Electrolytes and Concentration DRA copy-2

pdf

School

Kaplan University *

*We aren’t endorsed by this school

Course

MBA404

Subject

Chemistry

Date

Feb 20, 2024

Type

pdf

Pages

Uploaded by GeneralGalaxyLoris40

Name: ____________________________ Section # _________ Partner: ___________________________ TA: ___________________________ Electrolytes & Concentration Data Reduction and Analysis Show all work in the space provided. Part A Record the electrical conductivity values (in μS /cm) of each of the solutions. Based on the conductivity, identify the type of electrolyte (strong electrolyte, weak electrolyte, or nonelectrolyte). Write the dissociation equation for each chemical (depending on the type of electrolyte, use → or ). Solution Conductivity ( μS/cm) Type of electrolyte Dissociation Equation Distilled H 2 O N/A Tap H 2 O 0.0500 M NaCl 0.0500 M CH 3 CO 2 H 0.0500 M C 6 H 12 O 6 0.0500 M CaCl 2 0.0500 M CuSO 4 0.0500 M FeCl 3 Questions 1. Explain the difference in conductivity between distilled water and tap water. 60 1224 6923 370 60 11727 6589 18708 strong electrolyte strong electrolyte strong electrolyte strong electrolyte weak electrolyte nonelectrolyte NaCl (s) → Na+ (aq) + Cl- (aq) CH3COOH (aq) ⾮ CH3COO- (aq) + H3O+ (aq) CaCl2 (s) → Ca+ (aq) + 2Cl- (aq) CuSO4 (s) → Cu2+ (aq) + SO42- (aq) FeCl3 (s) → Fe3+ (aq) + 3Cl- (aq) The conductivity of distilled water is less than the conductivity of tap water. This is because tap water contain ions and salts while distilled water is a pure form of water. Ions pass electricity from one to the next, as a result, the more ions, the higher the conductivity. C6H12O6 (s) → C6H12O6 (s) Jenny Hao 518

2. For 0.0500 M NaCl, how many ions per unit molecule? ______________________ 3. Calculate the concentration of ions for 0.0500 M NaCl. Show your work General Equation: 4. Based on the type of electrolyte and the concentration of the solution, calculate the ion concentration in the remaining solutions and enter the calculated values in the table. Solution Ion Concentration (M) 0.0500 M NaCl 0.0500 M CH 3 CO 2 H 0.0500 M C 6 H 12 O 6 0.0500 M CaCl 2 0.0500 M CuSO 4 0.0500 M FeCl 3 Part B Mass of NaCl used: __________________________ Calculate the concentration of the initial NaCl solution ([NaCl] = ?). Show your work General Equation: 2 Molarity x number of ions per unit molecule 0.0500 M x 2 ions = 0.1000 M 0.100 M 0.150 M 0.100 M 0.200 M 0.0770 g M=moles/L 0.0100 M 0 M M=moles/L 0.0770 g NaCl x 1/58.44g NaCl =0.00131759 moles 0.00131759 moles/ 0.01 L= 0.132 M

Calculate the concentration of each dilution of NaCl. Show your work General Equation: 1 st Dilution 2 nd Dilution 3 rd Dilution 4 th Dilution Based on the type of electrolyte and the concentration of NaCl in the solution, calculate the ion concentration of each of the solutions and enter the calculated values in the table. Record the associated conductivity readings in the table as well. Solution [NaCl] (M) Ion Concentration (M) Conductivity ( μS/cm) Initial NaCl solution 1 st dilution of NaCl solution 2 nd dilution of NaCl solution 3 rd dilution of NaCl solution 4 th dilution of NaCl solution Questions 1. Plot ion concentration ( x -axis) vs. conductivity ( y -axis) for Part B. Best-fit line equation: ___________________________________________ R 2 : ___________________________________________ M1V1=M2V2 0.132M x 8mL =M2 x 10mL =0.106M 0.106M x 6.5mL =M2 x 10mL =0.0689M 0.0689M x 3mL =M2 x 10mL =0.0207M 0.0207M x 5.50mL =M2 x 10mL =0.0114M 0.132 0.106 0.0689 0.0207 0.0114 0.264 0.212 0.138 0.0414 0.0228 16160 13192 8869 2898 1546 y = 60442x + 334.68 0.9995

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

2. Based on your best-fit line equation, calculate the expected conductivity of the solutions in Part A. Show your sample calculation for CaCl 2 . General Equation: 3. Summarize your results and calculations in the table below Solution Ion Concentration (M) Expected Conductivity ( μS/cm) Actual Conductivity ( μS/cm) Percent Error 0.0500 M NaCl 0.0500 M CH 3 CO 2 H 0.0500 M C 6 H 12 O 6 0.0500 M CaCl 2 0.0500 M CuSO 4 0.0500 M FeCl 3 4. Calculate the percent error and enter the values in the table above. 5. What are some of the factors that may lead to the expected and actual conductivity values to be significantly different? Explain. y = 60442x + 334.68 y = 60442(0.150) + 334.68 = 9400.98 μS/cm 0.100 M 0.0100 M 0 M 0.150 M 0.100 M 0.200 M 9400.98 6378.88 6378.88 12,423.08 6923 370 60 11747 6589 18708 8.53% 60.6% 25.0% 3.29% 50.6% 939.10 334.68 82.07% The best fit line equation was done based off of the concentrations of Na+ and Cl- ions. We are comparing that to various different ions to predict conductivity. Therefore, it will not be accurate.