In the determination of the molar mass of a solid acid by titrating it with a st
ID: 1028572 • Letter: I
Question
In the determination of the molar mass of a solid acid by titrating it with a standardized base, which procedural error will yield a molar mass that is smaller than the actual value?
A) adding the standardized base to a buret containing drops of water
B) dissolving the weighted solid acid in twice the recommended volume of water
C) using half as many drops of indicator as suggested
D) weighing out half of the recommended mass of solid acid
The correct answer is A. Can someone please explain why?
Are you supposed to use VM = VM?
In that case, if the Concentration of base goes down, doesn't that mean that the volume of base titrated will increase, so the number of moles of acid will stay the same? What am I doing wrong?
Explanation / Answer
Ans : A) Adding the standardised base to a buret containing drops of water.
When the base gets diluted , more amount of it will be required to titrate the same amount of acid. Since the volume of the acid is fixed , so the molarity will come out large.
Molarity = no. of moles of acid / volume of solution in L
Since the molarity is more , therefore number of moles of acid will also come out to be more.
Number of moles = given mass / molar mass
So if the number of moles is more , so the molar mass determined will be less.