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Ashley Campoverde Ramsey Salcedo March 4, 2020 Lab Report #11 “Where is my Lewis Pair?” Introduction Many common chemical solutions can be divided into two broad classes, based on whether or not they are acidic or basic. The easiest way to determine if a solution is acidic or basic is by an acid-base indicator (pH indicator), where the substance changes color in an acid or base solution. In this lab we want to determine not only whether or not it is basic or acidic but the strength of that acid or base. The strength is measured by pH which is the concentration if the hydrogen ions (H+). The pH value ranges from 0 to 14, where the lower half of the spectrum being the acid, and the upper half being the bases. A pH of 7 means the solution is neutral. This is important because it means a weak acid could actually have a lower pH than a diluted strong acid. On the other hand, the pKa value is a constant for each type of solution, thus it is a better indicator of the exact substance we are dealing with, when compared to the pH. In lab, one of the methods used to determine whether a compound is an acid, or a base is a pH indicator, which will change color in an acidic or basic environment which would thus, give us a rough idea of a pH of a solution. Specifically, in this lab we are going to look at how to calculate the pH mathematically. In Part 1 of the lab we were given a variety of different solutions and their molarity’s, and we were tasked with calculating the pH using a couple relatively straightforward formulas, which would involve filling up a K equation with the values for molarity and the equilibrium constant Ka, and the working backwards to find the concentration of [H +] and eventually, the pH. In part 2 of the lab we will use these same equations to try to figure out the pK a of an unknown substance, and then try to figure out the chemical composition of the solution. Here, we are given the pH at the outset, essentially meaning we just needed to do the calculation in part 1, but in reverse. This, however, is considerably more time consuming, as it requires us to calculate [H +], [HA] and [A -] before figuring out the Ka value. Materials  Pencil  Lab Manual  Lab Notebook  Calculator Observations and Experimental Part 1:

Name and concentration of the substance

Predict: Acid, base or salt?

Calculated pH

Acid or Base? (based on calculated pH)

Nitric acid 0.00125 M

Acid

2.9

Strong Acid

Potassium hydroxide 0.00133

Base

11.1

Strong Base

Ashley Campoverde Ramsey Salcedo March 4, 2020

M

Ammonia 0.00120 M

Salt

10.2

Weak Base

Hypochlorous acid 0.00128 M

Acid

5.17

Weak acid

Sodium hypochlorite 0.00125 M

Salt

9.28

Basic Salt

Ammonium chloride 0.00142 M

Salt

6.05

Acidic Salt

Carbonic acid 0.00144 M

Acid

4.6

Weak acid

Hydrochloric acid 4.55 M

Acid

-0.66

Strong acid

Phosphoric acid 0.20 M

Acid

1.4

Strong acid

Sodium hydrogen carbonate 0.35 M

Salt

8.31

Basic Salt

Barium hydroxide 0.10 M

Base

13.30

Strong base

Sodium cyanide 0.0510 M

Salt

10.95

Amphoteric Basic Salt

Work: 1. Nitric acid: 0.00125 M HNO3: Strong acid HNO3 ↔ H+ + NO3[H+] = [HNO3] = 0.00125 M pH = -log[H+] = -log(0.00125) = 2.90

Ashley Campoverde Ramsey Salcedo March 4, 2020 2. Potassium hydroxide: 0.00133M KOH: Strong base KOH ↔ K+ + OHPOH = -log[OH-] = -log(0.00133) = 2.88 pH = 14 - pOH = 14 - 2.88 = 11.1 3. Ammonia: 0.00120 M NH3: Weak base NH3·H20 ↔ NH4+ + OHNH3·H20

initial

NH4+

OH-

0

0

+x

+x

Kb = x2/(0.00120 - x) ≈ x2/0.00120 Kb = Kw/Ka = (1*10-14)/(5.6*10-10) =

x

x

1.8 * 10-5 x = 1.46 * 10-4 M pOH = -log[OH-] = -log(1.46 * 10-4) =

0.000120 -x

change

equilibrium

0.00120 - x

3.83 pH = 14 - pOH = 14 - 3.83 = 10.2 4. Hypochlorous acid: 0.00128 M HClO: Weak acid HClO ↔ H+ + ClO-

HClO

H+ ClO-

initial

0.00128

0

change

-x

+x +x

equilibrium

0.00128 - x x

0

x

Ashley Campoverde Ramsey Salcedo March 4, 2020 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 ClO- + H2O↔HClO + OH-

ClO-

HClO OH-

initial

0.00125

0

0

change

-x

+x

+x x

equilibrium 0.00125 - x 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 6. Ammonium chloride: 0.00142 M NH4Cl NH4+ + H2O↔NH3 + H3O+

initial

NH4+

H2O H3O+

0.00142

0

0

+x

+x

-x change

Ashley Campoverde Ramsey Salcedo March 4, 2020

equilibriu x 0.00142 - x m

x

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 H2CO3 ↔ H+ + HCO3-

initial

change

H2CO3

H+ HCO3-

0.00144

0 0

-x

+x +x

equilibrium 0.00144 - x x 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 HCl ↔ H+ + Cl[H+] = [HCl] = 4.55 M pH = -log[H+] = -log(4.55) = -0.66

Ashley Campoverde Ramsey Salcedo March 4, 2020 9. Phosphoric acid: 0.20 M H3PO4: weak acid H3PO4 ↔ H+ + H2PO4-

H3PO4 H+ H2PO4-

initial

0.20

0

change

-x

+x +x

equilibrium 0.20 - x x

0

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 NaHCO3 (aq) ⇌ Na+ (aq) + HCO3- (aq) HCO3- (aq) ⇌ H+ (aq) + CO3 2- (aq) HCO3- (aq) + H2O (l) ⇌ OH- (aq) + H2CO3 (aq) pH= -log(4.3*10-7) + -log(5.6*10-11) /2 pH= 8.31 11. Barium hydroxide: 0.10 M Ba(OH)2: Strong base Ba(OH)2 ↔ Ba2+ + 2OH12. POH = -log[OH-] = -log(0.2) = 0.70 pH = 14 - pOH = 14 - 0.70 = 13.30 13. Sodium cyanide: 0.0510 M NaCN CN- + H2O ↔HCN + OH-

Ashley Campoverde Ramsey Salcedo March 4, 2020

CN-

HCN

OH-

initial

0.0510

0

0

change

-x

+x

+x

equilibrium

0.0510 - 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 Part 2:

n

A

B

C

pH

[H3O+]

3.21

6.2 x 10-4 M

3.66

4.01

[A-] eq

[HA] eq

1/[A-]

Ka

5.00

0.01 M

0.4 M

100

1.55 x 10-4

2.2 x 10-4 M

10.00

0.02 M

0.3 M

50

9.7 x 10-5 M

15.00

0.03 M

0.2 M

33.3

Solution A

mL 0.200 M NaOH

1.47 x 10-4

1.46 x 10-4

Ashley Campoverde Ramsey Salcedo March 4, 2020 [H

+ 3O ] -

= 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 Ka = [H3O+][A-]/[HA] = (6.2 * 10-4)(0.01)/(0.4) = 1.55 * 10-4 Solution B [H

+ 3O ] -

= 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 Ka = [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 [HA] = [HA0] - [A-] = 0.05 M - 0.03 M = 0.02 M 1/[A-] = 1/0.03 = 33.3 Ka = [H3O+][A-]/[HA] = (9.7 * 10-5)(0.03)/(0.02) = 1.46 * 10-4 Graph 1: Inverse of the Conjugate Acid and Hydronium Concentration

Inverse of the Conjugate Acid (M)

Inverse of the Conjugate Acid and Hydronium Concentration 120 100

f(x) = 126907.8 x + 21.46 R² = 1

80 60 40

[HA] + [A-] = [HA0] 0.04 + 0.01 = 0.05

20 0 0

0

0

0

0

Hydronium Concentration

pKa = -log(1.58 * 10-4) = 3.80

0

0

0

Ka = 1/slope[HA0] = 1/126908(0.05) = 1.58 x 10-4

Ashley Campoverde Ramsey Salcedo March 4, 2020

Discussion and Conclusion One rather interesting thing I noticed in this lab, was that the method by which one finds the pH of a solution will vary based on the type of solution itself. For very strong acids, such as HCl, which we looked at in this lab. For the pH calculation for HCl, there was no need to create and solve an ICE table, as Hydrochloric Acid is so strong, one can simply do the negative log of the given molarity and get the pH that way. Interestingly enough, this also works for very strong bases, although one must make sure to do 14 – pH in this case, to ensure that you get the answer in terms of acidity and not basicity. One of the most important factors we had to deal with in this lab was dissociation. Dissociation is the reaction by which compounds split, usually reversibly, into atoms or ions. A strong acid has a high tendency to give up its hydrogen ions while in solution, while a weak acid is more likely to hold on to its hydrogen. This means that strong acids dissociate more easily in water and have a higher dissociation constant. Weak acids are less likely to dissociate, meaning their pKa (and usually pH) would be larger in magnitude. I saw this concept affected calculations in Part 2 of the lab, where I needed to find Ka by first finding the concentrations of [H+], [HA] and [A-]. I noticed that although, pH and pKa are definitely not synonyms, they can both tell you similar and equally important information about the dissociation of a solution. If the value of pKa increases in magnitude that tells us that the Ka value, which is directly related to dissociation, has decreased, making the solution is more basic. In this lab we were able to calculate the pH mathematically. Focus Question: 1. The pH scale measures how acidic or basic a substance is. The pH scale ranges from 0 to 14. A pH of 7 is neutral. A pH less than 7 is acidic. A pH greater than 7 is basic.

2. To calculate the pH of an aqueous solution you need to know the concentration of the hydronium ion in moles per liter, or molarity. The pH is then calculated using the following expression pH = - log [H3O+]. If the calculated pH is lower than 7 it is considered to be acidic. If the pH is above 7 then it is basic. A solution with a pH value that is equal to or around 7 is considered to be a neutral solution. 3. The Ka for the acid in Part 2 is about 1.57*10-4, meaning that the pKa would be approximately 3.8. According to this it is likely that the solution contains Formic Acid (CH 2O2), which has a similar pKa value and pH.

References Smeureanu, G.; Geggier, S. General Chemistry Laboratory, Fall 2019 ed. pp. 81–87. Post-Lab Questions 1. The pH scale measures how acidic or basic a substance is. The pH scale ranges from 0 to 14. A pH of 7 is neutral. A pH less than 7 is acidic. A pH greater than 7 is basic. The definition of Arrhenius acid is that it produces hydrogen ions when dissolved in water.

Ashley Campoverde Ramsey Salcedo March 4, 2020 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 in chemistry in considered to be a solution with a high pH, which indicates that it has a low concentration of positively charged hydrogen ions. An Arrhenius base dissociates in water to form hydroxide ions, thus increasing the concentration of OH- ions in an aqueous solution. A Bronsted Base will accept protons (H+) when undergoing a chemical reaction. Bases are found on the higher end of the pH scale, above 7. 3. In part 1, my prediction matched with the conclusion I made based on the pH value. 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 -] = 10-2.63 = 0.0023 M [C3H6O3 -] = 0.045 – 0.0023= 0.043 M Ka= [H+][C3H6O3 -] / [C3H6O3 -] = (0.0023)2 / 0.043= 1.23 x 10-4 b) If I was not given the value of pH, I would need to measure all the necessary components myself and then fill in the Ka formula to ensure the most accurate value. This would require me to find [H+], [A-], [HA]. I would need to use something akin to a litmus test to obtain a general idea of the pH and solve for the [H+] from there. In order to find the concentration of [A-] and [HA] I would need to solve for the number of moles of Lactic Acid and the volume of the solution. I would not expect an improvement in accuracy through this method....