Experiment 11- Conductometry of Strong and Weak Electrolytes copy PDF

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Abstract: In this experiment we used conductometry to determine whether or not an electrolyte was weak or strong. In this experiment we prepared 5 solutions each of NaAc, HCl, NaCl, and CH 3OOH by weighing out and pipetting the amount according to what we calculated and then diluting it to the 100 mL line in the volumetric flask. Our results show a declining of accuracy in each solution we tested. According to graphs, our r-squared values gradually decrease. According to Graph 1A, the Λm0 (NaCl) =

0.136

S•cm2•mol-1,

−1

Λm0 ( HCl )=0.0971 S •cm • mol , Λ0m(NaAc) 2

=

0.133S•cm2•mol-1. In figure 2 the Ka of Acetic Acid is shown to be 2.866. All these errors can be taken because of not cleaning the equipment correctly, not using the correct amount in each solution, and adding to much water. Introduction: The main purpose of this experiment is to learn how use the conductimetric method to determine strong and weak electroltyes. In this experiment we are analyzing and comparing 4 solutions that we made that are HCl, NaCl, NaAc, and CH 3OOH. After all solutions are correctly prepared we will be using a conductivity probe to measure its molar conductancy and using computer software to achieve the data. After all the data has been collecting, you will be using this data to find the disassociation constant of the acetic acid. Theory: Conductometry is determination of the quantity of a material such as an element or salt, present in a mixture by measurement of its effect on the electrical conductivity of the mixture. When a potential difference is applied to two electrodes immersed in a solution contained ions, there is a flow of current to the mitigation of ions through the solution. One of the key electrical property of the solution is its resistance. This is expressed in Ohm’s ( ). One term we must define is

Ohm’s Law which states that for any circuit the electric current is directly proportional to the voltage and is inversely proportional to the resistance. Conductance is actually the inverse of the

resistance. This is expressed as

G=

1 , and the SI units are expressed as Siemens (1S = -1 = R

1 C V -1 s-1). The definition of conductivity is expressed is

G=κ

A . In this expression  is the l

length,  is the electrical conductivity. The next term to know is the molar conductivity (

∧m ¿ , this is expressed as

∧m =

κ . In this expression c is the molar concentration of the c

added electrolyte. The SI units for molar conductivity is expressed as (1 S m 2 mol-1). In Kohlrausch Law states that at low concentration the molar conductivities of strong electrolytes depend of the square root of the concentration. The equation is s follows. Kohlrausch Law there is a few more things we must understand. The

∧m=∧°m + K c 1 /2 . In °

∧m

is defined as the

limiting molar conductivity. The molar conductivity in the limit of zero concentration when the ions are so far part that they move independently of each other. Kohlrausch stated that when +

and - than the law of the independent mitigation follows that

−¿ −¿ λ¿ +¿+ v ¿ . Dubye-Hückel Theory + ¿ λ¿ ° ∧m=v ¿

is a homogeneous mixture of two or more components existing in a single phase. It is expressed

−¿ +¿ z ¿ as

1

z ¿ I2 log(γ ± )=− A ¿

Ionic Strength is the measure of the ions concentration ions solution.

I=

( )

b 1 z i2 ⊝i ∑ 2 b

. The

last topic to mention is the Degree of Disassociation. It is expressed in two forms:

x a=

2α 1−α ,x = . It is defined as is the fraction of a mole of the reactant that underwent 1+α b 1+ α

dissociation. Experimental: Equipment  Conductivity Probe  50 mL Beaker  20, 100 mL Volumetric Flask  Pippettes

Chemicals  Distilled Water (H2O)  0.1 N HCl  0.1 N Acetic Acid  Solid NaCl  Solid NaAc

Procedure: I: Preparation of the solutions: We are going to prepare 5 solutions in 100 mL volumetriv flasks of each of the following: Acetic Acid, HCl, NaCl, and NaAc. Each solution would be essential made the same with same method except for the amount of concentration we will be puttng in the 100 mLvolumetric flask. In HCl we will be putting the following amounts: 10 mL, 8 mL, 6 mL, 2 mL and 1mL and the dilute to the 100 mL mark with distilled water. In the acetic acid we will be putting the following amounts: 40 mL, 10 mL, 8 mL, 6 mL, and 4 mL Acetic acid and then diluting it to the 100 mL

mark with water. Lastly for the NaCl and NaAc they will be both using the same amounts to our volumetric flask. We will the following measurments to each of the 100 mL volumetric flasks: 30 mL, 15 mL, 6 mL, 3 mL and 1.5 mL. Once done pippeting all those the flash we will be diluting them with distilled water to the 100 mL mark. II: Measuring Conductance: In this part of the experiment we will be measurng 5 prepared solutions we made earlier in the lab. The first thing to is to connect the sensor the computer and set the range switch on the sensor to 0 - 2000 µS/cm. Then open the logger pro software and choose new from file. The first step to do is pour each of the the made soluton into the 50 mL beaker about 2/3 of filled and measure the conductivity. Next repeat the process until you measued all of your prepared solutions and begin the next trial the same for all your other prepared solutions. Make sure once your done with one set of prepared solution wash the beaker in order to begin next measurement of conductivity.

Results: Data/Plots Table 1A: Acetic Acid (CH3OOH):

Standard 1 Standard 2 Standard 3 Standard 4 Standard 5

Vi,mL

Ci,M

Vf,mL

Cf,M

40 10 8 6 4

0.1 0.1 0.1 0.1 0.1

100 100 100 100 100

0.040 0.010 0.008 0.006 0.004

Table 1B: Hydrochloric Acid (HCl) Vi,mL

Standard 1 10 Standard 2 8

Ci,M

Vf,mL

Cf,M

0.1 0.1

100 100

0.010 0.008

Standard 3 6 Standard 4 2 Standard 5 1

0.1 0.1 0.1

100 100 100

0.006 0.002 0.001

Table 1C: Sodium Chloride (NaCl) Vi,mL

Ci,M

Vf,mL

Cf,M

30 15 6 3 1.5

0.1 0.1 0.1 0.1 0.1

100 100 100 100 100

0.0300 0.0150 0.0060 0.0030 0.0015

Standard 1 Standard 2 Standard 3 Standard 4 Standard 5

Table 1D: Sodium Acetate (NaAc)

Standard 1 Standard 2 Standard 3 Standard 4 Standard 5

Vi,mL

Ci,M

Vf,mL

Cf,M

30 15 6 3 1.5

0.1 0.1 0.1 0.1 0.1

100 100 100 100 100

0.0300 0.0150 0.0060 0.0030 0.0015

Table 2A: Acetic Acid (CH3OOH): [CH3OOH], M

, µS/cm

√[ CH3 OOH ]

Λm, Scm2mol-1

M

0.040 0.010 0.008 0.006 0.004

103.3 127.5 149.6 165.7 270.3

0.0258 0.0213 0.0187 0.0166 0.0068

Table 2B: Hydrochloric Acid (HCl) , µS/cm Λm, Scm2mol-1 [HCl], M 0.010 289.0 0.0060 0.008 304.8 0.0184 0.006 317.3 0.0497

1/2

0.06 0.08 0.09 0.10 0.20

√ [ HCl]

M1/2

0.17 0.12 0.08

,

,

0.002 0.001

320.6 319.8

0.1038 0.2108

0.05 0.04

Table 2C: Sodium Chloride (NaCl) cm2mol-1 [NaCl], M , µS/cm Λm, S

√ [ NaCl]

,

M1/2

0.0300 0.0150 0.0060 0.0030 0.0015

180.9 275.8 298.2 311.3 316.2

0.0028 0.0115 0.0444 0.0999 0.2072

Table 2D: Sodium Acetate (NaAc) , µS/cm Λm, S cm2mol-1 [NaAc], M

0.17 0.12 0.08 0.05 0.04

√ [ NaAc]

M1/2

0.0300 0.0150 0.0060 0.0030 0.0015

84.9 172.3 266.2 299.7 310.8

2830.00 11486.67 44366.67 99900.00 207200.00

0.17 0.12 0.08 0.05 0.04

,

NaCl 0.20 0.18

0.17

0.16 0.14

M^1/2

0.12

f(x) =0.12 − 0.55 x + 0.14 R² = 0.7

0.10

Linear ()

0.08

0.08

0.06

0.05

0.04

0.04

0.02 0.00 0.0000

0.0500

0.1000

0.1500

Λm ( S×cm2×mol-1 )

Figure 1A: Acetic Acid Figure 1B: Hydrochloric Acid: Figure 1C: NaCl Figure 1D: NaAc

0.2000

0.2500

NaAc 0.18

0.17

0.16 0.14

Axis Title

0.12

f(x)0.12 = − 0.54 x + 0.13 R² = 0.69

0.10 0.08

Linear ()

0.08

0.06

0.05

0.04

0.04

0.02 0.00 0.0000

0.0500

0.1000

0.1500

0.2000

Λm ( S×cm2×mol-1 )

Table 3:  Acetic Acid (CH3OOH): [HAc], M 0.004 0.006 0.008 0.01 0.04 Figure 2:

� degree of dissociation

Apparent �c

logKc

2.744E-01 2.258E-01 1.987E-01 1.761E-01 7.181E-02

4.152E-04 3.952E-04 3.943E-04 3.763E-04 2.222E-04

-3.38 -3.4 -3.4 -3.42 -3.65

0.2500

Degree of Disassociation -3.20 3E-02 -3.25

4E-02

4E-02

5E-02

5E-02

6E-02

-3.30 -3.35

log(kc)

-3.40

f(x) = − 13.81 x − 2.89 R² = 0.9

-3.45

Linear ()

-3.50 -3.55 -3.60 -3.65 -3.70

(⍺ x C)^1/2

Results – Calculations/Interpretations Tabla 1A-1D Calculations: Acetic Acid (CH3OOH): Hydrochloric Acid (HCl): Cf V f 100(0.040) = =40 mL S 1: C i V i=C f V f :V i= (0.01) Ci CV 100(0.010) S 1:C i V i=C f V f :V i= f f = =10 mL Ci (0.01) C f V f 100(0.010) S 2: C i V i =C f V f :V i= = =10 mL Ci (0.01) C f V f 100(0.008) = =8 mL S 2:C i V i =C f V f :V i= (0.01) Ci C f V f 100(0.008) = S 3: C i V i=C f V f : V i= =8 mL Ci (0.01) C f V f 100(0.006) S 3:C i V i=C f V f : V i= = =6 mL (0.01) Ci C f V f 100(0.006) = =6 mL S 4 : C i V i =C f V f :V i= (0.01) Ci C V 100(0.002) S 4 : C i V i =C f V f :V i= f f = =2mL Ci (0.01)

C f V f 100(0.004) = =4 mL (0.01) Ci C f V f 100(0.001) = =1 mL S 5: C i V i =C f V f : V i= (0.01) Ci S 5:C i V i =C f V f : V i=

Sodium Chloride (NaCl): C f V f 100(0.030) = =30 mL S 1: C i V i=C f V f :V i= (0.01) Ci CV 100(0.030) S 1:C i V i=C f V f :V i= f f = =30 mL Ci (0.01) C f V f 100(0.015) =15 mL S 2: C i V i =C f V f :V i= = (0.01) Ci CV 100(0.015) =15 mL S 2:C i V i =C f V f :V i= f f = (0.01) Ci C f V f 100(0.006) = =6 mL S 3: C i V i=C f V f : V i= (0.01) Ci C f V f 100(0.006) S 3:C i V i=C f V f : V i= = =6 mL (0.01) Ci C f V f 100(0.003) S 4 : C i V i =C f V f :V i= =3 mL = (0.01) Ci C V 100(0.003) =3 mL S 4 : C i V i =C f V f :V i= f f = (0.01) Ci C f V f 100(0.0015) S 5: C i V i =C f V f : V i= = =1.5 mL Ci (0.01) C V 100(0.0015) =1.5 mL S 5:C i V i =C f V f : V i= f f = (0.01) Ci

Sodium Acetate (NaAc)

Table 2A-2D: Calculations: Λm, S cm2mol-1: CH3OOH 103.3 x 10−6 =0.0258∨√ C H 3 OOH =√ 0.004= 0.06 S 1: Λ m= 0.004 127.5 x 10−6 S 2: Λ m= =0.0213∨√ C H 3 OOH =√ 0.006=0.08 0.006 149.6 x 10−6 S 3: Λ m= =0.0187∨√ C H 3 OOH= √ 0.008 =0.09 0.008 165.8 x 10−6 =0.0166∨ √ C H 3 OOH =√ 0.010= 0.10 S 4 : Λ m= 0.010

S 5: Λ m=

270.3 x 10−6 =0.0068∨√ C H 3 OOH =√ 0.040= 0.20 0.040

HCl 298.9 x 10−6 =0.2890∨√ HCl= √ 0.001=0.03 0.001 −6 304.8 x 10 S 2: Λ m= =0.1524∨√ HCl=√ 0.002 =0.04 0.002 317.3 x 10−6 S 3: Λ m= =0.0529∨√ HCl=√ 0.006 =0.08 0.006 −6 320.6 x 10 S 4 : Λ m= =0.0401∨ √ HCl=√ 0.008= 0.09 0.008 319.3 x 10−6 S 5: Λ m= =0.0320∨√ HCl=√ 0.010 =0.10 0.010 S 1: Λ m=

NaCl: −6

180.9 x 10 =0.0060∨√ NaCl= √ 0.030= 0.17 0.030 275.8 x 10−6 =0.0184∨√ NaCl= √ 0.015 =0.12 S 2: Λ m= 0.015 298.2 x 10−6 =0.0497∨√ NaCl= √ 0.006 =0.08 S 3: Λ m= 0.006 −6 311.3 x 10 S 4 : Λ m= =0.1038∨ √ NaCl= √ 0.003 =0.05 0.003 316.3 x 10−6 =0.2108∨√ NaCl= √ 0.0015= 0.04 S 5: Λ m= 0.0015 S 1: Λ m=

NaAc

84.9 x 10−6 =0.0028∨ √ NaAc=√ 0.030= 0.17 0.030 −6 172.3 x 10 S 2: Λ m= =0.0115∨√ NaAc= √0.015 =0.12 0.015 266.2 x 10−6 S 3: Λ m= =0.0444∨√ NaAc= √ 0.006 =0.08 0.006 299.7 x 10−6 S 4 : Λ m= =0.0999∨√ NaAc= √ 0.003= 0.05 0.003 310.8 x 10−6 S 5: Λ m= =0.2072∨√ NaAc= √ 0.0015=0.04 0.0015 S 1: Λ m=

Limiting Molar Conductivity: CH3OOH: 0.2832 S•cm2•mol-1 HCl: 0.0971S•cm2•mol-1 NaCl: 0.136 S•cm2•mol-1 NaAc: 0.133S•cm2•mol-1 Calculations of Λ °m −1 −¿+ NaAc−NaC l ⋀ m° ( HAc)=HC l¿ λ°m ( HAc )=[ ( 0.0971)+ (0.133 ) ]−( 0.136 ) ° 2 −1 λm ( HAc)=0.0941 S ∙ c m ∙ mo l

Table 3: Calculations � degree of dissociation 0.0259 =2.744 x 10−1 0.004 :α= 0.0931 0.0213 −1 =2.258 x 10 0.006 :α= 0.0941 0.0186 =1.987 x 10−1 0.008 :α= 0.0941

0.0166 =1.761 x 10−1 0.0941 0.0068 −2 =7.181 x 10 0.040 :α= 0.0941

0.010 :α=

Apparent �c (0.004 )(2.744 x 10−1) =4.152 x 10−4 −1 (1−(2.744 x 10 ) ) (0.010 )(3.899 x 10−6) 0.010 : K c = =3.763 x 10−4 −1 ( 1−(1.761 x 10 ) ) 0.004 : K c =

0.006 : K c =

0.008 : K c =

(0.006 )(2.258 x 10−1) =3.952 x 10−4 −1 ( 1−(1.987 x 10 ) ) (0.008 )(4.400 x 10−6 )

( 1−(1.987 x 10 ) ) −1

Log(Kc): −4 0.004 : log ( 4.152 x 10 ) =3.38 0.006 :log ( 3.952 x 10− 4) =3.40 −4 0.008 :log ( 3.943 x 10 ) =3.40

7.181 x 10−2 1−(¿ ) ¿ ¿ (0.004 )(7.181 x 10−2) 0.040 : K c = ¿

=3.943 x 10− 4

0.010 :log ( 3.763 x 10 ) =3.42 0.040 :log ( 2.222 x 10−4 ) =3.65 −4

Discussion The main purpose of this lab was to learn how to use conductometry to find weak and strong electrolytes. In this experiment we made 5 solutions of each of the following HCl, CH 3OOH, NaCl and NaAc. Each of them was diluting with water and put in a certain amount according to what we calculated. According to Graph 1A, the

Λm0 (NaCl) = 0.136 S•cm2•mol-1,

2 −1 Λm0 ( HCl )=0.0971 S •cm • mol , Λ0m(NaAc) = 0.133S•cm2•mol-1. In figure 2 the Ka of Acetic

Acid is shown to be 2.866. This value seems to be a little off than what is expected to be. The Ka value should be 1.76 x 10-5. Our value comes to 1.36x10-3. The statistics in our experiment show little percent error according to r-squared value. Most of the r-squared values according to figures 1A is 0.94, 1B is 0.87, 1C is 0.70, 1D is 0.69. In each experiment our data becomes less and less accurate. All the error my lab partner and I obtained could have due to incorrect measurements or improper cleaning of our equipment. If this experiment could have done again, we would clean better the 100 mL flasks, and make sure to be very careful measuring out the increments needed for each dilution.

References Atkins, P. W., Paula, J. D., & Keeler, J. (2018). Atkins Physical Chemistry (11th ). New York, NY: Oxford University Press. Conductometry (n.d) In Merriam Webster Online. Retrieved September 29, 2019, from https://www.merriam-webster.com/dictionary/conductometry Ohm’s Law (2011). In Dictionary: Houghton Mifflin Harcourt Publishing Company Retrieved September 29, 2019 from https://www.dictionary.com/browse/ohm-s-law Libretexts. (2019, June 5). Debye-Hückel Theory. Retrieved from https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/ Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Physical_Properties_of_Matter/ Solutions_and_Mixtures/Nonideal_Solutions/Debye-Hückel. Young, H. D., Freedman, R. A., Ford, A. L., Sears, F. W., & Zemansky, M. W. (2014). Sears and Zemanskys university physics: with modern physics: technology update (13th ed.). Boston: Pearson....