PRE Name UESTIONS Buffer and Vampire Slayer to submitted before class Date _c4lo
ID: 934677 • Letter: P
Question
PRE Name UESTIONS Buffer and Vampire Slayer to submitted before class Date _c4lohs Read the experiment and complete this assignment before the lab 1. Predict the pH of the following solutions that you will prepare in this experiment. Assume 250 C a. 0.10 M HCI b. 0.10 M NaOH c.0.10 M CH3COOH d. 0.10 M CH3COONa e. 0.10 M Na2C4H204 . 10 mL 0.20 M HCI mixed with 10 mL 0.20 M NaOH & 10 mL 0.20 M CH3COOH mixed with 10 mL 0.20 M CH3COONa a pH 7 7b. pH-pH>7 a. pH 7 a. pH 7 a. pH 7 h. distilled water that has been exposed to the atmosphere. a. pH 7 2. a. Calculate the pH of a 0.20 M HCI solution 2. b. Calculate the pH of the solution prepared by mixing 10.0 mL of 0.20 M HCI and 10.0 ml of distilled water 3. a. Calculate the pH of a 0.31 M CH3COOH solution. Ka of CH3COOH 1.76 x 10 s 3. b. Calculate the pH of a 0.23 M CH3COONa solution. (Think about whether this is an acidic or basic solution.)Explanation / Answer
1.
a. As HCl is an acid, pH <7
b. As NaOH is a base, pH >7
c. As CH3COOH is an acid, pH <7
d. As CH3COONa is a salt of a weak acid, pH > 7
e. As Na2C4H2O4 is a salt of a weak acid, pH > 7
f. As the HCl and NaOH are equal in moles, pH =7
g. As, the moles of acetic acid and moles of sodium acetate are equal, pH = pKa of acetic acid
Thus, pH = 4.75
Thus, pH <7
h. Water has a neutral pH. pH = 7
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2.
a)
0.20 M HCl
[H+] = 0.20 M
pH = -log [H+] = -log [0.20]
pH = 0.69
b)
Given 10.0 mL of 0.20 M HCl
Moles = 0.20 x 0.010 = 2 x 10^-3 mol
Total volume = 10.0 + 10.0 = 20.0 mL = 0.020 L
Molarity = 2 x 10^-3 mol / 0.020 = 0.1 M
[H+] = 0.1 M
pH = -log[H+] = -log [0.1] = 1
Thus, pH = 1
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3.
a.
Given, 0.31 M CH3COOH solution
CH3COOH <-----------------> CH3COO- + H+
I 0.31 M 0 0
C -x +x +x
E 0.31 - x +x +x
Ks = [CH3COO-][H+] / [CH3COOH]
1.76 x 10^-5 = x^2 / (0.31-x)
x^2 = 5.45 x 10^-6
x = 2.34 x 10^-3
[H+] = x = 2.34 x 10^-3 M
pH = -log[H+]
pH = -log [2.34 x 10^-3]
pH = 2.63
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