5 = x^2/0.1
x = [H+] = 1.34 x 10^-3 M
pH = -log[H+] = 2.87
B : 5 ml of 0.1 M CH3COOH + 45 ml DI water
[CH3COOH] = 0.1 x 5/50 = 0.01 M
1.8 x 10^-5 = x^2/0.01
x = [H+] = 4.24 x 10^-4 M
pH = -log[H+] = 3.37
C : 5 ml B + 45 ml DI water
[H+] = 0.01 x 5/50 = 0.001 M
1.8 x 10^-5 = x^2/0.001
x = [H+] = 1.34 x 10^-4 M
pH = -log[H+] = 3.87
D: : 5 ml C + 45 ml DI water
[H+] = 0.001 x 5/50 = 0.0001 M
1.8 x 10^-5 = x^2/0.0001
x = [H+] = 4.24 x 10^-5 M
pH = -log[H+] = 4.37
For part 2,
pH = -log[H3O+]
pOH = 14 - pH
pOH = -log[OH-]
or, [OH-] = Kw/[H+]
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