NotTim
  • NotTim
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."
Chemistry
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
NotTim
  • NotTim
I'm thinking c1v1=c2v2
NotTim
  • NotTim
I have titration volumes of 11.72, 9.49, 10.27, 10.47, 10.27 mL.
NotTim
  • NotTim
MY Potassium Hydrogen Phthalate concentration is 0.1024 M

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

NotTim
  • NotTim
1. Freundlich Isotherm 2. Langmuir Isotherm are mentioned in the lab intro.
NotTim
  • NotTim
ITs like this lab: http://www.cpp.edu/~sjanz/physical_chemistry/chm_353_labs/adsorption_of_acetic_acid_by_a_solid.pdf
dan815
  • dan815
@hwyl
NotTim
  • NotTim
http://www.isca.in/rjcs/Archives/vol2/i9/9.ISCA-RJCS-2012-145.pdf The proper research article, surprisngly not helpful though.
anonymous
  • anonymous
is the acid ur titrant or ur base?
NotTim
  • NotTim
i think its the base
anonymous
  • anonymous
so for ur first one, 11.72 ml of ur base was used to neutralise the acid?
NotTim
  • NotTim
Wait base is the thing that is the beaker already right?
anonymous
  • anonymous
ive forgotten too haahah, its been a yr since i did titration in uni
anonymous
  • anonymous
i should know
anonymous
  • anonymous
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)
NotTim
  • NotTim
Alright, any ideas on figuring out concentrations?
NotTim
  • NotTim
test 1
NotTim
  • NotTim
test 2
anonymous
  • anonymous
best bet is dilution factor c1v1=c2v2
anonymous
  • anonymous
give me a sec, i'm working on a maths problem as we speak, i posted the question and need some help haha
NotTim
  • NotTim
i dunno what to use for v1 and v2 though i just know that i titrated 11.72 mL into 10mL KHP.
anonymous
  • anonymous
KHP is ur acid right?
NotTim
  • NotTim
yes
anonymous
  • anonymous
so you titrated 11.72ml of base into 10ml KHP?
NotTim
  • NotTim
fr my first sample, yes.
anonymous
  • anonymous
wait, dilution factor is unessecary, do you know the reaction equation?
NotTim
  • NotTim
no, i don't remember what that means
anonymous
  • anonymous
A+B>C+D etc
NotTim
  • NotTim
oh, no
NotTim
  • NotTim
I do not
anonymous
  • anonymous
i reckon that the key
anonymous
  • anonymous
as in do you know the stoichiometry is it 1:1?
NotTim
  • NotTim
I'm looking thru the lab manual, and it says nithing.
anonymous
  • anonymous
haha well....
anonymous
  • anonymous
http://www.ausetute.com.au/titrcalc.html
anonymous
  • anonymous
cause thats sorta what u wanna do, find moles etc then use stoichiometry to find ur base conc
NotTim
  • NotTim
you think i was suppose to deriive that from the equation?
anonymous
  • anonymous
what is ur base ur using?
NotTim
  • NotTim
sorry, was at bathrooom. its naoh
anonymous
  • anonymous
well that makes life so much easier ahah
anonymous
  • anonymous
CH3COOH + NaOH ----> CH3COONa + H2O
anonymous
  • anonymous
so stoichiometry is 1:1
anonymous
  • anonymous
C(Acetic)= 0.1024 M V(acetic)=10ml V(naoh)=11.72 C(naoh)=??
NotTim
  • NotTim
wait the acid is khp
NotTim
  • NotTim
acetic is the part after this section of the lab
anonymous
  • anonymous
oh
anonymous
  • anonymous
\[KHC _{8}H _{4}O _{4}+NaOH \rightarrow H _{2}O+NaKC _{8}H _{4}O _{4}\]
anonymous
  • anonymous
same idea C(KHP)= 0.1024 M VKHP)=10ml V(naoh)=11.72 C(naoh)=??
anonymous
  • anonymous
C(KHP)= 0.1024 M VKHP)=0.01L V(naoh)=0.01172L C(naoh)=??M
anonymous
  • anonymous
now, n=CV
NotTim
  • NotTim
lol im stupid i was trying to figure out how you got 0.01172 L...
anonymous
  • anonymous
hahaha
anonymous
  • anonymous
>.<
anonymous
  • anonymous
n(kHp)=0.1024*0.01=0.001024 moles
anonymous
  • anonymous
we know mole ratio is 1:1 therefore number of acid moles=number of base moles
anonymous
  • anonymous
n(naoh)=0.001024 moles
anonymous
  • anonymous
n=CV C=n/V C(NaOH)=0.001024/0.01172 C(NaOH)=0.087M
anonymous
  • anonymous
thats ur first test
NotTim
  • NotTim
i don't know where n=cv is coming from; as in, why its being applied to this scenario
anonymous
  • anonymous
because u want to find the base concentration...
anonymous
  • anonymous
and you can only do that using stoichiometry of what u you (acid)
NotTim
  • NotTim
alrihgt.
anonymous
  • anonymous
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
NotTim
  • NotTim
You just had to explain up to n=CV. But thanks!
anonymous
  • anonymous
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

Looking for something else?

Not the answer you are looking for? Search for more explanations.