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Alyssa Soto Chemistry 106, Section 19 Lab Report # 11: Acids, Bases and Ka “Where is my Lewis pair?” Part I: Name and concentration of the substance

Predict: Acid, base or Calculated pH salt?

Acid or Base? (based on calculated pH)

Nitric acid 0.00125 M Acid

2.9

Acid

Potassium hydroxide 0.00133 M

11.12

Base

Ammonia 0.00120 M Salt

10.17

Base

Hypochlorous acid 0.00128 M

Acid

5.17

Acid

Sodium hypochlorite 0.00125 M

Salt

9.28

Base

Ammonium chloride 0.00142 M

Salt

6.05

Acid

Carbonic acid 0.00144 M

Acid

4.6

Acid

Hydrochloric acid 4.55 M

Acid

-0.66

Acid

Phosphoric acid 0.20 Acid M

1.4

Acid

Sodium hydrogen carbonate 0.35 M

Salt

8.31

Base

Barium hydroxide 0.10 M

Base

13.70

Base

Sodium cyanide 0.0510 M

Salt

10.95

Base

Base

1.

Nitric acid: 0.00125 M HNO3: Strong acid

[H+] = [HNO3] = 0.00125 M pH = -log[H+] = -log(0.00125) = 2.90 2.

Potassium hydroxide: 0.00133M KOH: Strong base

POH = -log[OH-] = -log(0.00133) = 2.88 pH = 14 - pOH = 14 - 2.88 = 11.12 3.

Ammonia: 0.00120 M NH3: Weak base

initial change equilibrium

-

NH3·H20

NH4+

OH-

0.000120

0

0

x

0.00120 - x

Kb = x2/(0.00120 - x) ≈ x2/0.00120 Kb = Kw/Ka = (1*10-14)/(5.6*10-10) = 1.8 * 10-5 x = 1.46 * 10-4 M pOH = -log[OH-] = -log(1.46 * 10-4) = 3.83 pH = 14 - pOH = 14 - 3.83 = 10.17

x

+ x

x

+ x

4.

Hypochlorous acid: 0.00128 M HClO: Weak acid

initial change equilibrium

HClO

H+

ClO-

0.00128

0

0

-

x

+

0.00128 - x

x

+

x

x

x

ClO-

HClO

OH-

0.00125

0

0

Ka = x2/(0.00128 - x) ≈ x2/0.00128 Ka = 3.5 * 10-8 x = 6.7 * 10-6 M pH = -log[H+] = -log(6.7 * 10-6) = 5.17 5.

Sodium hypochlorite: 0.00125 M NaClO

initial change equilibrium

-

x

+

0.00125 - x

Kb = x2/(0.00125 - x) ≈ x2/0.00125 Kb = Kw/Ka = (1 * 10-14)/(3.5 * 10-8) = 2.9 * 10-7 x = 1.9 * 10-5 M pOH = -log[OH-] = -log(1.9 * 10-5) = 4.72 pH = 14 - pOH = 9.28

x x

+

x x

6.

Ammonium chloride: 0.00142 M NH4Cl

initial change equilibrium

-

NH4+

H2O

H3O+

0.00142

0

0

x

+

0.00142 - x

x

+

x

x

x

H2CO3

H+

HCO3-

0.00144

0

0

Ka = x2/(0.00142 - x) = x2/0.00142 Ka = 5.6 * 10-10 x = 8.9 * 10-7 M pH = -log[H3O+] = -log(8.9 * 10-7) = 6.05 7.

Carbonic acid: 0.00144 M H2CO3: weak acid H

initial change equilibrium

-

x

0.00144 - x

Ka = x2/(0.00144 - x) ≈ x2/0.00144 Ka = 4.3 * 10-7 x = 2.5 * 10-5 M pH = -log[H+] = -log(2.5 * 10-5) = 4.6 8.

Hydrochloric acid: 4.55 M HCl: Strong acid

+

x x

+

x x

[H+] = [HCl] = 4.55 M pH = -log[H+] = -log(4.55) = -0.66 9.

Phosphoric acid: 0.20 M H3PO4: weak acid H

initial change

H3PO4

H+

H2PO4-

0.20

0

0

-

equilibrium

x

+

0.20 - x

Ka = x2/(0.20 - x) = x2/0.20 Ka = 7.5 * 10-3 x = 3.9 * 10-2 M pH = -log[H+] = -log(3.9 * 10-2) = 1.4 10.

Sodium hydrogen carbonate: 0.35 M NaHCO3

+ H 2CO3 (aq) pH= -log(4.3*10-7) + -log(5.6*10-11)

2 pH= 8.31

x x

+

x x

11.

Barium hydroxide: 0.10 M Ba(OH)2: Strong base

POH = -log[OH-] = -log(0.2) = 0.70 pH = 14 - pOH = 14 - 0.70 = 13.30 12.

Sodium cyanide: 0.0510 M NaCN

CN-

HCN

OH-

initial

0.0510

0

0

change

-

equilibrium

x

0.0510 - x

+ x

+ x

x

x

Kb = x2/(0.0510-x) ≈ x2/0.0510 Kb = Kw/Ka = (1 * 10-14)/(6.2 * 10-10) = 1.6 * 10-5 x = 9 * 10-4 M pOH = -log[OH-] = -log(9 * 10-4) = 3.05 pH = 14 - pOH = 14 - 3.05 = 10.95 Solution pH

[H3O+]

mL 0.200 [A-] eq M NaOH

A

3.21

6.2 * 10-4 5.00 M

B

3.66

2.2 * 10-4 10.00 M

[HA] eq

1/[A-]

Ka

0.01 M

0.4 M

100

1.55 * 10-4

0.02 M

0.3 M

50

1.47 * 10-4

C

Solution A

4.01

9.7 * 10-5 15.00 M

0.03 M

0.2 M

33.3

[H3O+] = 10-pH = 10-3.21 = 6.2 * 10-4 M nOH- = 0.200 M * 5 mL * (1 L/103 mL) = 0.001 mol nA- = nOH- = 0.001 mol [A-] = nA-/Vt = 0.001 mol/.1 L= 0.01 M [HA] = [HA0] - [A-] = 0.05 M - 0.01M = 0.04 M 1/[A-] = 1/0.01 = 100 K a = [H3O+][A-]/[HA] = (6.2 * 10-4)(0.01)/(0.4) = 1.55 * 10-4

Solution B

[H3O+] = 10-pH = 10-3.66 = 2.2 * 10-4 M nOH- = 0.200 M * 10 mL * (1 L/103 mL) = 0.002 mol nA- = nOH- = 0.002 mol [A-] = nA-/Vt = 0.002 mol/0.1 L = 0.02 M [HA] = [HA0] - [A-] = 0.05 M - 0.02 M = 0.03 M 1/[A-] = 1/0.02 = 50 K a = [H3O+][A-]/[HA] = (2.2 * 10-4)(0.02)/(0.03) = 1.47 * 10-4

Solution C

[H3O+] = 10-pH = 10-4.01 = 9.7 * 10-5 M nOH- = 0.200 M * 15 mL * (1 L/103 mL) = 0.003 mol nA- = nOH- = 0.003 mol [A-] = nA-/Vt = 0.003 mol/0.1 L = 0.03 M

1.46 * 10-4

[HA] = [HA0] - [A-] = 0.05 M - 0.03 M = 0.02 M 1/[A-] = 1/0.03 = 33.3 K a = [H3O+][A-]/[HA] = (9.7 * 10-5)(0.03)/(0.02) = 1.46 * 10-4

Ka = 1/slope[HA0] = 1/127835(0.05) = 1.56 * 10-4 pKa = -log(1.56 * 10-4) = 3.81 Part II: 1. An acid is a chemical that releases hydrogen when added to an aqueous solution. The definition of Arrhenius acid is that it produces hydrogen ions when dissolved in water. While the Bronsted-Lowry acid is defined as the proton donor. They are found on the left side of a pH scale, below 7.

2. A base is a chemical that releases hydroxide ion when added to an aqueous solution. The definition of Arrhenius base is that it produces hydroxide ion when dissolved in water. While the Brønsted-Lowry base is defined as the proton acceptor. They are found on the right side of the pH scale, greater than 7. 3. In part 1, my prediction matched with the conclusion I made based on the pH value. The results were not surprising because when I predicted if a solution is acidic or basic, I wrote out the chemical equation to predict what would happen when the solution is mixed with water. If it produces hydroxide ion, I predicted it is a base, and if it produces hydrogen ion, I predicted it as an acid. 4. a) [H+] = [C3H6O3 -] = 10-2.75= 0.00178 M [C3H6O3 -] = 0.025 – 0.00178 = 0.0232 M Ka= [H+][C3H6O3 -] / [C3H6O3 -] = (0.00178)2 / 0.023= 1.36 x 10-4 b) I would expect the accuracy to decrease without the application of a pH probe because pH probe is a device designed to accurately measure the pH level of the solution. And as shown in the result of part 2 of the experiment, there is variability if pH probe is not used as a tool to measure the pH level....