3. A 0.7026 g sample of an unknown acid requires 40.96 mL of 0.1158 M NaOH...
2. A 0.2349 g sample of an unknown acid requires 33.66 mL of 0.1086 M NaOH for neutralization to a phenolphthalein end point. There are 0.42 mL of 0.1049 M HCI used for back-titration. a. How many moles of OH- are used? moles OH moles H b. How many moles of H+ from HCI? moles H c. How many moles of H+ are there ins solid acid? (Eq. 5) g/mol d. What is the molar mass of the unknown acid?...
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...
2. In an acid-base titration, 25.62 mL of an NaOH solution are needed to neutralize 26.23 mL of a 0.1036 M HCI solution. To find the molarity of the NaOH solution, we can use the following procedure: First note the value of M in the HCl solution. a. H* b. Find M Oн in the NAOH solution. (Use Eq.3.) from MOH NaOH c. Obtain M.
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?
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 titration of 25.00 mL of O.1550 M HCl solution with a solution of NaOH of unknown molarity starts at a buret reading for NaOH of 0.33 mL The phenolphthalein indicator turns light pink the acid solution for over 30 seconds at a buret reading of 24.19 mL 2) What was the volume of HCl you started with? 3) How many moles of HCl were in the original solution? 4) Write the balanced chemical equation for the titration reaction. 5)...
based on thw data can someone solve question number 1, 2, 3, 4 18 Acid-Base Reactions Titration Curve Consider a 10. mL sample of 0.10 M HCI. a) What is the pH of the solution? b)How many ml. of 0.10 M NaOH would be required to neutralize it? c) what is the pH of the neutralized solution? d) What would the pH of the solution be if you added 20. mL of NaOH? volume ofvolumef 0.10 M HCI 0.10 M...
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 =...
2. In an acid-base titration, 25.62 mL of an NaOH solution are needed to neutralize 26.23 mL of a 0.1036 M HCl solution. To find the molarity of the NaOH solution, we can use the following procedure: a. First note the value of M.. in the HCl solution. b. Find Mo- in the NaOH solution. (Use Eq. 3.) c. Obtain M Nach from Moh-- M