In this exercise, weighed samples of a solid unknown containing NaCI and NaHCO3
ID: 538412 • Letter: I
Question
In this exercise, weighed samples of a solid unknown containing NaCI and NaHCO3 react with hydrochloric acid. The volume of the CO2 liberated by the reaction in the gas phase is measured with a gas collection syringe. From the volume we can calculate the number of moles of CO2 in the gas phase using the ideal gas approximation. Since the CO2 is generated in an aqueous environment (aqueous HCI), some CO2 will dissolve in the liquid phase. The amount of CO2 dissolved in the liquid phase is computed using Henry's Law which requires knowing the partial pressure of CO2. As described in the exercise, this requires knowing the entire system gas volume in addition to the initial and final syringe readings. The entire system gas volume is the sum of the quantities labeled Vtube and Vsyr in the diagram below, corrected for the liquid volume (the aqueous HCI). To insure that your answers are properly interpreted, the following are the values of the basic quantities used by OSCER in calculating the answers. Value Units Gas Constant Absolute Zero (0 K) Atmosphere Molar Mass of NaHCO3 Henry's Law Constant for CO2 in water 0.0821 L-atm/mol K -273.15 760.0 84.01 oc Torr g/mol mol/L-atm 3.2 X 10-2Explanation / Answer
3) NaHCO3 + HCl ----------> NaCl + H2O + CO2
1 mole of NaHCO3 gives 1 mole of CO2
1 mole of NaHCO3 has 84.01 gms
volume occupied by 1 mol of CO2 ,
P = 751.1 torr = 0.9882 atm
V = 1 * 0.0821 * 303.15 / 0.9882 = 25.1858 L = 25185.8 mL
84.01 gms occupies 25185.8 mL
? gms occupy 32.5 mL
0.1084 gms = weight of the sample in grams.
6) weight of CO2 in gms = 19.1629 - 18.9719 = 0.191 gms
mol of CO2 = 0.191 gms / 44 gm/mol = 4.3409*10-3 mol
P = 4.3409*10-3 * 0.0821 * 303.15 / 0.06 = 1.80064 atm
1.80064 = PCO2 + PH2O
1.80064 = 0.0417 atm (31.7 torr) + PCO2
1.80064 - 0.0417 = PCO2
PCO2 = 1.7589 atm
7) KH,CO2 = Cco2 / Pco2
3.2*10-2 * 1.7589 = Cco2 = 0.05628 mol/L
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.