Lab report 11 PDF

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Allen 1

Tamoy Allen Bryant Muniz April 9. 2020 Experiment 11: Acids, Bases, and Ka “Where is my Lewis pair?”

Part 1: pH Calculations Name and Concentration of the substance Nitric Acid 0.00125 M (HNO3) Potassium hydroxide 0.00133 M (KOH) Ammonia 0.00120 M (NH3) Hypochlorous acid 0.00128 M (HOCl) Sodium hypochlorite 0.00125 M (NaOCl) Ammonium chloride 0.00142 M (NH4Cl) Carbonic acid 0.00144 M (CH2CO3) Hydrochloric acid 4.55 M (HCl) Phosphoric acid 0.20 M (H3PO4) Barium hydroxide 0.10 M (Ba [OH]2) Sodium cyanide 0.0510 M (NaCN) Equations used: +¿ ¿ H3O ¿ ¿ pH =−lo g ¿

Predict: Acid, base, or salt?

Calculated pH

Acid or base? (based on calculated pH)

ACID

2.90

STRONG ACID

BASE

11.1

STRONG BASE

BASE

10.16

WEAK BASE

ACID

5.17

WEAK ACID

SALT

9.28

BASIC SALT

SALT

6.05

ACIDIC SALT

ACID

4.60

WEAK ACID

ACID

0

STRONG ACID

ACID

1.47

WEAK ACID

BASE

13.3

STRONG BASE

SALT

10.96

BASIC SALT

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−¿ OH¿ ¿ ¿ pOH=−lo g ¿ pH + pOH =14 −14

K a x K b =1 x 10

Nitric Acid 0.00125 M (HNO3) – Strong Acid HNO3 ⇌ H+ + NO3pH =−log [ 0.00125 ]=2.90

Potassium hydroxide 0.00133 M (KOH) = Strong Base KOH ⇌ K+ + OHpOH=−log [ 0.00133 ]=2.88 pH =14 −2.88=11.1

Ammonia 0.00120 M (NH3) – Weak Base NH3 + H2O ⇌ NH4+ + OHNH4+

NH3 Initial Change Exchange

0.00120 - x 0.00120 - x

1 x 10−14 =1.78 x 10−5 K a=5.6 x 10−10 K b= −10 5.6 x 10 K b=¿

X2 √ X 2 = √ ( 1.78 x 10−5) ( .0012 ) .0012−x

−¿ ¿ Concentration of O H X=1.46 x 10−4 ¿

pOH=−log [ .000146 ]=3.83

pH =14 −3.83=10.16

Hypochlorous acid 0.00128 M (HOCl) – Weak Base

OH0 +x x

0 +x x

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HOCl + H2O ⇌ OCl- + H3OHOCl

OCl-

0.00128 - x 0.00128 - x

Initial Change Exchange

H3O0 +x x

0 +x x

−8

K a=3.5 x 10

3.5 x 10−8 =¿

X2 .0012 8−x

X =6.7 x 10−6 pH =−log [ 6.7 x 10

−6

]=2.90

Sodium hypochlorite 0.00125 M (NaOCl) -Salt OCl- + H2O ⇌ HOCl+ + OHOCl0.00125 - x 0.00125 - x

Initial Change Exchange K a=3.5 x 10−8 K b= K b=¿

HOCl+

OH0 +x x

0 +x x

1 x 10−14 =2 . 86 x 10−7 −8 3.5 x 10

−7 X2 X 2= √ ( 2.86 x 10 ) ( .00125 ) √ .00125−x

−¿ ¿ Concentration of O H X=1. 90 x 10−5 ¿

pOH=−log [ .0 0001 9 ]= 4.72

pH=14 −4.72 =9.28

Ammonium chloride 0.00142 M (NH4Cl)- Salt NH4 + H2O ⇌ NH3- + H3O+ NH3-

NH4 Initial Change Exchange

0.00142 - x 0.00142 - x

H3O+ 0 +x x

0 +x x

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−10

K a=5.6 x 10 −10

5.6 x 10

2

=¿

X .001 42−x

−7

X =8.92 x 10

pH=−log [ 8.92 x 10

−7

] =6.50

Carbonic acid 0.00144 M (CH2CO3) – Weak Acid H2CO3 + H2O ⇌ HCO3- + H3O+ H2CO3 0.00144 - x 0.00144 - x

Initial Change Exchange

HCO3-

H3O+ 0 +x x

0 +x x

K a=4.3 x 10−7 4.3 x 10−7=¿

X2 .001 44−x

−5

X =2.5 x 10

pH=−log [ 2.5 x 10−5 ]=4.60

Hydrochloric acid 4.55 M (HCl)- Strong Acid HCl ⇌ H+ + ClpH=−log [ 4.55 ]=−.66=0

Phosphoric acid 0.20 M (H3PO4)- Weak Acid H3PO4 + H2O ⇌ H2PO3- + H3O+ H2PO3-

H3PO4 Initial Change Exchange K a=7.5 x 10−3

0.20 - x 0.20 - x

H3O+ 0 +x x

0 +x x

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−3

7.5 x 10 =¿

2

X .001 44−x

X =0.03386

pH=−log [ 0.03386 ]=1.47

Barium hydroxide 0.10 M (Ba [OH]2)- Strong Base Ba [OH]2 ⇌ Ba2+ + 2OHpOH=−log [ 0.10 ]=.70 pH=14 −0.70 =13.3

Sodium cyanide 0.0510 M (NaCN) – Salt CN- + H2O ⇌ HCN+ + OHCN0.0510 - x 0.0510 - x

Initial Change Exchange

K b=

HCN+

OH0 +x x

1 x 10−14 −5 =1.6 x 10 −10 6 . 2 x 10

K b=¿

X2 √ X 2 =√ ( 1.6 x 10−5 ) (. 0510 ) .0 510−x

−¿ ¿ Concentration of O H −4 X=9.03 x 10 ¿

pOH=−log [ 9.03 x 10− 4 ]=3 . 04

pH=14 −3.04 =10.96

Part 2: Determining a Ka Value

0 +x x

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Solutio n

pH

[H3O+]

mL 0.200 M NaOH

[A-] eq

[HA] eq

1/[A-]

Ka

A

3.21

6.7 X 10-4

5.00

.01 M

.04 M

100

1.54 X 10-4

B

3.66

2.19 X 10-4

10.00

.02 M

.03 M

50

1.46 X 10-4

C

4.01

9.77 X 10-5

15.00

.03 M

.02 M

33.33

1.47 X 10-4

Solution A To calculate

+¿ ¿ H3O ¿

pH=3.14 +¿ ¿ H3O ¿ ¿

 H+

To Calculate [A- ] V T =100.00 ml=.1 L

Moles WA before diluting: ( .2 ) x ( .005 )=0.001 Moles after diluting:

To calculate

(.2 ) x ( .005 ) =.01 NaOH [A-] .1

[ HA]

Moles WA=.01 x .5=.005 mols HA eq =.05−.01 =.04

To calculate (1/[A- ])

mol

 HA0

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[A-] = 1/.01=100 + ¿¿ H ¿ −¿ ¿ A ¿ −¿ ¿ A ¿ [ HA0 ]−¿ ¿ K a=¿

Solution B To calculate

+¿ ¿ H3O ¿

pH=3. 66

+¿ ¿ H3O ¿ ¿

M

To Calculate [A- ] V T =100.00 ml=.1 L Moles WA before diluting: ( .1 L ) x ( .5 M ) =0.005 mol Moles after diluting [A-]: To calculate HA=

[ HA ]

.005 =.05 .01

HA eq =.05−.02 =.03

To calculate (1/[A- ]) [A-] = 1/.02=50

( .01 ) x ( . 2 ) =.0 2 M NaOH .1

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+ ¿¿ H ¿ −¿ ¿ A ¿ −¿ ¿ A ¿ HA [ 0 ]−¿ ¿ K a=¿

Solution C To calculate

+¿ ¿ H3O ¿

pH=4.01

+¿ ¿ H3O ¿ ¿

 H+

To Calculate [A- ] V T =100.00 ml=.1 L Moles WA before diluting: ( .1 ) x ( .5 )=0.005 Moles after diluting: To calculate HA=

[ HA ]

.005 =.05 .01

HA eq =.05−.03 =.02

To calculate (1/[A- ]) [A-] = 1/.03=33.33

(.015 ) x (.2 ) =.03 M NaOH [A-] .1

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+ ¿¿ H ¿ −¿ ¿ A ¿ −¿ ¿ A ¿ HA [ 0 ]−¿ ¿ K a=¿

Inverse conjugate acid concentration vs Hydronium concentration 120 100

f(x) = 114960.57 x + 23.3

1/[A-]

80 60 40 20 0

0

0

0

0

0

0

[H3O+] 1/[A-]

Average Ka = 1.49 x 10-4

( Slope ) K a= 11 4961=

1 K a [ HA0]

1 K a( .05 )

K a=1. 73 x 10 −4

Linear (1/[A-])

0

0

0

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Based on the dissociation constants that were found, the acid that was used in the solution was lactic acid. The average Ka value found from the solutions in the chart was 1.49 x 10-4. The Ka value that was found using the graph was 1.73 x 10-4. Both of these values were close to the Ka value of lactic acid, 1.38 x 10-4 and formic acid 1.77 x 10-4. Discussion and Conclusion In this lab, we learned how to calculate the pH value of a substance using the concentrations given that were used to determine whether the substance was an acid, base, or salt. Also, we learned how to calculate the Ka value of a substance using by neutralizing it. In part 1 of the lab, we first predicted whether the substance was an acid, base, or salt. After using the concentrations that were given, we used an abundance of formulas to calculate the pH. For the substance that was a strong acid or base, the formula pH= -log [H+], or pOH= -log [OH-], respectively. With strong bases, after the concentration of OH- was found, you subtracted that value from 14 to get the pH of the substance. For weak acids and bases, water was added to the solution to help it dissociate. There was an “ICE” chart used to figure out the concentrations using a Ka or Kb value. After the concentrations were found, the values were plugged into the equations mentioned before, and the pH was calculated. For part 2 of the lab, the Ka of a substance had to be determined. To find the Ka, three solutions were partly neutralized (titrated with a base). Multiple steps and calculations led up to finding the Ka value for each solution. The first step was to find all the moles of the solution before and after the solution was neutralized. After, all the values were put into an equation that was given to find the Ka value. A graphing method was also used to find the Ka value of the solution. With all the values that were obtained, there was a graph made (inverse conjugate acid concentration vs. Hydronium concentration). By making this graph, the slope was used to determine the Ka value of the substance. From all the data that was gathered, a conclusion was made that the substance that was used in part 2 of the lab was either Lactic acid or formic acid.

Focus Questions: 1. What is the pH scale? The pH scales measure how acidic or basic a substance is. The pH ranges from 0 to 14 where & is neutral. A substance that has a pH higher than seven is considered basic, and a substance with a pH less than seven are considered acidic. 2. How is the pH of a solution calculated? What information is required? How can you tell whether the solution is acidic, basic, or neutral? To calculate the pH of a solution varies on if the solution is a strong/weak acid or base. For +¿ ¿ H3O strong acids and bases, the equation used to find the pH is the pH is or ¿ ¿ pH =−lo g ¿

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−¿ OH¿ , respectively. For bases after the pOH is found, you subtract that value from ¿ ¿ pOH=−lo g ¿ 14 to get the pH (the value found from calculating the pOH is the concentration of hydroxide ions). For solutions that have weak acids and bases, the solution usually has water in it to help the acid and bases dissolve. An “ICE chart” is used to find the concentrations using a Ka or Kb value. After finding the concentrations, the value depending on the active species in the solution is plugged into the equations mentioned before, and the pH is calculated. A solution is acidic if it has a pH level of less than 7. A solution is basic if it has a pH level higher than 7. Lastly, if the pH level of a solution is 7, the solution is neutral. 3. What is the Ka for the acid provided in part 2? The average Ka value for the acid provided in part 2 was 1.49 x 10-4. pKa value of 3.83 References: "Arrhenius, Bronsted-Lowry, and Lewis Acids and Bases in Organic Chemistry.", http://leah4sci.com/arrhenius-bronsted-lowry-and-lewis-acids-and-bases-in-organicchemistry/.

Post Lab Questions: 1. What is an acid? Distinguish Arrhenius acids and Bronsted-Lowry acids. Where are they found on the pH scale? An acid is a solution that is found on the lower half of the pH scale (1-7). This indicates that there is a high concentration of H+ ions. An Arrhenius acid is a substance that, when dissociates in water, donates an H+ in the solution. In other words, an Arrhenius acid is a proton donor. A Bronsted-Lowry acid is an acid, is very similar to an Arrhenius acid. A Bronsted-Lowry acid dissociates to give H+ in a solution. The only difference is that the solution does not need to be water. 2. What is a base? Distinguish Arrhenius bases and Bronsted-Lawry bases. Where are they found on the pH scale? A base is a solution that is found on the upper half of the pH scale (7-14). This indicates that there is a low concentration of H+ ions and a high concentration of OH-. An Arrhenius base is a substance that, when dissociates in water, would eject an OH- in the solution. A Bronsted-Lowry base refers to an atom or ion capable of accepting a proton in the solution. 3. Compare your prediction with your conclusion based on the calculated pH in part 1. Were any of your results surprising? Why or why not?

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The predictions that were made in part 1 of the experiment matched the conclusions based on the pH levels. Knowing the strong acids and bases before doing any calculations helped figured what solution would have a high or low pH. For example, HCl is known to be a very stung acid, so, therefore, it would have a very low pH. Also, NH4 is known to be a weak base, so one can conclude that the pH level is going to range somewhere between 7 to 11 on the pH scale. 4. Lactic acid C3H6O3 is found in sour milk where is it produced by the action of lactobacilli in lactose or the sugar in milk. The pH of a 0.045 M solution of lactic acid was determined using a pH probe and found to be 2.63. a. Calculate the equilibrium constant for this acid C3H6O3 ⇌ C3H5O3- + H+ +¿¿ H ¿ +¿ ¿ H ¿ −¿ ¿ C3 H 5 O 3 ¿ ¿ [ C3 H 6 O3] =0.045−0.00234 =0.04266 M −¿ ¿ C3 H 5 O 3 ¿ +¿ ¿ H ¿ ¿ K a =¿ b. Had you not been given the pH of the acid and you had to measure it yourself, how would the method in part 2 be applied to the determination of Ka? Would you expect an improvement in the accuracy of your result with the application of the method of this experiment? Explain why or why not. The same process would be used from part 2 to calculate Ka if there was not an application of the pH probe. The lactic acid would be titrated with a strong base to figure out the Ka value. The pH = pKa when the number of moles of the lactic acid is equal to the number of moles of the titrated base. From here, Ka can be derived using Ka= 10-pKa. The accuracy would decrease without using the application of the pH probe. The pH probe is designed to measure the pH accurately. Using the method in part 2 leaves room for error due to the lengthy calculations that have to be formed....