Question

Hi! Can someone PLEASE help me out with titrations?

Answer 1

So there's a titration between 250 L of the weak BASE C6H5NHCH3 with a concentration of 2.53% and between 101.3 L of the strong ACID HCF3SO3.
Ka(C6H5NHCH3)= 2.5*10^-10

Answer 2

What is the concentration of the strong ACID? The answer is 8.75%

Answer 3

@nincompoop do you know it by any chance?

Answer 4

I am not sure i was looking to see if I could help you but i can't

Answer 5

Sorry

Answer 6

Yeah i only wish I could help you

Answer 7

@Kainui

Answer 8

Thanks @Kainui !

Answer 9

The Ka= 4*10^-5

Answer 10

So when they say the concentration of the weak base is 2.53% what does that mean, is that percent by volume (I'm assuming) or by mass?

Answer 11

By mass, the question is about the concentration before the titration itself

Answer 12

we were taught that percentage os gram/0.1 L

Answer 13

is*

Answer 14

so that means 2.53 gr/0.1 L

Answer 15

So when they say the concentration of the weak base is 2.53% what does that mean, is that percent by volume (I'm assuming) or by mass?

Answer 16

by mass I think

Answer 17

Sorry my computer crashed and sent the same message again weird

Answer 18

yeah mine did it today as well

Answer 19

Ok so let's get into it, how far are you able to take the problem on your own, what have you tried?

Answer 20

I think a good place to start would be to write the chemical reaction out.

Answer 21

I tried to convert the percentage to grams, and got grams in total, then I've found the moles which are 59

Answer 22

Have you heard of ICE tables or used something like that before maybe?

Answer 23

then found the molarity which is before the titration

Answer 24

no...

Answer 25

Yes I've heard! but we haven't used them

Answer 26

initial is for I

Answer 27

Keep explaining what you tried so I can see what you're doing.

Answer 28

then I did X^2/0.236=4*10^-10

Answer 29

sorry 4*10^-5

Answer 30

and X got 3.072*10^-3

Answer 31

that's the concentration of the hydronium before?

Answer 32

So wait, do you know where in the titration we ended, before,at, or after the equivalence point?

Answer 33

and no matter how many times I calculate the percentage it doesn't get the answer

Answer 34

that's before the end point

Answer 35

the question asks before the end point

Answer 36

Should I have divided the moles by the total volume?

Answer 37

I guess I'm just a little confused by what your definition of concentration means. I am used to using molarity, and I'm not entirely sure I see what you're dealing with here which is giving me trouble.

Answer 38

% is another way to determine concentration

Answer 39

I don't know why we study this, but we have 3 questions like that out of almost 40 questions in our gen chem final exam.

Answer 40

Fair enough, so really what we're looking for is how many grams per .1 L of acid we have before the equivalence point.

Answer 41

yes! and I have no idea how to find it, I thought I knew

Answer 42

So in effect, we're really just dumping 101.3L of acid into 250L of base, right?

Answer 43

yes

Answer 44

but how can we find grams?

Answer 45

So we can set up the formula with the rate constant to solve for the concentration which is good. I don't think we'll need to find the grams of acid we just need to find the concentration which is slightly different.

Answer 46

I'm trying

Answer 47

I got 0.168M of the weak base

Answer 48

and concentration of the H3O+ is 2.594*10^-3

Answer 49

and got 136 gr

Answer 50

hmmmmmm

Answer 51

I guess I'm just too tired, this seems simple but I'm just sort of stumped. Have you tried the henderson hasselbach equation or M1V1=M2V2?

Answer 52

yes

Answer 53

go to sleep/rest it's more important I'll try to work more on it

Answer 54

I haven't really played with titrations in a while, in practice I've never really needed them, I just use a pH probe lol.

Answer 55

lol makes sense

Answer 56

Thanks allot though! you're truly the BEST!

Answer 57

I gtg to class now anyways

Answer 58

hope you get some rest :)

Answer 59

You wouldn't use henderson-hasselbalch here. Why don't you start by writing out the chemical equation to see what you're doing first of all.

Answer 60

@abb0t like that?

\[C6H5NCH3+HCF3SO3 \rightarrow HC6H5NCH3+ -CF3SO3\]

Answer 61

and then hydrolysis:

\[HC6H5NHCH3+ H2O \rightarrow C6H5NHCH3+ H3O\]

Answer 62

is that a good start? @abb0t

Answer 63

Wait, what? Are you reacting the N-methyl aniline with trifluoromethanesulfonic acid or water?

Answer 64

I thought that's how you do titrations. first you react the reagents and then hydrolysis

Answer 65

now I'm completely confused

Answer 66

trifluoromethanesulfonic acid is a VERY strong acid, I think this is one of the superacids, if I remember correctly.
Aniline, with a pKa of about 4 reacting with an acid acid with a pKa of about -14, which is about 10\(^6\) times stronger than even concentrated sulfuric acid, which has a pKa of about -2?

I'm making rough estimates here, from what I recall.

Answer 67

I've found the Ka of C6H5NHCH3 if that helps which is 4.0*10^-5

Answer 68

Either way, let's tread this as if we didn't know this was a superacid, and let's just assume you have a weak base reacting with a very very strong acid.

Given 2.53%, which means that you have 2.53/100 mL of solution

Answer 69

yes

Answer 70

should I find the mass in grams in 250L?

Answer 71

So what I think you were doing is this, when you wrote water:
\(\sf \color{blue}{B + H^+ \rightarrow } BH + H_2O\)

A strong acid will react with a weak base to form an acidic solution.

Answer 72

but how can I get further from here? I guess I can find the grams of the weak base in 250 L which gives 6325 grams

Answer 73

from there I can find the moles and the molarity, right?

Answer 74

@abb0t ?

Answer 75

@ganeshie8

Answer 76

@iambatman

Answer 77

I really need you help with this one

Answer 78

your