2. A 0.2349 g sample of an unknown acid requires 33.66 mL of 0.1086 M NaOH...
3. A 0.7026 g sample of an unknown acid requires 40.96 mL of 0.1158 M NaOH for neutralization to a phenolphthalein end point. There are 1.22 mL of 0.1036 M HCl used for back-titration. a. How many moles of OH are used? How many moles of Ht from HCI? moles OH moles H+ b. How many moles of Ht are there in the solid acid? (Use Eq. 5.) moles H+ in solid c. What is the molar mass of the...
Experiment 6: The Standardization of a Basic Solution and the Determination of the MM of an Acid *Note: This experiment is relatively long unless you know precisely what to do. Read the experiment before coming to class and arrive on time. After reading through the experiment answer the following question: 1. 7.0 mL of 6.0 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...
0.422 gram sample of an unknown acid (HX) required 26.75 mL of 0.244 M NaOH for neutralization to a phenolphthalein end point. What is the molar mass of the acid?
A 0.860 −g sample of an unknown acid requires 33.0 mL of a 0.190 M barium hydroxide solution for neutralization. Part A Assuming the acid is diprotic, calculate the molar mass of the acid. M = g/mol
A 0.864 −g sample of an unknown acid requires 32.7 mL of a 0.183 M barium hydroxide solution for neutralization. Assuming the acid is diprotic, calculate the molar mass of the acid.
A 0.5240 −g sample of an unknown monoprotic acid was titrated with 9.94×10−2 M NaOH. The equivalence point of the titration occurs at 23.90 mL . Determine the molar mass of the unknown acid. Use the Ksp values to calculate the molar solubility of each of the following compounds in pure water. PART A---- MX (Ksp = 8.57×10−36)Express your answer in moles per liter. PART B----PbCl2 (Ksp = 1.17×10−5) Express your answer in moles per liter. PART C-----Ni(OH)2 (Ksp =...
The student dissolved 2.050 grams of acid in 50 mL of distilled water. He then titrated the unknown acid with a standardized NaOH solution. Unfortunately, the student over shot the end point and had to back titrate the solution with 2.17 mL of 0.98861 M HCl. Once the final end point was determined, the student had added a total of 29.15 mL of a 1.0324 M NaOH solution. A) How many moles of OH- were used to reach the end...
The student dissolved 2.050 grams of acid in 50 mL of distilled water. He then titrated the unknown acid with a standardized NaOH solution. Unfortunately, the student over shot the end point and had to back titrate the solution with 2.17 mL of 0.98861 M HCl. Once the final end point was determined, the student had added a total of 29.15 mL of a 1.0324 M NaOH solution. A) How many moles of OH- were used to reach the end...
A 0.5220 −g sample of an unknown monoprotic acid was titrated with 9.94×10−2 M NaOH. The equivalence point of the titration occurs at 23.86 mL . Determine the molar mass of the unknown acid.
ratra 1z 1 t B: Detes Name Date inoe Acid-Base Titration Data and Calculations Part A: Standardization of NaOH solution, Data: Trial 1 Trial 3 Trial 2 Initial level, HCI buret 26-60 mL Final level, HCI buret 33.85 mL 43.15 2u.95 mL mL Initial level, NaOH buret 00mL Final level, NaoH buret 23.45 mL 44.35 mL 24.25 mL Calculations: Trial 1 Trial 2 Trial 3 Total volume HCI solution 24.45 mL 33-86 mL 43.95 mL 0.1030 Molarity of standardized HCI...