SUSB-054 - Gasometric
 1. Preliminaries In this exercise, you will weigh a precise amount of a sample of a mixture of NaHCO3 and NaCl, and react it with 10 mL of aqueous hydrogen chloride (hydrochloric acid) in a closed system consisting of a syringe, with a moveable plunger, attached to a large test tube. The reaction produces CO2 gas. The system now contains :         1.   a mixture of CO2 gas and water vapor in the empty space         space of the syringe and large test tube. (The water vapor comes from the aqueous solution of acid)         2.  CO2 gas dissovled in the aqueous acid You must first find the total amount of CO2 produced (gaseous + dissolved), then using the stoichiometry of the reaction determine the mass of NaHCO3 in the mixture sample 2. Stoichiometry and Limiting Reagent: The stoichiometric relationship between the reagents is given by the overall equation: NaHCO3(s) +  H+(aq)     =   Na+(aq) + H2O(l) + CO2(g) 1 mole of bicarbonate ion,  HCO3-  reacts with 1 mole of hydrogen ion, H+ The concentration of the hydrochloric acid is 1.00 M. The unknown sample that your TA will give you will have a mass percent of NaHCO3 between  50% and 100%   Suppose a sample of the unknown weighs 0.2000 g (200.0 mg). A 50% sample would contain   0.500 x 200 mg = 100 mg, and a 100.0% sample   1.000 X 200 mg = 200 mg of NaHCO3 The molar mass of NaHCO3 is 84.0 g/mol (or, mg/mmol), so the sample is known to contain between: 100/84.0 = 1.19 mmol   and   200/84.0 = 2.38 mmol of NaHCO3 If we use 10 mL of 1.0 M HCl to react with the NaHCO3, the 10 mL contain 10 mL X 1.0 mmol/mL = 10 mmol HCl. This is clearly in excess of the amount required to react completely with the NaHCO3. THE UNKNOWN IS THE LIMITING REAGENT! 3. Moles of CO2 in the Gas Phase The moles of CO2 gas can be calculated using the ideal gas law and Dalton's Law of Partial Pressures. When the plunger of the syringe stops moving (reaction complete) the pressure outside the system (Patm) equals the pressure inside the system (PCO2 + PH2O).         PCO2   =   Patm   −   PH2O       nCO2   =   (Patm   −   PH2O)(Vfsyr − Visyn) / RT   Eq 3 Lab Manual 4. Henry's Law for the Solubility of CO2 in HCl (aq) Henry's Law is an equation that relates the concentration (mol/L) of a dissolved gas to the partial pressue of that gas above the surface of the liquid :             Si   =   kH P i Where S is the solubility in moles / L,   kH is the Henry's Law Constant for the solvent, and   P is the partial pressure of the gas (CO2) above the liquid.             SCO2   =   kH P CO2       [ Eq 4 Lab Manual ]             PCO2   =   nCO2 RT / VfTOT       [ Eq 5 Lab Manual] VfTOT   =   VTUBE   +   Vfsyr   −   VHCL       [ Eq 6 Lab Manual]                     where   VHCl   =   10 mL nCO2 dissolved in HC (aq)   =   VHCl (10 mL) X SCO2

Sample Calculation

 Sample Weight Initial    14.3772 g Final    14.1836 g Net        0.1936 g = 193.6 mg Volume of Gaseous CO2 Final Syringe Reading     47.5 mL Initial Syringe Reading     5.0 mL                   V    = 42.5 mL Laboratory and System Conditions   Atmospheric Pressure = 765 mm Hg   P = 765 / 760 = 1.007 atm   Room Temperature = 23oC   T = 23 + 273 = 296 K   Volume of the Tube and Elbow   Vtube = 80.0 mL Vapor Pressure of Water at 23oC (from Posted Table) PH2O = 21.1 mm Hg = 21.1 / 760 = 0.0278 atm mmol of CO2 gas captured nCO2(gas) = ( P - PH2O ) V / RT = (1.007-0.0278) 42.5/0.0821*296 = 1.71 mmol Henry's Law S = kH PCO2 kH = 3.2 X 10-2 mol/L-atm VfTOT  = 80.0 + 47.5 -10.0 = 117.5 mL     PCO2 = nCO2 RT / VfTOT         1.71 x 0.0821 x 296 / 117.5 = 0.354 atm S = 3.2 X 10-2 X 0.354 = 1.1 X 10-2 mol/L But we have 10 mL of solution (the volume of HCl used), so, nCO2(soln) = 1.1 X 10-2 X 0.010 L     = 1.1 X 10-4 mol = 0.11 mmol So, the total number of mmol of CO2 liberated by the reaction was 1.71 + 0.11 = 1.82 mmol. This must have come from 1.82 mmol of NaHCO3, which weighed 1.82 mmol X 84.0 mg/mmol = 153 mg. % NaHCO3 in the original sample = 100 X 153 / 193.6 = 79.0 %