Porta Experiment 13 Advance Study Assignment: The Standardization of a Basic Sol
ID: 563729 • Letter: P
Question
Porta Experiment 13 Advance Study Assignment: The Standardization of a Basic Solution and the Determination of the Equivalent Mass of an Acid 1. 70 mL of 6.0 M NaOH are diluted with water to a volume of 400 mL. You are asked to find the molarity a. First find out how many moles of NaOH there are in 7.0 mL of 6.0 M NaOH. Use Equation 1. Note moles b. Since the total number of moles of NaOH is not changed on dilution, the molarity after dilution can of the resulting solution that the volume must be in liters. also be found by Equation 1, using the final volume of the solution. Calculate that molarity. 2. In an acid-base titration, 22.13 mL of an NaOH solution are needed to neutralize 24.65 mL of a 0.1094 MHCI solution. To find the molarity of the NaOH solution, we can use the following procedure a. First note the value of My, in the HCI solution b. Find Mon in the NaOH solution. (Use Eq 3) c. Obtain Mo, from Mot . 3. A 0.2678 g sample of an unknown acid requires 27.21 mL of 0.1164 M NaOH for neutralization to a phenolphthalein end point. There are 0.35 mL of 0.1012 M HCl used for back-titration. a. How many moles of OH are used? How many moles of H from HC12 moles OH moles H b. How many moles of H are ther in th solid acd? (Use Eg 5) moles H in solid c. What is the molar mass of the unknown acid? (Use Eq.4.)Explanation / Answer
1. a. Volume of NaOH = 7.0 ml = 0.007 L
Molarity of NaOH = 6.0 M
Moles of NaOH = molarity*volume = 6.0*0.007 = 0.042 mol
b. Moles of NaOH = 0.042 mol
Total volume = 400 ml = 0.400 L
Molarity of NaOH = moles/volume = 0.042/0.400 = 0.105 M
2. a. Value of MH in the HCl solution = Molarity of HCl solution = 0.1094 M
b. To find MOH in the solution = MH*volume of HCl/volume of NaOH = 0.1094*24.65/22.13 = 0.1219 M
c. MNaOH = MOH = 0.1219 M
3. a. Moles of OH- used = molarity of NaOH*volume of NaOH in litres = 0.1164*0.02721 = 0.00317 mol
Moles of H+ used = molarity of HCl*volume of HCl in litres = 0.1012*0.00035 = 3.54*10-5 mol = 0.00003 mol
b. Moles of H+ in solid acid = (moles of OH- used) - (moles of H+ in HCl) = 0.00317-0.00003 mol = 0.00314 mol
c. Molar mass of unknown acid = mass of unknown acid/moles of H+ in solid acid = 0.2678/0.00314 = 85.29 g/mol