Physician × Physician x .] Enc × \\ Inbox (2.0 Class Not X E 17 S1 Ge x-D 5 secu
ID: 696690 • Letter: P
Question
Physician × Physician x .] Enc × Inbox (2.0 Class Not X E 17 S1 Ge x-D 5 secure https://wkingsnorthpark my sharepoint com :o/r/persona rienstr kracote northpark edu/ ayout 15 WopFrame.aspx?sourcedoc-67B40AB5726- G a. Place the magnetic stirrer on the base of the ring stand j. Insert the Conductivity Probe through the large hole in the Drop Counter k. Adjust the positions of the Drop Counter and burette so they are both lined up with the center of the magnetic stirrer. Drain any excess water from the burette into a waste beaker. Add 15 mL of the 0.1 M H2SO4 to the burette, Lift up the Conductivity Probe, and slide the beaker containing the Ba(OH)2 solution onto the magnetic stirrer. Lower the Conductivity Probe into the beaker. Adjust the position of the Drop Counter so that the tip of the Conductivity Probe is submerged Adjust the burette so its tip is just above the Drop Counter slot Turn on the magnetic stirrer so that the stir bar is stirring at a slow rate. 1. m. o. p. 7. You are now ready to perform the titration Start data collection. No data will be collected until the first drop goes through the Drop Counter slot. a. 5:24 PM 12/11/2017Explanation / Answer
1) Mole(s) of H2SO4 needed to reach the equivalence point = (volume of H2SO4 needed in L)*(molarity of H2SO4) = (6.619 mL)*(1 L/1000 mL)*(0.1 M) = 6.619*10-4 mole (ans).
2) Need the volume of Ba(OH)2 taken to answer the question.
3) Molar mass of BaSO4 = (1*137.327 + 1 *32.065 + 4*15.9994) g/mol = 233.3896 g/mol.
Mole(s) of BaSO4 corresponding to 0.104 g BaSO4 precipitated = (0.104 g)/(233.3896 g/mol) = 4.4561*10-4 mole (ans).
4) The balanced chemical equation for the formation of BaSO4 is given below.
Ba(OH)2 (aq) + H2SO4 (aq) -------> BaSO4 (s) + 2 H2O (l)
As per the stoichiometric equation,
1 mole BaSO4 = 1 mole Ba(OH)2.
Therefore, 4.4561*10-4 mole BaSO4 = 4.4561*10-4 mole Ba(OH)2.
Molarity of Ba(OH)2 = (moles of Ba(OH)2)/(volume of Ba(OH)2 in L) = (4.4561*10-4 mole)/[(10 mL)*(1 L/1000 mL)] = 0.044561 mol/L 0.0446 M (ans).
5) Need the data for answering (2) above.