Name Section Advance Study Assignment: Molar Mass of an Acid 1. 7.0 ml of 60 M N
ID: 1034801 • Letter: N
Question
Name Section Advance Study Assignment: Molar Mass of an Acid 1. 7.0 ml of 60 M NaOH are diluted with water to a volume of 400 mL. You are asked to find the molarity of the resulting solution. a. First find out how many moles of NaOH there are in 7.0 mL of 6.0 M NaOH. Use Equation 1. Note that the volume must be in liters. _ moles b. Since the total number of moles of NaOH is not changed on dilution, the molarity after dilution can also be found by Equation 1, using the final volume of the solution. Calculate that molarity. 2. In an acid-base titration, 23.91 mL of an NaOH solution are needed to neutralize 24.58 mL. of a 0.1002 M HCI solution. To find the molarity of the NaOH solution, we can use the following procedure: a. First note the value of M in the HCI solution. b Find Moer in the NaOH solution. (Use Eq. 3) c. Obtain MNwou from Moar 3. A 0.3174 g sample of an unknown acid requires 26.23 mL of 0.1056 M NaOH for neutralization to a phenolphthalein end point. There are 0.27 ml of 0.1096 M HCl used for back-titration. a. How many moles of OH are used? How many moles of H' from HCI? -moles OHh- moles H. b. How many moles of H' are there in the solid acid? (Use Eq. 5.) moles H" in solid c. What is the molar mass of the unknown acid? (Use Eq.4.) -g/mol custom page 151Explanation / Answer
1 a. Moles of NaOH= Molarity of NaOH* Volume in litres = 6 * (7/1000) = 0.042 moles.
b. Molarity = Moles / (Volume in litre)= 0.042 / (400/1000) = (0.042*1000)/400 = 0.105 M.
2 a. Molartiy of H+ = Molarity of HCl = 0.1002 M
b. Molarity of OH- = ( Molarity of H+ * Volume of HCl ) / Volume of NaOH = (0.1002 * 24.58) / 23.91 = 0.103 M
c. Molarity of NaOH= Molarity of OH- = 0.103 M (Ans)
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