Solution Found!
Solved: (a) A 0.1044-g sample of an unknown monoprotic
Chapter 10, Problem 109IE(choose chapter or problem)
Problem 109IE
(a) A 0.1044-g sample of an unknown monoprotic acid requires 22.10 mL of 0.0500 M NaOH to reach the end point. What is the molecular weight of the unknown? (b) As the acid is titrated, the pH of the solution after the addition of 11.05 mL of the base is 4.89. What is the Ka for the acid? (c) Using Appendix D, suggest the identity of the acid.
Questions & Answers
QUESTION:
Problem 109IE
(a) A 0.1044-g sample of an unknown monoprotic acid requires 22.10 mL of 0.0500 M NaOH to reach the end point. What is the molecular weight of the unknown? (b) As the acid is titrated, the pH of the solution after the addition of 11.05 mL of the base is 4.89. What is the Ka for the acid? (c) Using Appendix D, suggest the identity of the acid.
ANSWER:
Step 1 of 3
As the unknown acid is mono-protic acid that means the one proton released by the acid will neutralize the one hydroxide ion (-OH) at a time during the titration. Therefore, we can conclude that the reaction is equimolar reaction.
Moles of NaOH = Moles of Unknown acid
Given, molarity of unknown acid = molarity of NaOH = 0.0500M
Therefore, number of moles = Molarity Volume of the solvent in liters
= 0.0500M 0.02210 L = 0.001105 moles
Given mass of the unknown sample of monoprotic acid = 0.1044 g
The molecular weight of unknown acid = Given mass of the sample/ Number of moles
= 0.1044g /0.001105 moles = 94.48 g