Question

ADSORPTION OF ACETIC ACID BY ACTIVATED CHARCOAL LAB-"Calculate the exact concentration of your base. To do this, use the volume of base added to 10 mL of your standardized acid solution. Calculate the base concentration for each titration performed then determine the mean concentration."

Answer 1

I'm thinking c1v1=c2v2

Answer 2

I have titration volumes of 11.72, 9.49, 10.27, 10.47, 10.27 mL.

Answer 3

MY Potassium Hydrogen Phthalate concentration is 0.1024 M

Answer 4

1. Freundlich Isotherm 2. Langmuir Isotherm are mentioned in the lab intro.

Answer 5

ITs like this lab: http://www.cpp.edu/~sjanz/physical_chemistry/chm_353_labs/adsorption_of_acetic_acid_by_a_solid.pdf

Answer 6

@hwyl

Answer 7

http://www.isca.in/rjcs/Archives/vol2/i9/9.ISCA-RJCS-2012-145.pdf The proper research article, surprisngly not helpful though.

Answer 8

is the acid ur titrant or ur base?

Answer 9

i think its the base

Answer 10

so for ur first one, 11.72 ml of ur base was used to neutralise the acid?

Answer 11

Wait base is the thing that is the beaker already right?

Answer 12

ive forgotten too haahah, its been a yr since i did titration in uni

Answer 13

i should know

Answer 14

but i think you can have either acid or base as ur titrant, it just depends what ur experiment tells u, i think the norm is to have the base in the beaker thought (from my vauge memory)

Answer 15

Alright, any ideas on figuring out concentrations?

Answer 16

test 1

Answer 17

test 2

Answer 18

best bet is dilution factor c1v1=c2v2

Answer 19

give me a sec, i'm working on a maths problem as we speak, i posted the question and need some help haha

Answer 20

i dunno what to use for v1 and v2 though
i just know that i titrated 11.72 mL into 10mL KHP.

Answer 21

KHP is ur acid right?

Answer 22

yes

Answer 23

so you titrated 11.72ml of base into 10ml KHP?

Answer 24

fr my first sample, yes.

Answer 25

wait, dilution factor is unessecary, do you know the reaction equation?

Answer 26

no, i don't remember what that means

Answer 27

A+B>C+D etc

Answer 28

oh, no

Answer 29

I do not

Answer 30

i reckon that the key

Answer 31

as in do you know the stoichiometry is it 1:1?

Answer 32

I'm looking thru the lab manual, and it says nithing.

Answer 33

haha well....

Answer 34

http://www.ausetute.com.au/titrcalc.html

Answer 35

cause thats sorta what u wanna do, find moles etc then use stoichiometry to find ur base conc

Answer 36

you think i was suppose to deriive that from the equation?

Answer 37

what is ur base ur using?

Answer 38

sorry, was at bathrooom. its naoh

Answer 39

well that makes life so much easier ahah

Answer 40

CH3COOH + NaOH ----> CH3COONa + H2O

Answer 41

so stoichiometry is 1:1

Answer 42

C(Acetic)= 0.1024 M
V(acetic)=10ml

V(naoh)=11.72
C(naoh)=??

Answer 43

wait the acid is khp

Answer 44

acetic is the part after this section of the lab

Answer 45

oh

Answer 46

\[KHC _{8}H _{4}O _{4}+NaOH \rightarrow H _{2}O+NaKC _{8}H _{4}O _{4}\]

Answer 47

same idea
C(KHP)= 0.1024 M
VKHP)=10ml

V(naoh)=11.72
C(naoh)=??

Answer 48

C(KHP)= 0.1024 M
VKHP)=0.01L

V(naoh)=0.01172L
C(naoh)=??M

Answer 49

now, n=CV

Answer 50

lol im stupid i was trying to figure out how you got 0.01172 L...

Answer 51

hahaha

Answer 52

>.<

Answer 53

n(kHp)=0.1024*0.01=0.001024 moles

Answer 54

we know mole ratio is 1:1

therefore number of acid moles=number of base moles

Answer 55

n(naoh)=0.001024 moles

Answer 56

n=CV
C=n/V
C(NaOH)=0.001024/0.01172
C(NaOH)=0.087M

Answer 57

thats ur first test

Answer 58

i don't know where n=cv is coming from; as in, why its being applied to this scenario

Answer 59

because u want to find the base concentration...

Answer 60

and you can only do that using stoichiometry of what u you (acid)

Answer 61

alrihgt.

Answer 62

so 11.72, 9.49, 10.27, 10.47, 10.27 mL.
Test 1:
C(naoh)=0.087M

Test 2:
C(naoh)= 0.001024/(9.49/1000)=0.108M

Test 3:
C(naoh)= 0.001024/(10.27/1000)=0.0997M

Test 4:
C(naoh)= 0.001024/(10.47/1000)=0.0978M

test 5:
C(naoh)= 0.001024/(10.27/1000)=0.0997M

Therefore mean concentration is the average right.

C(naoh)= test 1+test2+test3+test4+test5/5runs

C(naoh)=0.098M (to be honest, no lab supervisor would make up batches of 0.098M, we can safely say that the sodium hydroxide concentration is 0.1M

Answer 63

You just had to explain up to n=CV. But thanks!

Answer 64

its because moles of a reaction always is constant, its just concentration changes. so we use stoichiometry (which is in effect molar ratio) so we use a comparison of moles to determine unknown concentrations